Book Insights

5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?

5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6] 4– (ii) [FeF6] 3– (iii) [Co(C2O4)3]3– (iv) [CoF6] 3–

5.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii) [Co(NH3)Cl(en)2] 2+ (iii) [Co(NH3)2Cl2(en)]+

5.29 Amongst the following ions which one has the highest magnetic moment value? (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6] 2+ (iii) [Zn(H2O)6]2+ 5.30 Amongst the following, the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6] 3+ (iii) [Fe(C2O4)3]3– (iv) [FeCl6] 3– 5.31 What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6] 4–, [Ni(NH3)6] 2+, [Ni(H2O)6] 2+ ? Answers to Some Intext Questions 5.1 (i) [Co(NH3)4(H2O)2]Cl3 (iv) [Pt(NH3)BrCl(NO2)]– (ii) K2[Ni(CN)4] (v) [PtCl2(en)2](NO3)2 (iii) [Cr(en)3]Cl3 (vi) Fe4[Fe(CN)6]3 5.2 (i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanidoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methanamine)platinum(II) chloride 5.3 (i) Both geometrical (cis-, trans-) and optical isomers for cis can exist. (ii) Two optical isomers can exist. (iii) There are 10 possible isomers. (Hint: There are geometrical, ionisation and linkage isomers possible). (iv) Geometrical (cis-, trans-) isomers can exist. 5.4 The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ ® BaSO4 (s) [Co(NH3)5SO4]Br + Ba2+ ® No reaction [Co(NH3)5Br]SO4 + Ag+ ® No reaction [Co(NH3)5SO4]Br + Ag+ ® AgBr (s) 5.6 In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl42–, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons. 5.7 In presence of CN–, (a strong ligand) the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d 2sp 3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp 3d 2 forming an outer orbital complex containing five unpaired electrons, it is strongly paramagnetic. 5.8 In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2sp3 hybridisation forming inner orbital complex in case of [Co(NH3)6]3+. In Ni(NH3)6 2+, Ni is in +2 oxidation state and has d 8 configuration, the hybridisation involved is sp 3d 2 forming outer orbital complex. 5.9 For square planar shape, the hybridisation is dsp 2. Hence the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp2 hybridisation. Thus there is no unpaired electron. Chemistry 140 Reprint 2025-26

5.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes: (i) K3[Co(C2O4)3] (iii) (NH4)2[CoF4] (ii) cis-[CrCl2(en)2]Cl (iv) [Mn(H2O)6]SO4

5.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field.

5.28 How many ions are produced from the complex Co(NH3)6Cl2 in solution? (i) 6 (ii) 4 (iii) 3 (iv) 2 139 Coordination Compounds Reprint 2025-26

5.19 [Cr(NH3)6] 3+ is paramagnetic while [Ni(CN)4] 2– is diamagnetic. Explain why? 5.20 A solution of [Ni(H2O)6] 2+ is green but a solution of [Ni(CN)4] 2– is colourless. Explain. 5.21 [Fe(CN)6] 4– and [Fe(H2O)6] 2+ are of different colours in dilute solutions. Why?

5.18 What is crystal field splitting energy? How does the magnitude of Do decide the actual configuration of d orbitals in a coordination entity?

5.24 Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex: (i) K[Cr(H2O)2(C2O4)2].3H2O (iii) [CrCl3(py)3] (v) K4[Mn(CN)6] (ii) [Co(NH3)5Cl-]Cl2 (iv) Cs[FeCl4]

5.1 Thermodynamic terms the system from the surroundings is called We are interested in chemical reactions and boundary. This is designed to allow us to control and keep track of all movements ofthe energy changes accompanying them. For matter and energy in or out of the system.this we need to know certain thermodynamic terms. These are discussed below. 5.1.2 Types of the System 5.1.1 The System and the Surroundings We, further classify the systems according A system in thermodynamics refers to that to the movements of matter and energy in or part of universe in which observations are out of the system. made and remaining universe constitutes 1. Open Systemthe surroundings. The surroundings include everything other than the system. System In an open system, there is exchange of energy and the surroundings together constitute the and matter between system and surroundings universe. [Fig. 5.2 (a)]. The presence of reactants in an open beaker is an example of an open system*.The universe = The system + The surroundings Here the boundary is an imaginary surface However, the entire universe other than enclosing the beaker and reactants. the system is not affected by the changes taking place in the system. Therefore, for all 2. Closed System practical purposes, the surroundings are that In a closed system, there is no exchange of portion of the remaining universe which can matter, but exchange of energy is possible interact with the system. Usually, the region between system and the surroundings of space in the neighbourhood of the system [Fig. 5.2 (b)]. The presence of reactants in a constitutes its surroundings. closed vessel made of conducting material For example, if we are studying the e.g., copper or steel is an example of a closed reaction between two substances A and B system. kept in a beaker, the beaker containing the reaction mixture is the system and the room where the beaker is kept is the surroundings (Fig. 5.1). Fig. 5.1 System and the surroundings Note that the system may be defined by physical boundaries, like beaker or test tube, or the system may simply be defined by a set of Cartesian coordinates specifying a particular volume in space. It is necessary to think of the system as separated from the surroundings by some sort of wall which may be real or imaginary. The wall that separates Fig. 5.2 Open, closed and isolated systems. * We could have chosen only the reactants as system then walls of the beakers will act as boundary. Reprint 2025-26 138 chemIstry 3. Isolated System a quantity which represents the total energy of the system. It may be chemical, electrical,In an isolated system, there is no exchange mechanical or any other type of energy youof energy or matter between the system and may think of, the sum of all these is the energythe surroundings [Fig. 5.2 (c)]. The presence of the system. In thermodynamics, we call itof reactants in a thermos flask or any other the internal energy, U of the system, whichclosed insulated vessel is an example of an may change, whenisolated system. • heat passes into or out of the system, 5.1.3 The State of the System • work is done on or by the system, The system must be described in order to • matter enters or leaves the system. make any useful calculations by specifying quantitatively each of the properties such as These systems are classified accordingly its pressure (p), volume (V), and temperature as you have already studied in section 5.1.2. (T ) as well as the composition of the system. (a) WorkWe need to describe the system by specifying it before and after the change. You would Let us first examine a change in internal energy recall from your Physics course that the by doing work. We take a system containing state of a system in mechanics is completely some quantity of water in a thermos flask or in an insulated beaker. This would notspecified at a given instant of time, by the allow exchange of heat between the systemposition and velocity of each mass point of and surroundings through its boundary andthe system. In thermodynamics, a different we call this type of system as adiabatic. Theand much simpler concept of the state of a manner in which the state of such a systemsystem is introduced. It does not need detailed may be changed will be called adiabaticknowledge of motion of each particle because, process. Adiabatic process is a process inwe deal with average measurable properties of which there is no transfer of heat betweenthe system. We specify the state of the system the system and surroundings. Here, the wallby state functions or state variables. separating the system and the surroundings The state of a thermodynamic system is is called the adiabatic wall (Fig. 5.3).described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T ), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. In order to completely define the state of a system it is not necessary to define all the properties of the system; as only a certain number of properties can be varied independently. This number depends on the nature of the system. Once these minimum number of macroscopic properties are fixed, Fig. 5.3 An adiabatic system which does not others automatically have definite values. permit the transfer of heat through its The state of the surroundings can never boundary. be completely specified; fortunately it is not Let us bring the change in the internal necessary to do so. energy of the system by doing some work on it. Let us call the initial state of the system 5.1.4 The Internal Energy as a State as state A and its temperature as TA. Let Function the internal energy of the system in state A When we talk about our chemical system be called UA. We can change the state of the losing or gaining energy, we need to introduce system in two different ways. Reprint 2025-26 THERMODYNAMICS 139 One way: We do some mechanical work, say the route taken. Volume of water in a pond, for 1 kJ, by rotating a set of small paddles and example, is a state function, because change thereby churning water. Let the new state in volume of its water is independent of the be called B state and its temperature, as route by which water is filled in the pond, TB. It is found that TB > TA and the change either by rain or by tubewell or by both. in temperature, ∆T = TB–TA. Let the internal (b) Heat energy of the system in state B be UB and the We can also change the internal energychange in internal energy, ∆U =UB– UA. of a system by transfer of heat from theSecond way: We now do an equal amount surroundings to the system or vice-versa(i.e., 1kJ) electrical work with the help of an without expenditure of work. This exchangeimmersion rod and note down the temperature of energy, which is a result of temperaturechange. We find that the change in temperature difference is called heat, q. Let us consideris same as in the earlier case, say, TB – TA. bringing about the same change in temperature In fact, the experiments in the above (the same initial and final states as before manner were done by J. P. Joule between in section 5.1.4 (a) by transfer of heat 1840–50 and he was able to show that a through thermally conducting walls instead given amount of work done on the system, of adiabatic walls (Fig. 5.4). no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system. So, it seems appropriate to define a quantity, the internal energy U, whose value is characteristic of the state of a system, whereby the adiabatic work, wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ∆U i.e., ∆U =U2 –U1= wad Fig. 5.4 A system which allows heat transfer Therefore, internal energy, U, of the through its boundary. system is a state function. We take water at temperature, TA in a By conventions of IUPAC in chemical container having thermally conducting walls,thermodynamics. The positive sign expresses say made up of copper and enclose it in athat wad is positive when work is done on the huge heat reservoir at temperature, TB. Thesystem and the internal energy of system heat absorbed by the system (water), q can beincreases. Similarly, if the work is done by the measured in terms of temperature difference,system, wad will be negative because internal energy of the system decreases. TB – TA. In this case change in internal energy, ∆U = q, when no work is done at constant Can you name some other familiar state volume.functions? Some of other familiar state By conventions of IUPAC in chemicalfunctions are V, p, and T. For example, if we thermodynamics. The q is positive, when heatbring a change in temperature of the system is transferred from the surroundings to thefrom 25°C to 35°C, the change in temperature system and the internal energy of the systemis 35°C–25°C = +10°C, whether we go straight increases and q is negative when heat isup to 35°C or we cool the system for a few transferred from system to the surroundingsdegrees, then take the system to the final resulting in decrease of the internal energy oftemperature. Thus, T is a state function and the change in temperature is independent of the system. * Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention. Reprint 2025-26 140 chemIstry (c) The general case SolutionLet us consider the general case in which a change of state is brought about both by (i) ∆ U = w ad, wall is adiabatic doing work and by transfer of heat. We write (ii) ∆ U = – q, thermally conducting change in internal energy for this case as: walls ∆U = q + w (5.1) (iii) ∆ U = q – w, closed system. For a given change in state, q and w can 5.2 Applications vary depending on how the change is carried Many chemical reactions involve the generation out. However, q +w = ∆U will depend only on of gases capable of doing mechanical work or initial and final state. It will be independent the generation of heat. It is important for us of the way the change is carried out. If there to quantify these changes and relate them is no transfer of energy as heat or as work to the changes in the internal energy. Let us (isolated system) i.e., if w = 0 and q = 0, then see how! ∆ U = 0. The equation 5.1 i.e., ∆U = q + w is 5.2.1 Work mathematical statement of the first law of First of all, let us concentrate on the nature of thermodynamics, which states that work a system can do. We will consider only The energy of an isolated system is mechanical work i.e., pressure-volume work. constant. For understanding pressure-volume work, let us consider a cylinder whichIt is commonly stated as the law of conservation contains one mole of an ideal gas fitted withof energy i.e., energy can neither be created a frictionless piston. Total volume of the gasnor be destroyed. is Vi and pressure of the gas inside is p. IfNote: There is considerable difference between external pressure is pex which is greater thanthe character of the thermodynamic property p, piston is moved inward till the pressureenergy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ∆U of the system. Problem 5.1 Express the change in internal energy of a system when (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ? (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have? Fig. 5.5 (a) Work done on an ideal gas in a (iii) w amount of work is done by the cylinder when it is compressed by system and q amount of heat is a constant external pressure, pex supplied to the system. What type (in single step) is equal to the shaded of system would it be? area. Reprint 2025-26 THERMODYNAMICS 141 inside becomes equal to pex. Let this change If the pressure is not constant but be achieved in a single step and the final changes during the process such that it volume be Vf . During this compression, is always infinitesimally greater than the suppose piston moves a distance, l and pressure of the gas, then, at each stage of is cross-sectional area of the piston is A compression, the volume decreases by an [Fig. 5.5(a)]. infinitesimal amount, dV. In such a case we then, volume change = l × A = ∆V = (Vf – Vi ) can calculate the work done on the gas by the relationWe also know, pressure = f Therefore, force on the piston = pex . A V ex dV (5.3) w pIf w is the work done on the system by Vi movement of the piston then w = force × distance = pex . A .l Here, pex at each stage is equal to (pin + dp) in case of compression [Fig. 5.5(c)]. In an = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi ) (5.2) expansion process under similar conditions, The negative sign of this expression is the external pressure is always less than the required to obtain conventional sign for w, pressure of the system i.e., pex = (pin– dp). In which will be positive. It indicates that in case general case we can write, pex = (pin + dp). Such of compression work is done on the system. processes are called reversible processes. Here (Vf – Vi ) will be negative and negative A process or change is said to bemultiplied by negative will be positive. Hence reversible, if a change is brought out in such athe sign obtained for the work will be positive. way that the process could, at any moment, If the pressure is not constant at every be reversed by an infinitesimal change. stage of compression, but changes in number A reversible process proceeds infinitely of finite steps, work done on the gas will be slowly by a series of equilibrium states summed over all the steps and will be equal such that system and the surroundings are to – Σ р ∆V [Fig. 5.5 (b)] always in near equilibrium with each other. Fig. 5.5 (c) pV-plot when pressure is not constant Fig. 5.5 (b) pV-plot when pressure is not constant and changes in infinite steps (reversible and changes in finite steps during conditions) during compression from compression from initial volume, Vi to initial volume, Vi to final volume, Vf . final volume, Vf . Work done on the gas Work done on the gas is represented is represented by the shaded area. by the shaded area. Reprint 2025-26 142 chemIstry Processes other than reversible processes Isothermal and free expansion of an are known as irreversible processes. ideal gas In chemistry, we face problems that can For isothermal (T = constant) expansion of be solved if we relate the work term to the an ideal gas into vacuum; w = 0 since pex = 0. internal pressure of the system. We can Also, Joule determined experimentally that relate work to internal pressure of the system q = 0; therefore, ∆U = 0 under reversible conditions by writing Equation 5.1, can beequation 5.3 as follows: V f V f expressed for isothermal irreversible and reversible changes as follows: wrev  p ex dV  p in  dp ) dV  ( 1. For isothermal irreversible change V V i i q = – w = pex (Vf – Vi ) Since dp × dV is very small we can write 2. For isothermal reversible change V f in dV (5.4) q = – w = nRT ln wrev p Vi Now, the pressure of the gas (pin which we V f can write as p now) can be expressed in terms = 2.303 nRT log Vi of its volume through gas equation. For n mol For adiabatic change, q = 0, of an ideal gas i.e., pV =nRT ∆U = wad nRT  p  Problem 5.2 V Two litres of an ideal gas at a pressure of 10 Therefore, at constant temperature atm expands isothermally at 25 °C into a (isothermal process), vacuum until its total volume is 10 litres. V f How much heat is absorbed and how dV V f much work is done in the expansion ? RT n RT ln wrev  n V Vi V i Solution V f We have q = – w = pex (10 – 2) = 0(8) = 0= – 2.303 nRT log (5.5) No work is done; no heat is absorbed. Vi Problem 5.3 Free expansion: Expansion of a gas in Consider the same expansion, butvacuum (pex = 0) is called free expansion. this time against a constant externalNo work is done during free expansion of an pressure of 1 atm. ideal gas whether the process is reversible or irreversible (equation 5.2 and 5.3). Solution Now, we can write equation 5.1 in number We have q = – w = pex (8) = 8 litre-atm of ways depending on the type of processes. Problem 5.4 Let us substitute w = – pex∆V (eq. 5.2) in Consider the expansion given in problemequation 5.1, and we get 5.2, for 1 mol of an ideal gas conducted U  q  p ex V reversibly. If a process is carried out at constant volume Solution (∆V = 0), then V We have q = – w = 2.303 nRT log f s ∆U = qV V the subscript V in qV denotes that heat is = 2.303 × 1 × 0.8206 × 298 × log 10 supplied at constant volume. 2 Reprint 2025-26 THERMODYNAMICS 143 Remember ∆H = qp, heat absorbed by the = 2.303 x 0.8206 x 298 x log 5 system at constant pressure. = 2.303 x 0.8206 x 298 x 0.6990 ∆H is negative for exothermic reactions = 393.66 L atm which evolve heat during the reaction and ∆H is positive for endothermic reactions5.2.2 Enthalpy, H which absorb heat from the surroundings. (a) A Useful New State Function At constant volume (∆V = 0), ∆U = qV,We know that the heat absorbed at constant therefore equation 5.8 becomes volume is equal to change in the internal ∆H = ∆U = qVenergy i.e., ∆U = qV. But most of chemical reactions are carried out not at constant The difference between ∆H and ∆U is not volume, but in flasks or test tubes under usually significant for systems consisting constant atmospheric pressure. We need to of only solids and / or liquids. Solids and define another state function which may be liquids do not suffer any significant volume suitable under these conditions. changes upon heating. The difference, however, becomes significant when gases are We may write equation (5.1) as involved. Let us consider a reaction involving ∆U = qp – p∆V at constant pressure, where gases. If VA is the total volume of the gaseousqp is heat absorbed by the system and –p∆V reactants, VB is the total volume of the gaseousrepresent expansion work done by the system. products, nA is the number of moles of gaseous Let us represent the initial state by reactants and nB is the number of moles ofsubscript 1 and final state by 2 gaseous products, all at constant pressure We can rewrite the above equation as and temperature, then using the ideal gas U2–U1 = qp – p (V2 – V1) law, we write, On rearranging, we get pVA = nART qp = (U2 + pV2) – (U1 + pV1) (5.6) and pVB = nBRT Now we can define another thermodynamic Thus, pVB – pVA = nBRT – nART = (nB–nA)RTfunction, the enthalpy H [Greek word enthalpien, to warm or heat content] as : or p (VB – VA) = (nB – nA) RT H = U + pV (5.7) or p ∆V = ∆ngRT (5.9) so, equation (5.6) becomes qp= H2 – H1 = ∆H Here, ∆ng refers to the number of moles of gaseous products minus the number of moles Although q is a path dependent function, of gaseous reactants.H is a state function because it depends on U, p and V, all of which are state functions. Substituting the value of p∆V from Therefore, ∆H is independent of path. Hence, equation 5.9 in equation 5.8, we get qp is also independent of path. ∆H = ∆U + ∆ngRT (5.10) For finite changes at constant pressure, The equation 5.10 is useful for calculating we can write equation 5.7 as ∆H from ∆U and vice versa. ∆H = ∆U + ∆pV Problem 5.5 Since p is constant, we can write If water vapour is assumed to be a ∆H = ∆U + p∆V (5.8) perfect gas, molar enthalpy change for It is important to note that when heat is vapourisation of 1 mol of water at 1bar absorbed by the system at constant pressure, and 100°C is 41kJ mol–1. Calculate the we are actually measuring changes in the internal energy change, when enthalpy. Reprint 2025-26 144 chemIstry 1 mol of water is vapourised at 1 bar pressure and 100°C. Solution (i) The change H2O (l) → H2O (g) ∆H = ∆U + ∆ngRT Fig. 5.6(a) A gas at volume V and temperature T or ∆U = ∆H – ∆ngRT, substituting the values, we get ∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 Fig. 5.6 (b) Partition, each part having half the volume of the gas (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measure between extensive properties and intensive heat transferred to a system. This heat appears as a rise in temperature of the systemproperties. An extensive property is a in case of heat absorbed by the system.property whose value depends on the quantity or size of matter present in the system. For The increase of temperature is proportional example, mass, volume, internal energy, to the heat transferred enthalpy, heat capacity, etc. are extensive q  coeff Tproperties. The magnitude of the coefficient depends Those properties which do not depend on the size, composition and nature of the on the quantity or size of matter present system. We can also write it as q = C ∆T are known as intensive properties. For The coefficient, C is called the heatexample temperature, density, pressure etc. capacity.are intensive properties. A molar property, Thus, we can measure the heat suppliedχm, is the value of an extensive property χ of the system for 1 mol of the substance. If n is by monitoring the temperature rise, provided  we know the heat capacity.  mthe amount of matter, n is independent When C is large, a given amount of heat of the amount of matter. Other examples are results in only a small temperature rise. Water molar volume, Vm and molar heat capacity, has a large heat capacity i.e., a lot of energy Cm. Let us understand the distinction is needed to raise its temperature. between extensive and intensive properties by C is directly proportional to amount ofconsidering a gas enclosed in a container of substance. The molar heat capacity of avolume V and at temperature T [Fig. 5.6(a)]. Let us make a partition such that volume substance, Cm=  C  is the heat capacityis halved, each part [Fig. 5.6 (b)] now has  n , V for one mole of the substance and isone half of the original volume, , but the 2 the quantity of heat needed to raise the temperature will still remain the same i.e., T. temperature of one mole by one degree It is clear that volume is an extensive property celsius (or one kelvin). Specific heat, also and temperature is an intensive property. called specific heat capacity is the quantity Reprint 2025-26 THERMODYNAMICS 145 of heat required to raise the temperature of i) at constant volume, qV one unit mass of a substance by one degree ii) at constant pressure, qp celsius (or one kelvin). For finding out the heat, q, required to raise the temperatures (a) ∆U Measurements of a sample, we multiply the specific heat For chemical reactions, heat absorbed at of the substance, c, by the mass m, and constant volume, is measured in a bomb temperatures change, ∆T as calorimeter (Fig. 5.7). Here, a steel vessel (the bomb) is immersed in a water bath. The whole q  c  m  T  C T (5.11) device is called calorimeter. The steel vessel is (d) The Relationship between Cp and CV immersed in water bath to ensure that no heat for an Ideal Gas is lost to the surroundings. A combustible At constant volume, the heat capacity, C is denoted by CV and at constant pressure, this is denoted by Cp . Let us find the relationship between the two. We can write equation for heat, q at constant volume as qV = C V T  U at constant pressure as qp = C pT  H The difference between Cp and CV can be derived for an ideal gas as: For a mole of an ideal gas, ∆H = ∆U + ∆(pV ) = ∆U + ∆(RT ) = ∆U + R∆T  H  U  R T (5.12) On putting the values of ∆H and ∆U, we have C p T  C V T RT C p  C V R Fig. 5.7 Bomb calorimeter Cp – CV = R (5.13) substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the5.3 Measurement of ∆U and ∆H: reaction is transferred to the water around the Calorimetry bomb and its temperature is monitored. Since We can measure energy changes associated the bomb calorimeter is sealed, its volume with chemical or physical processes by an does not change i.e., the energy changes experimental technique called calorimetry. associated with reactions are measured at In calorimetry, the process is carried out in a constant volume. Under these conditions, no vessel called calorimeter, which is immersed work is done as the reaction is carried out in a known volume of a liquid. Knowing at constant volume in the bomb calorimeter. the heat capacity of the liquid in which Even for reactions involving gases, there is no calorimeter is immersed and the heat capacity work done as ∆V = 0. Temperature change of of calorimeter, it is possible to determine the the calorimeter produced by the completed heat evolved in the process by measuring reaction is then converted to qV, by using the temperature changes. Measurements are known heat capacity of the calorimeter with made under two different conditions: the help of equation 5.11. Reprint 2025-26 146 chemIstry (b) ∆H Measurements of the bomb calorimeter is 20.7kJ/K, Measurement of heat change at constant what is the enthalpy change for the above pressure (generally under atmospheric pressure) reaction at 298 K and 1 atm? can be done in a calorimeter shown in Fig. 5.8. SolutionWe know that ∆Η = qp (at constant p) and, therefore, heat absorbed or evolved, qp at Suppose q is the quantity of heat from constant pressure is also called the heat of the reaction mixture and CV is the reaction or enthalpy of reaction, ∆rH. heat capacity of the calorimeter, then the quantity of heat absorbed by the In an exothermic reaction, heat is evolved, calorimeter. and system loses heat to the surroundings. q = CV × ∆TTherefore, qp will be negative and ∆rH will also be negative. Similarly in an endothermic Quantity of heat from the reaction will reaction, heat is absorbed, qp is positive and have the same magnitude but opposite ∆rH will be positive. sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter. q = –CV × ∆T = – 20.7 kJ/K × (299 – 298) K = – 20.7 kJ (Here, negative sign indicates the exothermic nature of the reaction) Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK–1 For combustion of 1 mol of graphite, 1 20 .7 kJ  12 .0 g mol  = 1 g = – 2.48 ×102 kJ mol–1 , Since ∆ ng = 0, ∆ H = ∆ U = – 2.48 ×102 kJ mol–1 5.4 Enthalpy change, ∆rH of a reaction – Reaction Enthalpy In a chemical reaction, reactants are converted into products and is represented by,Fig. 5.8 Calorimeter for measuring heat changes at constant pressure (atmospheric Reactants → Products pressure). The enthalpy change accompanying a reaction is called the reaction enthalpy. The Problem 5.6 enthalpy change of a chemical reaction, is 1g of graphite is burnt in a bomb given by the symbol ∆rH calorimeter in excess of oxygen at 298 K ∆rH = (sum of enthalpies of products) – (sum and 1 atmospheric pressure according of enthalpies of reactants) to the equation  H products b i H reactants → ai  (5.14) C (graphite) + O2 (g) CO2 (g) i i During the reaction, temperature rises Here symbol (sigma) is used for ∑ from 298 K to 299 K. If the heat capacity summation and ai and bi are the stoichiometric Reprint 2025-26 THERMODYNAMICS 147 coefficients of the products and reactants ethanol at 298 K is pure liquid ethanol at respectively in the balanced chemical 1 bar; standard state of solid iron at 500 K equation. For example, for the reaction is pure iron at 1 bar. Usually data are taken CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) at 298 K. b i H reac tan ts Standard conditions are denoted by∆r H = ∑ a i H Pr oducts − ∑ i i adding the superscript  to the symbol ∆H, = [Hm (CO2, g) + 2Hm (H2O, l)]– [Hm (CH4, g) e.g., ∆H + 2Hm (O2, g)] where Hm is the molar enthalpy. (b) Enthalpy Changes during Phase Transformations Enthalpy change is a very useful quantity. Knowledge of this quantity is required when Phase transformations also involve energy one needs to plan the heating or cooling changes. Ice, for example, requires heat for required to maintain an industrial chemical melting. Normally this melting takes place at reaction at constant temperature. It is also constant pressure (atmospheric pressure) and required to calculate temperature dependence during phase change, temperature remains of equilibrium constant. constant (at 273 K). H2O(s) → H2O(l); ∆fusH  = 6.00 kJ moI–1(a) Standard Enthalpy of Reactions Here ∆fusH is enthalpy of fusion in standardEnthalpy of a reaction depends on the state. If water freezes, then process is reversedconditions under which a reaction is carried and equal amount of heat is given off to theout. It is, therefore, necessary that we surroundings.must specify some standard conditions. The standard enthalpy of reaction is the The enthalpy change that accompanies enthalpy change for a reaction when all melting of one mole of a solid substance in the participating substances are in their standard state is called standard enthalpy standard states. of fusion or molar enthalpy of fusion, ∆fusH . The standard state of a substance at a specified temperature is its pure form at Melting of a solid is endothermic, so 1 bar. For example, the standard state of liquid all enthalpies of fusion are positive. Water Table 5.1 Standard Enthalpy Changes of Fusion and Vaporisation (Tf and Tb are melting and boiling points, respectively) Reprint 2025-26 148 chemIstry requires heat for evaporation. At constant Solutiontemperature of its boiling point Tb and at constant pressure: We can represent the process of evaporation asH2O(l) → H2O(g); ∆vapH = + 40.79 kJ moI–1 ∆vapH is the standard enthalpy of vaporisation. H 2 O(1) →vaporisation H 2 O(g) 1mol 1mol Amount of heat required to vaporize one mole of a liquid at constant temperature No. of moles in 18 g H2O(l) is 18g and under standard pressure (1bar) is called = –1 =1 mol its standard enthalpy of vaporization or 18g mol molar enthalpy of vaporization, ∆vapH . Heat supplied to evaporate18g water at Sublimation is direct conversion of a 298 K = n × ∆vap H  solid into its vapour. Solid CO2 or ‘dry ice’ = (1 mol) × (44.01 kJ mol–1) sublimes at 195K with ∆subH=25.2 kJ mol–1; = 44.01 kJ naphthalene sublimes slowly and for this (assuming steam behaving as an ideal ∆sub H = 73.0 kJ mol–1 . gas). Standard enthalpy of sublimation, ∆vapU = ∆vapH – p∆V = ∆vapH – ∆ngRT∆subH is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard ∆vapHV – ∆ngRT = 44.01 kJ pressure (1bar). –(1)(8.314 JK–1mol–1)(298K)(10–3kJ J–1) The magnitude of the enthalpy change ∆vapUV = 44.01 kJ – 2.48kJ depends on the strength of the intermolecular = 41.53 kJinteractions in the substance undergoing the phase transfomations. For example, Problem 5.8 the strong hydrogen bonds between water Assuming the water vapour to be a perfectmolecules hold them tightly in liquid phase. gas, calculate the internal energy changeFor an organic liquid, such as acetone, the when 1 mol of water at 100°C and 1 bar intermolecular dipole-dipole interactions are pressure is converted to ice at 0°C. Given significantly weaker. Thus, it requires less the enthalpy of fusion of ice is 6.00 kJ heat to vaporise 1 mol of acetone than it does mol-1 heat capacity of water is 4.2 J/g°C to vaporize 1 mol of water. Table 5.1 gives The change take place as follows:  1values of standard enthalpy changes of fusion Step - 1 1 mol H2O (l, 100°C) and vaporisation for some substances. mol (l, 0°C) Enthalpy change ∆H1 Problem 5.7  1 mol Step - 2 1 mol H2O (l, 0°C) A swimmer coming out from a pool is H2O( S, 0°C) Enthalpy covered with a film of water weighing change ∆H2 about 18g. How much heat must be Total enthalpy change will be - supplied to evaporate this water at ∆H = ∆H1 + ∆H2 298 K ? Calculate the internal energy of vaporisation at 298K. ∆H1 = - (18 x 4.2 x 100) J mol-1 ∆vap H  for water = - 7560 J mol-1 = - 7.56 k J mol-1 at 298K= 44.01kJ mol–1 ∆H2 = - 6.00 kJ mol-1 Reprint 2025-26 THERMODYNAMICS 149 Table 5.2 Standard Molar Enthalpies of Formation (∆f H ) at 298K of a Few Selected Substances of aggregation (also known as reference Therefore, states) is called Standard Molar Enthalpy ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1) of Formation. Its symbol is ∆fH , where the = -13.56 kJ mol-1 subscript ‘ f ’ indicates that one mole of the There is negligible change in the volume compound in question has been formed in its during the change form liquid to solid standard state from its elements in their most state. stable states of aggregation. The reference Therefore, p∆v = ∆ng RT = 0 state of an element is its most stable state ∆H = ∆U = - 13.56kJ mol-1 of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen (c) Standard Enthalpy of Formation is H2 gas and those of dioxygen, carbon The standard enthalpy change for the and sulphur are O2 gas, Cgraphite and Srhombic formation of one mole of a compound from respectively. Some reactions with standard its elements in their most stable states molar enthalpies of formation are as follows. Reprint 2025-26 150 chemIstry H2(g) + ½O2 (g) → H2O(1); Here, we can make use of standard enthalpy of formation and calculate the enthalpy∆f H = –285.8 kJ mol–1 change for the reaction. The following general C (graphite, s) + 2H2(g) → Ch4 (g); equation can be used for the enthalpy change ∆f H = –74.81 kJ mol–1 calculation. 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1); ∆rH = i∑ai ∆f H (products) – i∑bi ∆f H (reactants) ∆f H  = – 277.7kJ mol–1 (5.15) where a and b represent the coefficients of It is important to understand that a the products and reactants in the balancedstandard molar enthalpy of formation, ∆fH , equation. Let us apply the above equation foris just a special case of ∆rH , where one mole decomposition of calcium carbonate. Here,of a compound is formed from its constituent coefficients ‘a’ and ‘b’ are 1 each. Therefore,elements, as in the above three equations, where 1 mol of each, water, methane and ∆rH = ∆f H  = [CaO(s)]+ ∆f H  [CO2(g)] ethanol is formed. In contrast, the enthalpy – ∆f H  = [CaCO3(s)] change for an exothermic reaction: =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1) CaO(s) + CO2(g) → CaCo3(s); –1(–1206.9 kJ mol–1) ∆rH  = – 178.3kJ mol–1 = 178.3 kJ mol–1 is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been Thus, the decomposition of CaCO3 (s) is formed from other compounds, and not from an endothermic process and you have to heat its constituent elements. Also, for the reaction it for getting the desired products. given below, enthalpy change is not standard (d) Thermochemical Equations enthalpy of formation, ∆fH  for HBr(g). A balanced chemical equation together with H2(g) + Br2(l) → 2HBr(g); the value of its ∆rH is called a thermochemical ∆r H  = – 178.3kJ mol–1 equation. We specify the physical state Here two moles, instead of one mole of the (alongwith allotropic state) of the substance product is formed from the elements, i.e., in an equation. For example: ∆r H  = 2∆f H C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); Therefore, by dividing all coefficients in ∆rH = – 1367 kJ mol–1 the balanced equation by 2, expression for The above equation describes the enthalpy of formation of HBr (g) is written as combustion of liquid ethanol at constant ½H2(g) + ½Br2(1) → HBr(g); temperature and pressure. The negative sign ∆f H = – 36.4 kJ mol–1 of enthalpy change indicates that this is an exothermic reaction. Standard enthalpies of formation of some common substances are given in Table 5.2. It would be necessary to remember the following conventions regarding thermo- By convention, standard enthalpy for chemical equations.formation, ∆fH , of an element in reference 1. The coefficients in a balanced thermo-state, i.e., its most stable state of aggregation chemical equation refer to the number ofis taken as zero. moles (never molecules) of reactants and Suppose, you are a chemical engineer and products involved in the reaction.want to know how much heat is required to decompose calcium carbonate to lime and 2. The numerical value of ∆rH  refers to the carbon dioxide, with all the substances in number of moles of substances specified their standard state. by an equation. Standard enthalpy change CaCO3(s) → CaO(s) + CO2(g); ∆r H = ? ∆rH will have units as kJ mol–1. Reprint 2025-26 THERMODYNAMICS 151 To illustrate the concept, let us consider (e) Hess’s Law of Constant Heat the calculation of heat of reaction for the Summation following reaction : We know that enthalpy is a state function, Fe 2 O 3 s 3 H 2 g 2 Fe s 3 H 2 O ,l therefore the change in enthalpy is independent of the path between initial state (reactants) From the Table (5.2) of standard enthalpy of and final state (products). In other words, formation (∆f H ), we find : enthalpy change for a reaction is the same ∆f H (H2O,l) = –285.83 kJ mol–1; whether it occurs in one step or in a series of steps. This may be stated as follows in the∆f H (Fe2O3,s) = – 824.2 kJ mol–1; form of Hess’s Law. Also ∆f H (Fe, s) = 0 and If a reaction takes place in several ∆f H (H2, g) = 0 as per convention steps then its standard reaction enthalpy Then, is the sum of the standard enthalpies of ∆f H1 = 3(–285.83 kJ mol–1) the intermediate reactions into which the overall reaction may be divided at the same – 1(– 824.2 kJ mol–1) temperature. = (–857.5 + 824.2) kJ mol–1 Let us understand the importance of this = –33.3 kJ mol–1 law with the help of an example. Note that the coefficients used in these Consider the enthalpy change for thecalculations are pure numbers, which reactionare equal to the respective stoichiometric C (graphite,s) + O2 (g) → CO (g); ∆r H  = ?coefficients. The unit for ∆rH  is kJ mol–1, which means per mole of reaction. Although CO(g) is the major product, some Once we balance the chemical equation in a CO2 gas is always produced in this reaction. particular way, as above, this defines the mole Therefore, we cannot measure enthalpy of reaction. If we had balanced the equation change for the above reaction directly. differently, for example, However, if we can find some other reactions 1 3 3 involving related species, it is possible to Fe 2 O 3 s H 2 g Fe s H 2 O l2 2 2 calculate the enthalpy change for the above reaction.then this amount of reaction would be one mole of reaction and ∆rH  would be Let us consider the following reactions:  3 C (graphite,s) + O2 (g) → CO2 (g);∆f H 2 = (–285.83 kJ mol–1)  = – 393.5 kJ mol–1 (i) 2 ∆r H 1 – (–824.2 kJ mol–1) 1 2 CO (g) + O2 (g) → CO2 (g) 2  = (– 428.7 + 412.1) kJ mol–1 ∆r H = – 283.0 kJ mol–1 (ii) = –16.6 kJ mol–1 = ½ ∆r H 1 We can combine the above two reactions It shows that enthalpy is an extensive in such a way so as to obtain the desired quantity. reaction. To get one mole of CO(g) on the 3. When a chemical equation is reversed, right, we reverse equation (ii). In this, heat the value of ∆rH  is reversed in sign. For is absorbed instead of being released, so we example change sign of ∆rH  value N2(g) + 3H2 (g) → 2NH3 (g); CO2 (g) → CO (g) + O2 (g); ∆r H  = – 91.8 kJ. mol–1 ∆r H  = + 283.0 kJ mol–1 (iii) 2NH3(g) → N2(g) + 3H2 (g); ∆r H  = + 91.8 kJ mol–1 Reprint 2025-26 152 chemIstry Adding equation (i) and (iii), we get the Similarly, combustion of glucose gives out desired equation, 2802.0 kJ/mol of heat, for which the overall equation is : 1 C  graphite, s  O 2 g CO g ; 2 C 6 H12 O 6 ( g )  6 O 2 ( g )  6 CO 2 ( g )  6 H 2 O(1);  = (– 393.5 + 283.0) ∆C H = – 2802.0 kJ mol–1 for which ∆r H  Our body also generates energy from food = – 110.5 kJ mol–1 by the same overall process as combustion, In general, if enthalpy of an overall although the final products are produced after reaction A→B along one route is ∆rH and a series of complex bio-chemical reactions ∆rH1, ∆rH2, ∆rH3..... representing enthalpies involving enzymes. of reactions leading to same product, B along another route, then we have Problem 5.9 ∆rH = ∆rH1 + ∆rH2 + ∆rH3 ... (5.16) The combustion of one mole of benzene It can be represented as: takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are ∆rH produced and 3267.0 kJ of heat is A B liberated. Calculate the standard ∆H1 ∆rH3 enthalpy of formation, ∆f H  of benzene. Standard enthalpies of formation of C D CO2(g) and H2O(l) are –393.5 kJ mol–1 ∆rH2 and – 285.83 kJ mol–1 respectively. Solution5.5 Enthalpies for different types of reactions The formation reaction of benezene is given by :It is convenient to give name to enthalpies specifying the types of reactions. 6 C  graphite  3 H 2 g C 6 H 6 ;l  = ? ... (i)(a) Standard Enthalpy of Combustion ∆f H (symbol : ∆cH ) The enthalpy of combustion of 1 mol Combustion reactions are exothermic in of benzene is : nature. These are important in industry, 15rocketry, and other walks of life. Standard C 6 H 6 l O 2  6 CO 2 g 3 H 2 O ;l 2enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) ∆C H  = – 3267 kJ mol–1... (ii) of a substance, when it undergoes combustion The enthalpy of formation of 1 mol of and all the reactants and products being CO2(g) : in their standard states at the specified C  graphite  O 2 g CO 2 g ; temperature. ∆f H  = – 393.5 kJ mol–1... (iii) Cooking gas in cylinders contains mostly butane (C4H10). During complete combustion The enthalpy of formation of 1 mol of of one mole of butane, 2658 kJ of heat is H2O(l) is : released. We can write the thermochemical 1 H 2 g O 2 g H 2 O ;lreactions for this as: 2 13 C 4 H10 ( g )  O 2 ( g )  4 CO 2 ( g )  5 H 2 O(1); ∆C H = – 285.83 kJ mol–1... (iv) 2 multiplying eqn. (iii) by 6 and eqn. (iv) ∆C H  = – 2658.0 kJ mol–1 by 3 we get: Reprint 2025-26 THERMODYNAMICS 153 In this case, the enthalpy of atomization is same as the enthalpy of sublimation. 6 C  graphite  6 O 2 g 6 CO 2 ;g ∆f H  = – 2361 kJ mol–1 (c) Bond Enthalpy (symbol: ∆bondH ) 3 Chemical reactions involve the breaking and 3 H 2 g O 2 g 3 H 2 O ;1 making of chemical bonds. Energy is required 2  to break a bond and energy is released when a ∆f H = – 857.49 kJ mol–1 bond is formed. It is possible to relate heat of Summing up the above two equations : reaction to changes in energy associated with 6 C  graphite  3 H 2 g 15 O 2 g 6 CO 2 g breaking and making of chemical bonds. With 2 reference to the enthalpy changes associated  3 H 2 O ;l with chemical bonds, two different terms are  used in thermodynamics. ∆f H = – 3218.49 kJ mol–1... (v) (i) Bond dissociation enthalpy Reversing equation (ii); (ii) Mean bond enthalpy 15 6 CO 2 g 3 H 2 O l C 6 H 6 l O 2 ; Let us discuss these terms with reference 2 to diatomic and polyatomic molecules. ∆f H  = – 3267.0 kJ mol–1... (vi) Diatomic Molecules: Consider the following Adding equations (v) and (vi), we get process in which the bonds in one mole of 6 C  graphite  3 H 2 g C 6 H 6 ;l dihydrogen gas (H2) are broken: H2(g) → 2H(g); ∆H–HH = 435.0 kJ mol–1 ∆f H  = – 48.51 kJ mol–1... (iv) The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond. The bond dissociation enthalpy is the (b) Enthalpy of Atomization change in enthalpy when one mole of covalent (symbol: ∆aH ) bonds of a gaseous covalent compound is Consider the following example of atomization broken to form products in the gas phase. of dihydrogen Note that it is the same as the enthalpy of H2(g) → 2H(g); ∆aH  = 435.0 kJ mol–1 atomization of dihydrogen. This is true for all diatomic molecules. For example:You can see that H atoms are formed by breaking H–H bonds in dihydrogen. The Cl2(g) → 2Cl(g); ∆Cl–ClH = 242 kJ mol–1 enthalpy change in this process is known O2(g) → 2O(g); ∆O=OH = 428 kJ mol–1as enthalpy of atomization, ∆aH . It is the enthalpy change on breaking one mole of In the case of polyatomic molecules, bond bonds completely to obtain atoms in the gas dissociation enthalpy is different for different bonds within the same molecule.phase. Polyatomic Molecules: Let us now consider In case of diatomic molecules, like a polyatomic molecule like methane, CH4.dihydrogen (given above), the enthalpy of The overall thermochemical equation for itsatomization is also the bond dissociation atomization reaction is given below:enthalpy. The other examples of enthalpy of atomization can be CH 4 (g) → C(g) + 4H(g);  = 1665 kJ mol–1CH4(g) → C(g) + 4H(g); ∆aH  = 1665 kJ mol–1 ∆a H Note that the products are only atoms of C In methane, all the four C – H bonds are and H in gaseous phase. Now see the following identical in bond length and energy. However, reaction: the energies required to break the individual Na(s) → Na(g); ∆aH = 108.4 kJ mol–1 C – H bonds in each successive step differ : Reprint 2025-26 154 chemIstry CH4(g) → CH3(g)+H(g);∆bond H = +427 kJ mol–1 given in Table 5.3. The reaction enthalpies are very important quantities as these arise fromCH3(g) → CH2(g)+H(g);∆bond H = +439 kJ mol–1 the changes that accompany the breaking of CH2(g) → CH(g)+H(g);∆bond H = +452 kJ mol–1 old bonds and formation of the new bonds. CH(g) → C(g)+H(g);∆bond H = +347 kJ mol–1 We can predict enthalpy of a reaction in gas phase, if we know different bond enthalpies.Therefore, The standard enthalpy of reaction, ∆rH isCH4(g) → C(g)+4H(g);∆a H = 1665 kJ mol–1 related to bond enthalpies of the reactants In such cases we use mean bond enthalpy and products in gas phase reactions as: of C – H bond. bond enthalpiesreactants ∆r H  For example in CH4, ∆C–HH  is calculated as: bond enthalpies products∆C–HH = ¼ (∆a H) = ¼ (1665 kJ mol–1)   (5.17)** = 416 kJ mol–1 This relationship is particularly more We find that mean C–H bond enthalpy useful when the required values of ∆f H are in methane is 416 kJ/mol. It has been not available. The net enthalpy change of a found that mean C–H bond enthalpies differ reaction is the amount of energy required slightly from compound to compound, as to break all the bonds in the reactant in CH3CH2Cl, CH3NO2, etc., but it does not molecules minus the amount of energy differ in a great deal*. Using Hess’s law, bond required to break all the bonds in the product enthalpies can be calculated. Bond enthalpy molecules. Remember that this relationship is values of some single and multiple bonds are approximate and is valid when all substances Table 5.3(a) Some Mean Single Bond Enthalpies in kJ mol–1 at 298 K H C N O F Si P S Cl Br I 435.8 414 389 464 569 293 318 339 431 368 297 H 347 293 351 439 289 264 259 330 276 238 C 159 201 272 - 209 - 201 243 - N 138 184 368 351 - 205 - 201 O 155 540 490 327 255 197 - F 176 213 226 360 289 213 Si 213 230 331 272 213 P 213 251 213 - S 243 218 209 CI 192 180 Br 151 I Table 5.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1 at 298 K N = N 418 C = C 611 O = O 498 N ≡ N 946 C ≡ C 837 C = N 615 C = O 741 C ≡ N 891 C ≡ O 1070 * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. ** If we use enthalpy of bond formation, (∆f H bond), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ∆f H  = ∑∆f Hbonds of products – ∑∆f H bonds of reactants Reprint 2025-26 THERMODYNAMICS 155 (reactants and products) in the reaction are  1 2. Na( g )  Na ( g )  e ( g ) , the ionization ofin gaseous state. sodium atoms, ionization enthalpy (d) Lattice Enthalpy ∆iH = 496 kJ mol–1 1 The lattice enthalpy of an ionic compound is 3. Cl 2 ( g ) → Cl( g ), the dissociation of 2the enthalpy change which occurs when one mole of an ionic compound dissociates into chlorine, the reaction enthalpy is half the its ions in gaseous state. bond dissociation enthalpy.     1Na Cl s Na ( g )  Cl g ; ∆bondH = 121 kJ mol–1  2 ∆latticeH = +788 kJ mol–1 4. Cl( g )  e 1 ( g )  Cl( g ) electron gained Since it is impossible to determine lattice by chlorine atoms. The electron gainenthalpies directly by experiment, we use  enthalpy, ∆egH = – 348.6 kJ mol–1.an indirect method where we construct an You have learnt about ionization enthalpyenthalpy diagram called a Born-Haber Cycle and electron gain enthalpy in Unit 3.(Fig. 5.9). In fact, these terms have been taken Let us now calculate the lattice enthalpy from thermodynamics. Earlier terms, of Na+Cl–(s) by following steps given below : ionization energy and electron affinity 1. Na ( s ) → Na( g ) , sublimation of sodium were in practice in place of the above metal, ∆subH  = 108.4 kJ mol–1 terms (see the box for justification). Ionization Energy and Electron Affinity Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account. Enthalpies of reactions for M(g) → M+(g) + e– (for ionization) M(g) + e– → M–(g) (for electron gain) at temperature, T is T ∆rH(T ) = ∆rH(0) + ∆rCPdT 0∫ The value of Cp for each species in the above reaction is 5/2 R (CV = 3/2R) So, ∆rCp = + 5/2 R (for ionization) ∆rCp = – 5/2 R (for electron gain) Therefore, ∆rH (ionization enthalpy) = E0 (ionization energy) + 5/2 RT ∆rH (electron gain enthalpy) = – A( electron affinity) – 5/2 RT 5. Na + (g) + Cl − (g) → Na + Cl − (s) Fig. 5.9 Enthalpy diagram for lattice enthalpy of The sequence of steps is shown in NaCl Fig. 5.9, and is known as a Born-Haber Reprint 2025-26 156 chemIstry cycle. The importance of the cycle is that, The enthalpy of solution of AB(s), ∆solH, the sum of the enthalpy changes round a in water is, therefore, determined by the cycle is zero. Applying Hess’s law, we get, selective values of the lattice enthalpy, ∆latticeH ∆latticeH = 411.2 + 108.4 + 121 + 496 – 348.6 and enthalpy of hydration of ions, ∆hydH as ∆sol H  = ∆latticeH  + ∆hydH ∆latticeH = + 788kJ For most of the ionic compounds, ∆solfor NaCl(s)  Na+(g) + Cl–(g) H is positive and the dissociation processInternal energy is smaller by 2RT (because ∆ng is endothermic. Therefore the solubility of= 2) and is equal to + 783 kJ mol–1. most salts in water increases with rise of Now we use the value of lattice enthalpy temperature. If the lattice enthalpy is very to calculate enthalpy of solution from the high, the dissolution of the compound may expression: not take place at all. Why do many fluorides tend to be less soluble than the corresponding∆solH = ∆latticeH + ∆hydH chlorides? Estimates of the magnitudes ofFor one mole of NaCl(s), enthalpy changes may be made by usinglattice enthalpy = + 788 kJ mol–1  tables of bond energies (enthalpies) and latticeand ∆hydH = – 784 kJ mol–1( from the energies (enthalpies). literature) ∆sol H = + 788 kJ mol–1 – 784 kJ mol–1 (f) Enthalpy of Dilution = + 4 kJ mol–1 It is known that enthalpy of solution is the The dissolution of NaCl(s) is accompanied enthalpy change associated with the addition by very little heat change. of a specified amount of solute to the specified amount of solvent at a constant temperature(e) Enthalpy of Solution (symbol : ∆solH ) and pressure. This argument can be appliedEnthalpy of solution of a substance is the to any solvent with slight modification.enthalpy change when one mole of it dissolves Enthalpy change for dissolving one mole ofin a specified amount of solvent. The enthalpy gaseous hydrogen chloride in 10 mol of waterof solution at infinite dilution is the enthalpy can be represented by the following equation.change observed on dissolving the substance For convenience we will use the symbol aq.in an infinite amount of solvent when the for waterinteractions between the ions (or solute molecules) are negligible. HCl(g) + 10 aq. → HCl.10 aq. When an ionic compound dissolves in a ∆H = –69.01 kJ / mol solvent, the ions leave their ordered positions Let us consider the following set of on the crystal lattice. These are now more free in enthalpy changes: solution. But solvation of these ions (hydration (S-1) HCl(g) + 25 aq. → HCl.25 aq.in case solvent is water) also occurs at the ∆H = –72.03 kJ / molsame time. This is shown diagrammatically, for an ionic compound, AB (s) (S-2) HCl(g) + 40 aq. → HCl.40 aq. ∆H = –72.79 kJ / mol (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq. ∆H = –74.85 kJ / mol The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3). Reprint 2025-26 THERMODYNAMICS 157 If we subtract the first equation (equation many years without observing any perceptible S-1) from the second equation (equation S-2) change. Although the reaction is taking place in the above set of equations, we obtain– between them, it is at an extremely slow rate. It is still called spontaneous reaction. HCl.25 aq. + 15 aq. → HCl.40 aq. So spontaneity means ‘having the potential ∆H = [ –72.79 – (–72.03)] kJ / mol to proceed without the assistance of external = – 0.76 kJ / mol agency’. However, it does not tell about the This value (–0.76kJ/mol) of ∆H is enthalpy rate of the reaction or process. Another aspect of dilution. It is the heat withdrawn from of spontaneous reaction or process, as we see the surroundings when additional solvent is is that these cannot reverse their direction on added to the solution. The enthalpy of dilution their own. We may summarise it as follows: of a solution is dependent on the original A spontaneous process is an concentration of the solution and the amount irreversible process and may only be of solvent added. reversed by some external agency. 5.6 spontaneity (a) Is Decrease in Enthalpy a Criterion The first law of thermodynamics tells us for Spontaneity ? about the relationship between the heat If we examine the phenomenon like flow of absorbed and the work performed on or water down hill or fall of a stone on to the by a system. It puts no restrictions on ground, we find that there is a net decrease the direction of heat flow. However, the in potential energy in the direction of change. flow of heat is unidirectional from higher By analogy, we may be tempted to state that temperature to lower temperature. In fact, a chemical reaction is spontaneous in a all naturally occurring processes whether given direction, because decrease in energy chemical or physical will tend to proceed has taken place, as in the case of exothermic spontaneously in one direction only. For reactions. For example: example, a gas expanding to fill the available 1 volume, burning carbon in dioxygen giving N2(g) + H2(g) = NH3(g); 2 carbon dioxide. ∆r H  = – 46.1 kJ mol–1 But heat will not flow from colder body to 1 1 H2(g) + Cl2(g) = HCl (g);warmer body on its own, the gas in a container 2 2 will not spontaneously contract into one ∆r H  = – 92.32 kJ mol–1 corner or carbon dioxide will not form carbon 1 H2(g) + O2(g) → H2O(l) ;and dioxygen spontaneously. These and many 2 ∆r H  = –285.8 kJ mol–1other spontaneously occurring changes show unidirectional change. We may ask ‘what is The decrease in enthalpy in passing from the driving force of spontaneously occurring reactants to products may be shown for any changes ? What determines the direction of a exothermic reaction on an enthalpy diagram spontaneous change ? In this section, we shall as shown in Fig. 5.10(a). establish some criterion for these processes Thus, the postulate that driving force for whether these will take place or not. a chemical reaction may be due to decrease Let us first understand what do we mean in energy sounds ‘reasonable’ as the basis of by spontaneous reaction or change ? You evidence so far ! may think by your common observation that Now let us examine the following reactions: spontaneous reaction is one which occurs 1 N2(g) + O2(g) → NO2(g);immediately when contact is made between 2 the reactants. Take the case of combination ∆r H = +33.2 kJ mol–1 of hydrogen and oxygen. These gases may C(graphite, s) + 2 S(l) → CS2(l); be mixed at room temperature and left for ∆r H = +128.5 kJ mol–1 Reprint 2025-26 158 chemIstry Fig. 5.10 (a) Enthalpy diagram for exothermic reactions These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 5.10(b). Fig. 5.11 Diffusion of two gases respectively and separated by a movable partition [Fig. 5.11 (a)]. When the partition is withdrawn [Fig. 5.11(b)], the gases begin to diffuse into each other and after a period of time, diffusion will be complete. Let us examine the process. Before partition, if we were to pick up the gas molecules from left container, we would be Fig. 5.10 (b) Enthalpy diagram for endothermic sure that these will be molecules of gas A reactions and similarly if we were to pick up the gas molecules from right container, we would Therefore, it becomes obvious that while be sure that these will be molecules of gasdecrease in enthalpy may be a contributory B. But, if we were to pick up molecules fromfactor for spontaneity, but it is not true for container when partition is removed, we areall cases. not sure whether the molecules picked are of (b) Entropy and Spontaneity gas A or gas B. We say that the system has Then, what drives the spontaneous process become less predictable or more chaotic. in a given direction ? Let us examine such a We may now formulate another postulate: case in which ∆H = 0 i.e., there is no change in in an isolated system, there is always a enthalpy, but still the process is spontaneous. tendency for the systems’ energy to become Let us consider diffusion of two gases more disordered or chaotic and this could be into each other in a closed container which a criterion for spontaneous change ! is isolated from the surroundings as shown At this point, we introduce another in Fig. 5.11. thermodynamic function, entropy denoted The two gases, say, gas A and gas B are as S. The above mentioned disorder is the represented by black dots and white dots manifestation of entropy. To form a mental Reprint 2025-26 THERMODYNAMICS 159 picture, one can think of entropy as a measure qrev of the degree of randomness or disorder in the ∆S = (5.18) T system. The greater the disorder in an isolated The total entropy change (∆Stotal) for the system, the higher is the entropy. As far as a system and surroundings of a spontaneous chemical reaction is concerned, this entropy process is given by change can be attributed to rearrangement of S total  S system  S surr 0 (5.19)atoms or ions from one pattern in the reactants to another (in the products). If the structure When a system is in equilibrium, the of the products is very much disordered than entropy is maximum, and the change in that of the reactants, there will be a resultant entropy, ∆S = 0. increase in entropy. The change in entropy We can say that entropy for a spontaneousaccompanying a chemical reaction may be process increases till it reaches maximumestimated qualitatively by a consideration of and at equilibrium the change in entropy isthe structures of the species taking part in the zero. Since entropy is a state property, we canreaction. Decrease of regularity in structure calculate the change in entropy of a reversiblewould mean increase in entropy. For a given process bysubstance, the crystalline solid state is the state of lowest entropy (most ordered), The qsys ,rev gaseous state is state of highest entropy. ∆Ssys = T Now let us try to quantify entropy. One way We find that both for reversible and to calculate the degree of disorder or chaotic irreversible expansion for an ideal gas, under distribution of energy among molecules isothermal conditions, ∆U = 0, but ∆Stotalwould be through statistical method which i.e., S sys  S surr is not zero for irreversible is beyond the scope of this treatment. Other process. Thus, ∆U does not discriminate way would be to relate this process to the between reversible and irreversible process, heat involved in a process which would make whereas ∆S does. entropy a thermodynamic concept. Entropy, like any other thermodynamic property such as internal energy U and enthalpy H is a state Problem 5.10 function and ∆S is independent of path. Predict in which of the following, entropy Whenever heat is added to the system, increases/decreases : it increases molecular motions causing (i) A liquid crystallizes into a solid. increased randomness in the system. Thus (ii) Temperature of a crystalline solidheat (q) has randomising influence on the is raised from 0 K to 115 K.system. Can we then equate ∆S with q ? Wait ! Experience suggests us that the distribution  iii  2 NaHCO 3 s Na 2 CO 3 s of heat also depends on the temperature at CO 2 g H 2 O gwhich heat is added to the system. A system at higher temperature has greater randomness (iv) H 2 g 2 H g in it than one at lower temperature. Thus, Solutiontemperature is the measure of average chaotic motion of particles in the system. (i) After freezing, the molecules attain an ordered state and therefore,Heat added to a system at lower temperature entropy decreases.causes greater randomness than when the same quantity of heat is added to it at higher (ii) At 0 K, the contituent particles are temperature. This suggests that the entropy static and entropy is minimum. change is inversely proportional to the If temperature is raised to 115 K, temperature. ∆S is related with q and T for a these begin to move and oscillate reversible reaction as : Reprint 2025-26 160 chemIstry = 4980.6 JK–1 mol–1 about their equilibrium positions This shows that the above reaction is in the lattice and system becomes spontaneous. more disordered, therefore entropy increases. (c) Gibbs Energy and Spontaneity (iii) Reactant, NaHCO3 is a solid and it has low entropy. Among products We have seen that for a system, it is the there are one solid and two gases. total entropy change, ∆Stotal which decides Therefore, the products represent a the spontaneity of the process. But most of condition of higher entropy. the chemical reactions fall into the category (iv) Here one molecule gives two atoms of either closed systems or open systems. i.e., number of particles increases Therefore, for most of the chemical reactions leading to more disordered state. there are changes in both enthalpy and Two moles of H atoms have entropy. It is clear from the discussion in higher entropy than one mole of previous sections that neither decrease in dihydrogen molecule. enthalpy nor increase in entropy alone can determine the direction of spontaneous Problem 5.11 change for these systems. For oxidation of iron, For this purpose, we define a new 4 Fe s 3O 2 g 2 Fe 2 O 3 s thermodynamic function the Gibbs energy or Gibbs function, G, as entropy change is – 549.4 JK–1 mol–1 at 298 K. Inspite of negative entropy G = H – TS (5.20) change of this reaction, why is the Gibbs function, G is an extensive property reaction spontaneous? and a state function. (∆rH for this reaction is The change in Gibbs energy for the –1648 × 103 J mol–1) system, ∆Gsys can be written as Solution G sys = H sys  T S sys  S sys T One decides the spontaneity of a reaction by considering At constant temperature,  G sys = H sys  T S sys S total  S sys  S surr . For calculating ∆Ssurr, we have to consider the heat Usually the subscript ‘system’ is dropped absorbed by the surroundings which and we simply write this equation as is equal to – ∆rH . At temperature T, G  H  T S (5.21) entropy change of the surroundings is Thus, Gibbs energy change = enthalpy change – temperature × entropy change, and is referred to as the Gibbs equation, one of 1648  10 3 J mol 1   the most important equations in chemistry.  Here, we have considered both terms together 298 K for spontaneity: energy (in terms of ∆H) = 5530 JK–1mol–1 and entropy (∆S, a measure of disorder) as Thus, total entropy change for this indicated earlier. Dimensionally if we analyse, reaction we find that ∆G has units of energy because, r S total  5530 JK –1 mol –1  both ∆H and the T∆S are energy terms, since –1 –1 T∆S = (K) (J/K) = J. 549 .4 JK mol   Now let us consider how ∆G is related to reaction spontaneity. Reprint 2025-26 THERMODYNAMICS 161 We know, be small. The former is one of the reasons ∆Stotal = ∆Ssys + ∆Ssurr why reactions are often carried out at high temperature. Table 5.4 summarises the effect If the system is in thermal equilibrium with of temperature on spontaneity of reactions.the surrounding, then the temperature of the surrounding is same as that of the system. (d) Entropy and Second Law of Also, increase in enthalpy of the surrounding Thermodynamics is equal to decrease in the enthalpy of the We know that for an isolated systemsystem. the change in energy remains constant. Therefore, entropy change of surroundings, Therefore, increase in entropy in such H surr H sys systems is the natural direction of a Ssurr =  T T spontaneous change. This, in fact is the second law of thermodynamics. Like first  H sys  law of thermodynamics, second law can also S total  S sys   T  be stated in several ways. The second law of thermodynamics explains why spontaneous Rearranging the above equation: exothermic reactions are so common. T∆Stotal = T∆Ssys – ∆Hsys In exothermic reactions heat released For spontaneous process, Stotal 0 , so by the reaction increases the disorder T∆Ssys – ∆Hsys > Ο of the surroundings and overall entropy change is positive which makes the reaction  0 spontaneous.  H sys  T S sys  Using equation 5.21, the above equation can (e) Absolute Entropy and Third Law of be written as Thermodynamics –∆G > O Molecules of a substance may move in a straight line in any direction, they mayG  H  T S 0 (5.22) spin like a top and the bonds in the ∆Hsys is the enthalpy change of a reaction, molecules may stretch and compress. T∆Ssys is the energy which is not available to do These motions of the molecule are called useful work. So ∆G is the net energy available translational, rotational and vibrational to do useful work and is thus a measure of the motion respectively. When temperature of ‘free energy’. For this reason, it is also known the system rises, these motions become as the free energy of the reaction. more vigorous and entropy increases. On the ∆G gives a criteria for spontaneity at other hand when temperature is lowered, the entropy decreases. The entropy of any pureconstant pressure and temperature. crystalline substance approaches zero as (i) If ∆G is negative (< 0), the process is the temperature approaches absolute zero. spontaneous. This is called third law of thermodynamics. (ii) If ∆G is positive (> 0), the process is non This is so because there is perfect order in spontaneous. a crystal at absolute zero. The statement is confined to pure crystalline solids becauseNote : If a reaction has a positive enthalpy theoretical arguments and practical evidenceschange and positive entropy change, it can have shown that entropy of solutions andbe spontaneous when T∆S is large enough to super cooled liquids is not zero at 0 K. The outweigh ∆H. This can happen in two ways; importance of the third law lies in the fact that (a) The positive entropy change of the system it permits the calculation of absolute values can be ‘small’ in which case T must be of entropy of pure substance from thermal large. (b) The positive entropy change of the data alone. For a pure substance, this can system can be ‘large’, in which case T may Reprint 2025-26 162 chemIstry q rev is minimum. If it is not, the system wouldbe done by summing increments from 0 T spontaneously change to configuration of K to 298 K. Standard entropies can be used lower free energy. to calculate standard entropy changes by a So, the criterion for equilibrium Hess’s law type of calculation. A + B  C + D; is ∆rG = 0 5.7 Gibbs energy change and Gibbs energy for a reaction in which all equilibrium reactants and products are in standard state, We have seen how a knowledge of the sign ∆rG  is related to the equilibrium constant of and magnitude of the free energy change of a the reaction as follows:  + RT ln Kchemical reaction allows: 0 = ∆rG (i) Prediction of the spontaneity of the or ∆rG = – RT ln K chemical reaction.  or ∆rG = – 2.303 RT log K (5.23) (ii) Prediction of the useful work that could We also know that be extracted from it. So far we have considered free energy (5.24) changes in irreversible reactions. Let us now For strongly endothermic reactions, the examine the free energy changes in reversible value of ∆rH may be large and positive. In reactions. such a case, value of K will be much smaller ‘Reversible’ under strict thermodynamic than 1 and the reaction is unlikely to sense is a special way of carrying out form much product. In case of exothermic a process such that system is at all reactions, ∆rH is large and negative, and ∆rG times in perfect equilibrium with its is likely to be large and negative too. In such surroundings. When applied to a chemical cases, K will be much larger than 1. We may reaction, the term ‘reversible’ indicates expect strongly exothermic reactions to have a that a given reaction can proceed in either large K, and hence can go to near completion. direction simultaneously, so that a dynamic ∆rG also depends upon ∆rS, if the changes equilibrium is set up. This means that the in the entropy of reaction is also taken into reactions in both the directions should account, the value of K or extent of chemical proceed with a decrease in free energy, which reaction will also be affected, depending upon seems impossible. It is possible only if at whether ∆rS is positive or negative. equilibrium the free energy of the system Using equation (5.24), Table 5.4 Effect of Temperature on Spontaneity of Reactions ∆rH ∆rS ∆rG Description* – + – Reaction spontaneous at all temperatures – – – (at low T ) Reaction spontaneous at low temperature – – + (at high T ) Reaction nonspontaneous at high temperature + + + (at low T ) Reaction nonspontaneous at low temperature + + – (at high T ) Reaction spontaneous at high temperature + – + (at all T ) Reaction nonspontaneous at all temperatures * The term low temperature and high temperature are relative. For a particular reaction, high temperature could even mean room temperature. Reprint 2025-26 THERMODYNAMICS 163 (i) It is possible to obtain an estimate of ∆G –13 .6  10 3 J mol –1   from the measurement of ∆H and ∆S, = –1 –1 2 .303 8 .314 JK mol 298 K    and then calculate K at any temperature for economic yields of the products. = 2.38 Hence K = antilog 2.38 = 2.4 × 102. (ii) If K is measured directly in the laboratory, value of ∆G at any other temperature Problem 5.14 can be calculated. At 60°C, dinitrogen tetroxide is 50 Using equation (5.24), per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. Problem 5.12 Calculate ∆rG for conversion of oxygen Solution to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp N2O4(g) 2NO2(g) for this conversion is 2.47 × 10–29. If N2O4 is 50% dissociated, the mole Solution fraction of both the substances is given We know ∆rG = – 2.303 RT log Kp and by R = 8.314 JK–1 mol–1 1  0 .5 2  0 .5 x  : x NO 2  N 2 O 4 Therefore, ∆rG = 1  0 .5 1  0 .5 – 2.303 (8.314 J K–1 mol–1) 0 .5 × (298 K) (log 2.47 × 10–29) p N 2 O 4   1 atm, p NO 2  1 .5 = 163000 J mol–1 1 = 163 kJ mol–1.  1 atm. 1 .5 Problem 5.13 The equilibrium constant Kp is given by Find out the value of equilibrium constant 2 for the following reaction at 298 K.  p NO 2  1 .5 Kp =  2 p N 2 O 4 (1 .5 ) ( 0 .5 ) = 1.33 atm. Standard Gibbs energy change, ∆rG at Since the given temperature is –13.6 kJ mol–1. ∆rG = –RT ln Kp Solution ∆rG = (– 8.314 JK–1 mol–1) × (333 K) We know, log K = × (2.303) × (0.1239) = – 763.8 kJ mol–1 Reprint 2025-26 164 chemIstry Summary Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. For these purposes, we divide the universe into the system and the surroundings. Chemical or physical processes lead to evolution or absorption of heat (q), part of which may be converted into work (w). These quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in internal energy, depends on initial and final states only and is a state function, whereas q and w depend on the path and are not the state functions. We follow sign conventions of q and w by giving the positive sign to these quantities when these are added to the system. We can measure the transfer of heat from one system to another which causes the change in temperature. The magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pex∆V, in case of expansion of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT. At constant volume, w = 0, then ∆U = qV , heat transfer at constant volume. But in study of chemical reactions, we usually have constant pressure. We define another state function enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at constant pressure, ∆H = qp. There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be determined by b i  f H reactions   r H    a i  f H products    f i and in gaseous state by ∆rH = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products First law of thermodynamics does not guide us about the direction of chemical reactions i.e., what is the driving force of a chemical reaction. For isolated systems, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change, while energy change does not. Entropy changes can be measured by the equation qrev qrev ∆S = for a reversible process. is independent of path. T T Chemical reactions are generally carried at constant pressure, so we define another state function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation: ∆rG = ∆rH – T ∆rS For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0. Standard Gibbs energy change is related to equilibrium constant by ∆rG = – RT ln K. K can be calculated from this equation, if we know ∆rG which can be found from . Temperature is an important factor in the equation. Many reactions which are non-spontaneous at low temperature, are made spontaneous at high temperature for systems having positive entropy of reaction. Reprint 2025-26 THERMODYNAMICS 165 Exercises 5.1 Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. 5.2 For the process to occur under adiabatic conditions, the correct condition is: (i) ∆T = 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0 5.3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element 5.4 ∆U  of combustion of methane is – X kJ mol–1. The value of ∆H is (i) = ∆U  (ii) > ∆U  (iii) < ∆U  (iv) = 0 5.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1. 5.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature 5.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? 5.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. 3 NH2CN(g) + O2(g) → N2(g) + CO2(g) + H2O(l) 2 Reprint 2025-26 166 chemIstry 5.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1. 5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 36.8 J mol–1 K–1 5.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. 5.12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) 5.13 Given N2(g) + 3H2(g) → 2NH3(g); ∆rH = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas? 5.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data: 3 CH3OH (l) + O2(g) → CO2(g) + 2H2O(l) ; ∆rH = –726 kJ mol–1 2 C(graphite) + O2(g) → CO2(g) ; ∆cH = –393 kJ mol–1 1 H2(g) + O2(g) → H2O(l); ∆f H = –286 kJ mol–1. 2 5.15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ∆vapH(CCl4) = 30.5 kJ mol–1. ∆fH (CCl4) = –135.5 kJ mol–1. ∆aH (C) = 715.0 kJ mol–1, where ∆aH is enthalpy of atomisation ∆aH (Cl2) = 242 kJ mol–1 5.16 For an isolated system, ∆U = 0, what will be ∆S ? 5.17 For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range. 5.18 For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ? 5.19 For the reaction 2 A(g) + B(g) → 2D(g) ∆U  = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously. Reprint 2025-26 THERMODYNAMICS 167 5.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K. 5.21 Comment on the thermodynamic stability of NO(g), given 1 1 N2(g) + O2(g) → NO(g); ∆rH = 90 kJ mol–1 2 2 1 NO(g) + O2(g) → NO2(g): ∆rH= –74 kJ mol–1 2 5.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1. Reprint 2025-26 Unit 6 Equilibrium Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O2 molecules and the protein hemoglobin play After studying this unit you will be a crucial role in the transport and delivery of O2 fromable to our lungs to our muscles. Similar equilibria involving CO • identify dynamic nature of molecules and hemoglobin account for the toxicity of CO. equilibrium involved in physical and chemical processes; When a liquid evaporates in a closed container, • state the law of equilibrium; molecules with relatively higher kinetic energy escape • explain characteristics of the liquid surface into the vapour phase and number of equilibria involved in physical liquid molecules from the vapour phase strike the liquid and chemical processes; surface and are retained in the liquid phase. It gives rise • write expressions for equilibrium to a constant vapour pressure because of an equilibrium in constants; which the number of molecules leaving the liquid equals the• establish a relationship between number returning to liquid from the vapour. We say that Kp and Kc; the system has reached equilibrium state at this stage.• explain various factors that affect the equilibrium state of a However, this is not static equilibrium and there is a lot of reaction; activity at the boundary between the liquid and the vapour. • classify substances as acids or Thus, at equilibrium, the rate of evaporation is equal to the bases according to Arrhenius, rate of condensation. It may be represented by Bronsted-Lowry and Lewis concepts; H2O (l) H2O (vap) • classify acids and bases as The double half arrows indicate that the processes weak or strong in terms of their in both the directions are going on simultaneously. The ionization constants; • explain the dependence of degree mixture of reactants and products in the equilibrium state of ionization on concentration is called an equilibrium mixture. of the electrolyte and that of the Equilibrium can be established for both physical common ion; processes and chemical reactions. The reaction may be• describe pH scale for representing hydrogen ion concentration; fast or slow depending on the experimental conditions and • explain ionisation of water and the nature of the reactants. When the reactants in a closed its duel role as acid and base; vessel at a particular temperature react to give products, • describe ionic product (Kw ) and the concentrations of the reactants keep on decreasing, pKw for water; while those of products keep on increasing for some time • appreciate use of buffer after which there is no change in the concentrations solutions; of either of the reactants or products. This stage of the• calculate solubility product constant. system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to Reprint 2025-26 EQUILIBRIUM 169 this dynamic equilibrium stage that there is characteristic features. We observe that the no change in the concentrations of various mass of ice and water do not change with species in the reaction mixture. Based on the time and the temperature remains constant. extent to which the reactions proceed to reach However, the equilibrium is not static. the state of chemical equilibrium, these may The intense activity can be noticed at the be classified in three groups. boundary between ice and water. Molecules from the liquid water collide against ice and(i) The reactions that proceed nearly adhere to it and some molecules of ice escape to completion and only negligible into liquid phase. There is no change of mass concentrations of the reactants are of ice and water, as the rates of transfer of left. In some cases, it may not be even molecules from ice into water and of reverse possible to detect these experimentally. transfer from water into ice are equal at(ii) The reactions in which only small atmospheric pressure and 273 K. amounts of products are formed and It is obvious that ice and water are in most of the reactants remain unchanged equilibrium only at particular temperature at equilibrium stage. and pressure. For any pure substance at(iii) The reactions in which the concentrations atmospheric pressure, the temperature at of the reactants and products are which the solid and liquid phases are at comparable, when the system is in equilibrium is called the normal melting point equilibrium. or normal freezing point of the substance. The The extent of a reaction in equilibrium system here is in dynamic equilibrium and we varies with the experimental conditions such can infer the following: as concentrations of reactants, temperature, (i) Both the opposing processes occur etc. Optimisation of the operational conditions simultaneously. is very important in industry and laboratory (ii) Both the processes occur at the same so that equilibrium is favorable in the rate so that the amount of ice and water direction of the desired product. Some remains constant. important aspects of equilibrium involving physical and chemical processes are dealt in 6.1.2 Liquid-Vapour Equilibrium this unit along with the equilibrium involving This equilibrium can be better understood if ions in aqueous solutions which is called as we consider the example of a transparent box ionic equilibrium. carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride6.1 E Q U I L I B R I U M I N P H Y S I C A L (or phosphorus penta-oxide) is placed for PROCESSES a few hours in the box. After removing the The characteristics of system at equilibrium drying agent by tilting the box on one side, a are better understood if we examine some watch glass (or petri dish) containing water physical processes. The most familiar examples is quickly placed inside the box. It will be are phase transformation processes, e.g., observed that the mercury level in the right solid liquid limb of the manometer slowly increases and liquid gas finally attains a constant value, that is, the solid gas pressure inside the box increases and reaches a constant value. Also the volume of water in6.1.1 Solid-Liquid Equilibrium the watch glass decreases (Fig. 6.1). Initially Ice and water kept in a perfectly insulated there was no water vapour (or very less) inside thermos flask (no exchange of heat between its the box. As water evaporated the pressure in contents and the surroundings) at 273K and the box increased due to addition of water the atmospheric pressure are in equilibrium molecules into the gaseous phase inside state and the system shows interesting the box. The rate of evaporation is constant. Reprint 2025-26 170 chemistry Fig. 6.1 Measuring equilibrium vapour pressure of water at a constant temperature However, the rate of increase in pressure vapour to liquid state is much less than the decreases with time due to condensation rate of evaporation. These are open systems of vapour into water. Finally it leads to an and it is not possible to reach equilibrium in equilibrium condition when there is no net an open system. evaporation. This implies that the number Water and water vapour are in equilibriumof water molecules from the gaseous state position at atmospheric pressure (1.013 bar)into the liquid state also increases till the and at 100°C in a closed vessel. The boilingequilibrium is attained i.e., point of water is 100°C at 1.013 bar pressure. rate of evaporation= rate of condensation For any pure liquid at one atmospheric H2O(l) H2O (vap) pressure (1.013 bar), the temperature At equilibrium the pressure exerted by at which the liquid and vapours are at the water molecules at a given temperature equilibrium is called normal boiling point of remains constant and is called the equilibrium the liquid. Boiling point of the liquid depends vapour pressure of water (or just vapour on the atmospheric pressure. It depends on pressure of water); vapour pressure of water the altitude of the place; at high altitude the increases with temperature. If the above boiling point decreases. experiment is repeated with methyl alcohol, acetone and ether, it is observed that different 6.1.3 Solid – Vapour Equilibrium liquids have different equilibrium vapour Let us now consider the systems where solids pressures at the same temperature, and the sublime to vapour phase. If we place solidliquid which has a higher vapour pressure is iodine in a closed vessel, after sometimemore volatile and has a lower boiling point. the vessel gets filled up with violet vapour If we expose three watch glasses containing and the intensity of colour increases with separately 1mL each of acetone, ethyl alcohol, time. After certain time the intensity of and water to atmosphere and repeat the colour becomes constant and at this stage experiment with different volumes of the equilibrium is attained. Hence solid iodine liquids in a warmer room, it is observed sublimes to give iodine vapour and the iodinethat in all such cases the liquid eventually vapour condenses to give solid iodine. Thedisappears and the time taken for complete equilibrium can be represented as,evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the I2(solid) I2 (vapour) temperature. When the watch glass is open Other examples showing this kind of to the atmosphere, the rate of evaporation equilibrium are, remains constant but the molecules are Camphor (solid) Camphor (vapour)dispersed into large volume of the room. As a consequence the rate of condensation from NH4Cl (solid) NH4Cl (vapour) Reprint 2025-26 EQUILIBRIUM 171 6.1.4 Equilibrium Involving Dissolution pressure of the gas above the solvent. of Solid or Gases in Liquids This amount decreases with increase of Solids in liquids temperature. The soda water bottle is sealed under pressure of gas when its solubility inWe know from our experience that we can water is high. As soon as the bottle is opened,dissolve only a limited amount of salt or some of the dissolved carbon dioxide gassugar in a given amount of water at room escapes to reach a new equilibrium conditiontemperature. If we make a thick sugar syrup required for the lower pressure, namely itssolution by dissolving sugar at a higher partial pressure in the atmosphere. This istemperature, sugar crystals separate out if we how the soda water in bottle when left open cool the syrup to the room temperature. We call to the air for some time, turns ‘flat’. It can be it a saturated solution when no more of solute generalised that: can be dissolved in it at a given temperature. (i) For solid liquid equilibrium, there isThe concentration of the solute in a saturated only one temperature (melting point) atsolution depends upon the temperature. In 1 atm (1.013 bar) at which the twoa saturated solution, a dynamic equilibrium phases can coexist. If there is noexits between the solute molecules in the solid exchange of heat with the surroundings,state and in the solution: the mass of the two phases remainsSugar (solution) Sugar (solid), and constant. the rate of dissolution of sugar = rate of (ii) For liquid vapour equilibrium, thecrystallisation of sugar. vapour pressure is constant at a given Equality of the two rates and dynamic temperature. nature of equilibrium has been confirmed with (iii) For dissolution of solids in liquids,the help of radioactive sugar. If we drop some the solubility is constant at a givenradioactive sugar into saturated solution of temperature.non-radioactive sugar, then after some time radioactivity is observed both in the solution (iv) For dissolution of gases in liquids, and in the solid sugar. Initially there were no the concentration of a gas in liquid radioactive sugar molecules in the solution is proportional to the pressure but due to dynamic nature of equilibrium, (concentration) of the gas over the liquid. there is exchange between the radioactive These observations are summarised in and non-radioactive sugar molecules between Table 6.1 the two phases. The ratio of the radioactive Table 6.1 Some Features of Physical Equilibria to non-radioactive molecules in the solution increases till it attains a constant value. Process Conclusion Liquid Vapour pH2Oconstant at givenGases in liquids H2O (l) H2O (g) temperature When a soda water bottle is opened, some of Solid Liquid Melting point is fixed atthe carbon dioxide gas dissolved in it fizzes constant pressure out rapidly. The phenomenon arises due H2O (s) H2O (l) to difference in solubility of carbon dioxide Solute(s) Solute Concentration of solute at different pressures. There is equilibrium (solution) in solution is constant between the molecules in the gaseous state Sugar(s) Sugar at a given temperature and the molecules dissolved in the liquid (solution) under pressure i.e., Gas(g) Gas (aq) [gas(aq)]/[gas(g)] is CO2 (gas) CO2 (in solution) constant at a given This equilibrium is governed by Henry’s temperature CO2(g) CO2(aq)law, which states that the mass of a gas [CO2(aq)]/[CO2(g)] is dissolved in a given mass of a solvent at constant at a given temperatureany temperature is proportional to the Reprint 2025-26 172 chemistry 6.1.5 General Characteristics of Equilibria Involving Physical Processes For the physical processes discussed above, following characteristics are common to the system at equilibrium: (i) Equilibrium is possible only in a closed system at a given temperature. (ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition. (iii) All measurable properties of the system remain constant. (iv) When equilibrium is attained for a Fig. 6.2 Attainment of chemical equilibrium. physical process, it is characterised by constant value of one of its parameters at a given temperature. Table 6.1 lists Eventually, the two reactions occur at the such quantities. same rate and the system reaches a state of equilibrium.(v) The magnitude of such quantities at any stage indicates the extent to which the Similarly, the reaction can reach the state physical process has proceeded before of equilibrium even if we start with only C and reaching equilibrium. D; that is, no A and B being present initially, as the equilibrium can be reached from either 6.2 EQUILIBRIUM IN CHEMICAL direction. PROCESSES – DYNAMIC The dynamic nature of chemical EQUILIBRIUM equilibrium can be demonstrated in the Analogous to the physical systems chemical synthesis of ammonia by Haber’s process. reactions also attain a state of equilibrium. In a series of experiments, Haber started These reactions can occur both in forward with known amounts of dinitrogen and and backward directions. When the rates of dihydrogen maintained at high temperature the forward and reverse reactions become and pressure and at regular intervals equal, the concentrations of the reactants determined the amount of ammonia present. and the products remain constant. This He was successful in determining also the is the stage of chemical equilibrium. This concentration of unreacted dihydrogen and equilibrium is dynamic in nature as it consists dinitrogen. Fig. 6.4 (page 174) shows that after of a forward reaction in which the reactants a certain time the composition of the mixture give product(s) and reverse reaction in which remains the same even though some of the product(s) gives the original reactants. reactants are still present. This constancy in For a better comprehension, let us consider composition indicates that the reaction has a general case of a reversible reaction, reached equilibrium. In order to understand the dynamic nature of the reaction, synthesis A + B C + D of ammonia is carried out with exactly the With passage of time, there is same starting conditions (of partial pressure accumulation of the products C and D and and temperature) but using D2 (deuterium) depletion of the reactants A and B (Fig. 6.2). in place of H2. The reaction mixtures starting This leads to a decrease in the rate of either with H2 or D2 reach equilibrium with forward reaction and an increase in the rate the same composition, except that D2 and of the reverse reaction, ND3 are present instead of H2 and NH3. After Reprint 2025-26 EQUILIBRIUM 173 Dynamic Equilibrium – A Student’s Activity Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the following activity. The activity can be performed in a group of 5 or 6 students. Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of 3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder (number 2) empty. Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you notice that the level of coloured water in both the cylinders becomes constant. If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning. 1 2 1 2 (a) (b) Fig.6.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. Reprint 2025-26 174 chemistry 2NH3(g) N2(g) + 3H2(g) Similarly let us consider the reaction, H2(g) + I2(g) 2HI(g). If we start with equal initial concentration of H2 and I2, the reaction proceeds in the forward direction and the concentration of H2 and I2 decreases while that of HI increases, until all of these become constant at equilibrium (Fig. 6.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the concentration of HI will decrease and concentration of H2 and I2 will increase until they all become constant when equilibrium is reached (Fig. 6.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we Fig. 6.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product. N 2 g 3 H 2 g 2 NH 3 g equilibrium is attained, these two mixtures (H2, N2, NH3 and D2, N2, ND3) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (NH3, NH2D, NHD2 and ND3) and dihydrogen and its deutrated forms (H2, HD and D2) are present. Thus one can conclude that scrambling of H and D atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped Fig. 6.5 Chemical equilibrium in the reaction H2(g)when they reached equilibrium, then there + I2(g) 2HI(g) can be attained fromwould have been no mixing of isotopes in either direction this way. 6.3 LAW OF CHEMICAL EQUILIBRIUM Use of isotope (deuterium) in the formation AND EQUILIBRIUM CONSTANT of ammonia clearly indicates that chemical A mixture of reactants and products in thereactions reach a state of dynamic equilibrium state is called an equilibriumequilibrium in which the rates of forward mixture. In this section we shall address aand reverse reactions are equal and there number of important questions about theis no net change in composition. composition of equilibrium mixtures: What is Equilibrium can be attained from both the relationship between the concentrations sides, whether we start reaction by taking, of reactants and products in an equilibrium H2(g) and N2(g) and get NH3(g) or by taking mixture? How can we determine equilibrium NH3(g) and decomposing it into N2(g) and H2(g). concentrations from initial concentrations? What factors can be exploited to alter the N2(g) + 3H2(g) 2NH3(g) Reprint 2025-26 EQUILIBRIUM 175 composition of an equilibrium mixture? The Six sets of experiments with varying initial last question in particular is important when conditions were performed, starting with only choosing conditions for synthesis of industrial gaseous H2 and I2 in a sealed reaction vessel chemicals such as H2, NH3, CaO etc. in first four experiments (1, 2, 3 and 4) and To answer these questions, let us consider only HI in other two experiments (5 and 6). a general reversible reaction: Experiment 1, 2, 3 and 4 were performed taking different concentrations of H2 and / or A + B C + D where A and B are the reactants, C and D I2, and with time it was observed that intensity of the purple colour remained constant andare the products in the balanced chemical equilibrium was attained. Similarly, forequation. On the basis of experimental studies of many reversible reactions, the Norwegian experiments 5 and 6, the equilibrium was chemists Cato Maximillian Guldberg and attained from the opposite direction. Peter Waage proposed in 1864 that the Data obtained from all six sets of concentrations in an equilibrium mixture experiments are given in Table 6.2. are related by the following equilibrium It is evident from the experiments 1, 2,equation, 3 and 4 that number of moles of dihydrogen C D Kc  (6.1) reacted = number of moles of iodine reacted A B = ½ (number of moles of HI formed). Also, (6.1) where Kc is the equilibrium constant and experiments 5 and 6 indicate that, the expression on the right side is called the [H2(g)]eq = [I2(g)]eqequilibrium constant expression. Knowing the above facts, in order The equilibrium equation is also known as the law of mass action because in the early to establish a relationship between days of chemistry, concentration was called concentrations of the reactants and products, “active mass”. In order to appreciate their several combinations can be tried. Let us work better, let us consider reaction between consider the simple expression, gaseous H2 and I2 carried out in a sealed vessel [HI(g)]eq / [H2(g)]eq [I2(g)]eqat 731K. H2(g) + I2(g)   2HI(g) It can be seen from Table 6.3 that if we put the equilibrium concentrations of the 1 mol 1 mol 2 mol reactants and products, the above expression Table 6.2 Initial and Equilibrium Concentrations of H2, I2 and HI Reprint 2025-26 176 chemistry Table 6.3 E x p r e s s i o n I n v o l v i n g t h e The equilibrium constant for a general Equilibrium Concentration of reaction, Reactants a A + b B c C + d D H2(g) + I2(g) 2HI(g) is expressed as, Kc = [C]c[D]d / [A]a[B]b (6.4) where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products. Equilibrium constant for the reaction, 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) is written as Kc = [NO]4[H2O]6 / [NH3]4 [O2]5 Molar concentration of different species is is far from constant. However, if we consider indicated by enclosing these in square bracket the expression, and, as mentioned above, it is implied that these are equilibrium concentrations. While [HI(g)]2 eq / [H2(g)]eq [I2(g)]eq writing expression for equilibrium constant, we find that this expression gives constant symbol for phases (s, l, g) are generally ignored. value (as shown in Table 6.3) in all the six Let us write equilibrium constant for thecases. It can be seen that in this expression the power of the concentration for reactants reaction, H2(g) + I2(g) 2HI(g) (6.5) and products are actually the stoichiometric as, Kc = [HI]2 / [H2] [I2] = x (6.6)coefficients in the equation for the chemical reaction. Thus, for the reaction H2(g) + I2(g) The equilibrium constant for the reverse 2HI(g), following equation 6.1, the equilibrium reaction, 2HI(g) H2(g) + I2(g), at the same constant Kc is written as, temperature is, Kc = [HI(g)]eq2 / [H2(g)]eq [I2(g)]eq (6.2) K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc (6.7) Thus, K′c = 1 / Kc (6.8) Generally the subscript ‘eq’ (used for equilibrium) is omitted from the concentration Equilibrium constant for the reverse terms. It is taken for granted that the reaction is the inverse of the equilibrium concentrations in the expression for Kc are constant for the reaction in the forward equilibrium values. We, therefore, write, direction. Kc = [HI(g)]2 / [H2(g)] [I2(g)] (6.3) If we change the stoichiometric coefficients in a chemical equation by multiplying The subscript ‘c’ indicates that Kc is throughout by a factor then we must makeexpressed in concentrations of mol L–1. sure that the expression for equilibrium At a given temperature, the product of constant also reflects that change. For concentrations of the reaction products example, if the reaction (6.5) is written as, raised to the respective stoichiometric coefficient in the balanced chemical ½ H2(g) + ½ I2(g) HI(g) (6.9) equation divided by the product of the equilibrium constant for the above concentrations of the reactants raised to reaction is given by their individual stoichiometric coefficients K″c = [HI] / [H2]1/2[I2]1/2 = {[HI]2 / [H2][I2]}1/2 has a constant value. This is known as 1/2 = x1/2 = Kc (6.10)the Equilibrium Law or Law of Chemical Equilibrium. On multiplying the equation (6.5) by n, we get Reprint 2025-26 EQUILIBRIUM 177 nH2(g) + nI2(g) D 2nHI(g) (6.11) Therefore, equilibrium constant for the N2(g) + O2(g) 2NO(g) reaction is equal to Kc n. These findings are Solution summarised in Table 6.4. It should be noted For the reaction equilibrium constant, Kcthat because the equilibrium constants Kc can be written as, and K′c have different numerical values, it is 2 important to specify the form of the balanced NO Kc =chemical equation when quoting the value of N 2 O 2 an equilibrium constant. –3 2 2.8 10 M Table 6.4 Relations between Equilibrium = 3 3 3 .0 10 M 4 .2 10 M Constants for a General Reaction and its Multiples. = 0.622 Chemical equation Equilibrium constant a A + b B c C + dD Kc 6.4 HOMOGENEOUS EQUILIBRIA c C + d D a A + b B K′c =(1/Kc ) In a homogeneous system, all the reactants and products are in the same phase. na A + nb B ncC + ndD K′″c = (Kcn) For example, in the gaseous reaction, N2(g) + 3H2(g) 2NH3(g), reactants and products are in the homogeneous phase. Problem 6.1 Similarly, for the reactions, The following concentrations were CH3COOC2H5 (aq) + H2O (l) CH3COOH (aq) obtained for the formation of NH3 from + C2H5OH (aq) N 2 and H 2 at equilibrium at 500K. and, Fe3+ (aq) + SCN–(aq) Fe(SCN)2+ (aq) [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and [NH3] = 1.2 ×10–2M. Calculate equilibrium all the reactants and products are in constant. homogeneous solution phase. We shall now consider equilibrium constant for some Solution homogeneous reactions. The equilibrium constant for the reaction, N2(g) + 3H2(g) 2NH3(g) can be written as, 6.4.1 Equilibrium Constant in Gaseous Systems  NH 3  g  2 So far we have expressed equilibrium Kc = 3 constant of the reactions in terms of molar  N 2  g   H 2  g  concentration of the reactants and products, 2 2 and used symbol, Kc for it. For reactions 1 .2  10   involving gases, however, it is usually more = 2 2 3 1 .5  10 3 .0  10 convenient to express the equilibrium     constant in terms of partial pressure. = 0.355 × 103 = 3.55 × 102 The ideal gas equation is written as, pV = nRT Problem 6.2 n At equilibrium, the concentrations of ⇒ p = RT V N2=3.0 × 10–3M, O2 = 4.2 × 10–3M and NO= 2.8 × 10–3M in a sealed vessel at 800K. Here, p is the pressure in Pa, n is the number What will be Kc for the reaction of moles of the gas, V is the volume in m3 and T is the temperature in Kelvin Reprint 2025-26 178 chemistry Therefore, n/V is concentration expressed in mol/m3 2 −2 RT ] −2  NH 3 ( g )[ If concentration c, is in mol/L or mol/dm3, = 3 = K c ( RT ) and p is in bar then  N 2 ( g )  H 2 ( g ) p = cRT, −2 or K p = K c ( RT ) (6.14)We can also write p = [gas]RT. Here, R= 0.0831 bar litre/mol K Similarly, for a general reaction At constant temperature, the pressure of a A + b B c C + d D the gas is proportional to its concentration i.e., c d c d (c + d ) p p T ) C )( D ) [ C ] [ D ] ( Rp ∝ [gas] ( K p = a b = a b (a + b ) p p For reaction in equilibrium T ) ( A )( B ) [ A ] [ B ] ( R H2(g) + I2(g) 2HI(g) We can write either [ C ]c [ D ]d ( c + d )− (a + b ) 2 = a b ( RT ) A ] [ B ]  HI ( g ) [Kc =  H 2 ( g )  I 2 ( g ) c d [ C ] [ D ] ∆n ∆n 2 = a b ( RT ) = K c ( RT ) (6.15) ( p HI ) [ A ] [ B ]or K c = (6.12) where ∆n = (number of moles of gaseous ( p H 2 )( p I 2 ) products) – (number of moles of gaseous reactants) in the balanced chemical equation. RTFurther, since p HI =  HI ( g ) It is necessary that while calculating the value RT p I 2 =  I 2 ( g ) of Kp, pressure should be expressed in bar because standard state for pressure is 1 bar. RT p H 2 =  H 2 ( g ) We know from Unit 1 that : Therefore, 1pascal, Pa=1Nm–2, and 1bar = 105 Pa RT ]2 Kp values for a few selected reactions at ( p HI )2  HI ( g )[2 K p = = different temperatures are given in Table 6.5 RT H 2 ( g ) RT .  I 2 ( g ) ( p H 2 )( p I 2 )  2 Table 6.5 Equilibrium Constants, Kp for  HI ( g ) a Few Selected Reactions = = Kc (6.13)  H 2 ( g )  I 2 ( g ) In this example, Kp = Kc i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction N2(g) + 3H2(g) 2NH3(g) 2 ( p NH 3 )K p = 3 ( p N 2 )( p H 2 ) 2 2 Problem 6.3 RT ]  NH 3 ( g )[ = 3 3 PCl5, PCl3 and Cl2 are at equilibrium at 500 RT ) K and having concentration 1.59M PCl3,  N 2 ( g ) RT .  H 2 ( g )( 1.59M Cl2 and 1.41 M PCl5. Reprint 2025-26 EQUILIBRIUM 179 Calculate Kc for the reaction, the value 0.194 should be neglected because PCl5 PCl3 + Cl2 it will give concentration of the reactant Solution which is more than initial concentration. Hence the equilibrium concentrations are, The equilibrium constant Kc for the above reaction can be written as, [CO2] = [H2-] = x = 0.067 M Cl 2  1 .59 2 [CO] = [H2O] = 0.1 – 0.067 = 0.033 M PCl 3  Kc    1 .79 1 .41 Problem 6.5  PCl 5   Problem 6.4 For the equilibrium, The value of Kc = 4.24 at 800K for the 2NOCl(g) 2NO(g) + Cl2(g) reaction, the value of the equilibrium constant, Kc is CO (g) + H2O (g) CO2 (g) + H2 (g) 3.75 × 10–6 at 1069 K. Calculate the Kp for Calculate equilibrium concentrations of CO2, the reaction at this temperature? H2, CO and H2O at 800 K, if only CO and H2O are present initially at concentrations Solution of 0.10M each. We know that, Solution Kp = Kc(RT)∆n For the reaction, For the above reaction, ∆n = (2+1) – 2 = 1 CO (g) + H2O (g) CO2 (g) + H2 (g) Kp = 3.75 ×10–6 (0.0831 × 1069) Initial concentration: Kp = 0.033 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. 6.5 HETEROGENEOUS EQUILIBRIA At equilibrium: Equilibrium in a system having more than one (0.1-x) M (0.1-x) M x M x M phase is called heterogeneous equilibrium. The equilibrium between water vapour and where x is the amount of CO2 and H2 at liquid water in a closed container is an equilibrium. example of heterogeneous equilibrium. Hence, equilibrium constant can be written as, H2O(l) H2O(g) Kc = x2/(0.1-x)2 = 4.24 In this example, there is a gas phase and x2 = 4.24(0.01 + x2-0.2x) a liquid phase. In the same way, equilibrium between a solid and its saturated solution, x2 = 0.0424 + 4.24x2-0.848x 3.24x2 – 0.848x + 0.0424 = 0 Ca(OH)2 (s) + (aq) Ca2+ (aq) + 2OH–(aq) a = 3.24, b = – 0.848, c = 0.0424 is a heterogeneous equilibrium. (for quadratic equation ax2 + bx + c = 0, Heterogeneous equilibria often involve 2 pure solids or liquids. We can simplify  b  b  4 ac   equilibrium expressions for the heterogeneous x  2 a equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure x = 0.848±√(0.848)2– 4(3.24)(0.0424)/ solid or liquid is constant (i.e., independent (3.24×2) of the amount present). In other words if a x = (0.848 ± 0.4118)/ 6.48 substance ‘X’ is involved, then [X(s)] and [X(l)] x1 = (0.848 – 0.4118)/6.48 = 0.067 are constant, whatever the amount of ‘X’ is x2 = (0.848 + 0.4118)/6.48 = 0.194 taken. Contrary to this, [X(g)] and [X(aq)] will Reprint 2025-26 180 chemistry vary as the amount of X in a given volume This shows that at a particular varies. Let us take thermal dissociation of temperature, there is a constant concentration calcium carbonate which is an interesting and or pressure of CO2 in equilibrium with CaO(s) important example of heterogeneous chemical and CaCO3(s). Experimentally it has been equilibrium. found that at 1100 K, the pressure of CO2 in equilibrium with CaCO3(s) and CaO(s), is CaCO3 (s) CaO (s) + CO2 (g) (6.16) 2.0 ×105 Pa. Therefore, equilibrium constant On the basis of the stoichiometric equation, at 1100K for the above reaction is: we can write, Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00  CaO  s   CO 2  g  Similarly, in the equilibrium between Kc   CaCO 3  s  nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel), Since [CaCO3(s)] and [CaO(s)] are both constant, therefore modified equilibrium Ni (s) + 4 CO (g) Ni(CO)4 (g), constant for the thermal decomposition of the equilibrium constant is written as calcium carbonate will be  Ni  CO 4  K´c = [CO2(g)] (6.17) Kc  4  CO  or Kp = pCO2 (6.18) It must be remembered that for the existence of heterogeneous equilibrium pure Units of Equilibrium Constant solids or liquids must also be present (however small the amount may be) at equilibrium, but The value of equilibrium constant Kc can be calculated by substituting the concentration their concentrations or partial pressures do terms in mol/L and for Kp partial pressure not appear in the expression of the equilibrium is substituted in Pa, kPa, bar or atm. This constant. In the reaction, results in units of equilibrium constant based Ag2O(s) + 2HNO3(aq) 2AgNO3(aq) +H2O(l) on molarity or pressure, unless the exponents of both the numerator and denominator are  AgNO3  2 same. Kc = 2 For the reactions,  HNO 3  H2(g) + I2(g) 2HI, Kc and Kp have no unit. N2O4(g) 2NO2 (g), Kc has unit mol/L and Kp has unit bar Problem 6.6 Equilibrium constants can also be The value of Kp for the reaction, expressed as dimensionless quantities if the CO2 (g) + C (s) 2CO (g) standard state of reactants and products is 3.0 at 1000 K. If initially PCO2 = 0.48 are specified. For a pure gas, the standard bar and PCO = 0 bar and pure graphite is state is 1bar. Therefore a pressure of 4 bar present, calculate the equilibrium partial in standard state can be expressed as 4 pressures of CO and CO2. bar/1 bar = 4, which is a dimensionless Solution number. Standard state (c0) for a solute is 1 molar solution and all concentrations can be For the reaction, measured with respect to it. The numerical let ‘x’ be the decrease in pressure of CO2, value of equilibrium constant depends on the then standard state chosen. Thus, in this system CO2(g) + C(s) 2CO(g) both Kp and Kc are dimensionless quantities Initial but have different numerical values due to pressure: 0.48 bar 0 different standard states. Reprint 2025-26 EQUILIBRIUM 181 5. The equilibrium constant K for a reaction At equilibrium: is related to the equilibrium constant (0.48 – x)bar 2x bar of the corresponding reaction, whose 2 equation is obtained by multiplying or pCO K p = dividing the equation for the original pCO 2 reaction by a small integer. Let us consider applications of equilibrium Kp = (2x)2/(0.48 – x) = 3 constant to: 4x2 = 3(0.48 – x) • predict the extent of a reaction on the 4x2 = 1.44 – x basis of its magnitude, 4x2 + 3x – 1.44 = 0 • predict the direction of the reaction, and a = 4, b = 3, c = –1.44 • calculate equilibrium concentrations.  b  b 2  4 ac   x  6.6.1 Predicting the Extent of a 2 a Reaction = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 The numerical value of the equilibrium constant for a reaction indicates the extent = (–3 ± 5.66)/8 of the reaction. But it is important to note = (–3 + 5.66)/8 (as value of x cannot be that an equilibrium constant does not give negative hence we neglect that value) any information about the rate at which x = 2.66/8 = 0.33 the equilibrium is reached. The magnitude The equilibrium partial pressures are, of Kc or Kp is directly proportional to the concentrations of products (as these appear pCO2= 2x = 2 × 0.33 = 0.66 bar in the numerator of equilibrium constant pCO2= 0.48 – x = 0.48 – 0.33 = 0.15 bar expression) and inversely proportional to the concentrations of the reactants (these appear

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

3.27 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? 3.28 The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1? 3.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s–1. Calculate k at 318K and Ea. 3.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Answers to Some Intext Questions 3.1 rav = 6.66 × 10–6 Ms–1 3.2 Rate of reaction = rate of diappearance of A = 0.005 mol litre–1min–1 3.3 Order of the reaction is 2.5 3.4 X ® Y Rate = k[X]2 The rate will increase 9 times 3.5 t = 444 s 3.6 1.925 × 10–4 s–1 3.8 Ea = 52.897 kJ mol–1 3.9 1.471 × 10–19 Chemistry 88 Reprint 2025-26 UnitUnitUnitUnit Unit44 TheThe dd-- andand f-f-Objectives After studying this Unit, you will beable to BlockBlock ElementsElements • learn the positions of the d– and f-block elements in the periodic table; Iron, copper, silver and gold are among the transition elements that • know the electronic configurations have played important roles in the development of human civilisation. of the transition (d-block) and the The inner transition elements such as Th, Pa and U are proving inner transition (f-block) elements; excellent sources of nuclear energy in modern times. • appreciate the relative stability of various oxidation states in terms of electrode potential values; The d-block of the periodic table contains the elements of the groups 3-12 in which the d orbitals are• describe the preparation, progressively filled in each of the four long periods. properties, structures and uses of some important compounds The f-block consists of elements in which 4 f and 5 f such as K2Cr2O7 and KMnO4; orbitals are progressively filled. They are placed in a • understand the general separate panel at the bottom of the periodic table. The characteristics of the d– and names transition metals and inner transition metals f–block elements and the general are often used to refer to the elements of d-and horizontal and group trends in f-blocks respectively. them; There are mainly four series of the transition metals, • describe the properties of the 3d series (Sc to Zn), 4d series (Y to Cd), 5d series (La f-block elements and give a and Hf to Hg) and 6d series which has Ac and elements comparative account of the from Rf to Cn. The two series of the inner transition lanthanoids and actinoids with metals; 4f (Ce to Lu) and 5f (Th to Lr) are known as respect to their electronic lanthanoids and actinoids respectively. configurations, oxidation states Originally the name transition metals was derived and chemical behaviour. from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions. Zinc, cadmium and mercury of group 12 have full d10 configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively, their chemistry is studied along with the chemistry of the transition metals. The presence of partly filled d or f orbitals in their atoms makes transition elements different from that of Reprint 2025-26 the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non- transition elements can be applied successfully to the transition elements also. Various precious metals such as silver, gold and platinum and industrially important metals like iron, copper and titanium belong to the transition metals series. In this Unit, we shall first deal with the electronic configuration, occurrence and general characteristics of transition elements with special emphasis on the trends in the properties of the first row (3d) transition metals along with the preparation and properties of some important compounds. This will be followed by consideration of certain general aspects such as electronic configurations, oxidation states and chemical reactivity of the inner transition metals. THE TRANSITION ELEMENTS (d-BLOCK) 4.14.14.14.14.1 PositionPositionPositionPositionPosition ininininin thethethethethe The d–block occupies the large middle section of the periodic table PeriodicPeriodicPeriodicPeriodicPeriodic TableTableTableTableTable flanked between s– and p– blocks in the periodic table. The d–orbitals of the penultimate energy level of atoms receive electrons giving rise to four rows of the transition metals, i.e., 3d, 4d, 5d and 6d. All these series of transition elements are shown in Table 4.1. 4.24.24.24.24.2 ElectronicElectronicElectronicElectronicElectronic In general1– the electronic configuration of outer orbitals of these elements is (n-1)d 10ns1–2except for Pd where its electronic configuration is 4d105s0. ConfigurationsConfigurationsConfigurationsConfigurationsConfigurations The (n–1) stands for the inner d orbitals which may have one to ten ofofofofof thethethethethe d-Blockd-Blockd-Blockd-Blockd-Block electrons and the outermost ns orbital may have one or two electrons. ElementsElementsElementsElementsElements However, this generalisation has several exceptions because of very little energy difference between (n-1)d and ns orbitals. Furthermore, half and completely filled sets of orbitals are relatively more stable. A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d series. For example, consider the case of Cr, which has 3d 5 4s 1 configuration instead of 3d44s 2; the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals. Similarly in case of Cu, the configuration is 3d104s 1 and not 3d 94s2. The ground state electronic configurations of the outer orbitals of transition elements are given in Table 4.1. Table 4.1: Electronic Configurations of outer orbitals of the Transition Elements (ground state) 1st Series Sc Ti V Cr Mn Fe Co Ni Cu Zn Z 21 22 23 24 25 26 27 28 29 30 4s 2 2 2 1 2 2 2 2 1 2 3d 1 2 3 5 5 6 7 8 10 10 Chemistry 90 Reprint 2025-26 2nd Series Y Zr Nb Mo Tc Ru Rh Pd Ag Cd Z 39 40 41 42 43 44 45 46 47 48 5s 2 2 1 1 1 1 1 0 1 2 4d 1 2 4 5 6 7 8 10 10 10 3rd Series La Hf Ta W Re Os Ir Pt Au Hg Z 57 72 73 74 75 76 77 78 79 80 6s 2 2 2 2 2 2 2 1 1 2 5d 1 2 3 4 5 6 7 9 10 10 4th Series Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Z 89 104 105 106 107 108 109 110 111 112 7s 2 2 2 2 2 2 2 2 1 2 6d 1 2 3 4 5 6 7 8 10 10 The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n-1)d 10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. The d orbitals of the transition elements protrude to the periphery of an atom more than the other orbitals (i.e., s and p), hence, they are more influenced by the surroundings as well as affect the atoms or molecules surrounding them. In some respects, ions of a given dn configuration (n = 1 – 9) have similar magnetic and electronic properties. With partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. All these characteristics have been discussed in detail later in this Unit. There are greater similarities in the properties of the transition elements of a horizontal row in contrast to the non-transition elements. However, some group similarities also exist. We shall first study the general characteristics and their trends in the horizontal rows (particularly 3d row) and then consider some group similarities. On what ground can you say that scandium (Z = 21) is a transition ExampleExampleExampleExampleExample 4.14.14.14.14.1 element but zinc (Z = 30) is not? On the basis of incompletely filled 3d orbitals in case of scandium atom SolutionSolutionSolutionSolutionSolution in its ground state (3d1), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d10) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element. 91 The d- and f- Block Elements Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? We will discuss the properties of elements of first transition series only in the following sections. 4.34.34.34.34.3 GeneralGeneralGeneralGeneralGeneral 4.3.1 Physical Properties PropertiesPropertiesPropertiesPropertiesProperties ofofofofof Nearly all the transition elements display typical metallic properties thethethethethe TransitionTransitionTransitionTransitionTransition such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. With the exceptions of Zn, ElementsElementsElementsElementsElements Cd, Hg and Mn, they have one or more typical metallic structures at (d-Block)(d-Block)(d-Block)(d-Block)(d-Block) normal temperatures. Lattice Structures of Transition Metals Sc Ti V Cr Mn Fe Co Ni Cu Zn hcp hcp bcc bcc X bcc ccp ccp ccp X (bcc) (bcc) (bcc, ccp) (hcp) (hcp) (hcp) Y Zr Nb Mo Tc Ru Rh Pd Ag Cd hcp hcp bcc bcc hcp hcp ccp ccp ccp X (bcc) (bcc) (hcp) La Hf Ta W Re Os Ir Pt Au Hg hcp hcp bcc bcc hcp hcp ccp ccp ccp X (ccp,bcc) (bcc) 4 (bcc = body centred cubic; hcp = hexagonal close packed; ccp = cubic close packed; X = a typical metal structure). W The transition metals (with the exception Re Ta of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are 3 Mo Os high. Fig. 4.1 depicts the melting points of Nb Ru transition metals belonging to 3d, 4d and 5d Ir series. The high melting points of these metals Hf Tc K are attributed to the involvement of greater 3 Cr Rh number of electrons from (n-1)d in addition to Zr V Pt 2 the ns electrons in the interatomic metallic bonding. In any row the melting points of these M.p./10 Ti Fe Co Pd 5 metals rise to a maximum at d except for Ni anomalous values of Mn and Tc and fall Mn Cu regularly as the atomic number increases. Au Ag They have high enthalpies of atomisation which 1 are shown in Fig. 4.2. The maxima at about Atomic number the middle of each series indicate that one Fig. 4.1: Trends in melting points of unpaired electron per d orbital is particularly transition elements Chemistry 92 Reprint 2025-26 favourable for strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Since the enthalpy of atomisation is an important factor in determining the standard electrode potential of a metal, metals with very high enthalpy of atomisation (i.e., very high boiling point) tend to be noble in their reactions (see later for electrode potentials). Another generalisation that may be drawn from Fig. 4.2 is that the metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series; this is an important factor in accounting for the occurrence of much more frequent metal – metal bonding in compounds of the heavy transition metals. –1 mol V/kJ DaH Fig. 4.2 Trends in enthalpies of atomisation of transition elements 4.3.2 Variation in In general, ions of the same charge in a given series show progressive Atomic and decrease in radius with increasing atomic number. This is because the Ionic Sizes new electron enters a d orbital each time the nuclear charge increases of by unity. It may be recalled that the shielding effect of a d electron is Transition not that effective, hence the net electrostatic attraction between the Metals nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. The curves in Fig. 4.3 show an increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected 93 The d- and f- Block Elements Reprint 2025-26 increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship. 19 The factor responsible for the lanthanoid 18 contraction is somewhat similar to that observed in an ordinary transition series and is attributed 17 to similar cause, i.e., the imperfect shielding of 16 one electron by another in the same set of orbitals. However, the shielding of one 4f electron by 15 Radius/nm another is less than that of one d electron by 14 another, and as the nuclear charge increases 13 along the series, there is fairly regular decrease in the size of the entire 4f n orbitals. 12 Sc Ti V Cr Mn Fe Co Ni Cu Zn The decrease in metallic radius coupled with Y Zr Nb Mo Tc Ru Rh Pd Ag Cd increase in atomic mass results in a general increase in the density of these elements. Thus, La Hf Ta W Re Os Ir Pt Au Hg from titanium (Z = 22) to copper (Z = 29) the Fig. 4.3: Trends in atomic radii of significant increase in the density may be noted transition elements (Table 4.2). Table 4.2: Electronic Configurations and some other Properties of the First Series of Transition Elements Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic number 21 22 23 24 25 26 27 28 29 30 Electronic configuration M 3d 14s 2 3d 24s 2 3d 34s 2 3d 54s 1 3d 54s 2 3d 64s 2 3d 74s 2 3d 84s 2 3d 104s 1 3d 104s 2 M + 3d 14s 1 3d 24s 1 3d 34s 1 3d 5 3d 54s 1 3d 64s 1 3d 74s 1 3d 84s 1 3d 10 3d 104s 1 M 2+ 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10 M 3+ [Ar] 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 – – Enthalpy of atomisation, DaH o/kJ mol–1 326 473 515 397 281 416 425 430 339 126 Ionisation enthalpy/DiH o/kJ mol –1 DiHo I 631 656 650 653 717 762 758 736 745 906 DiHo II 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734 DiHo III 2393 2657 2833 2990 3260 2962 3243 3402 3556 3837 Metallic/ionic M 164 147 135 129 137 126 125 125 128 137 radii/pm M 2+ – – 79 82 82 77 74 70 73 75 M 3+ 73 67 64 62 65 65 61 60 – – Standard electrode M 2+/M – –1.63 –1.18 –0.90 –1.18 –0.44 –0.28 –0.25 +0.34 -0.76 potential Eo/V M 3+/M 2+ – –0.37 –0.26 –0.41 +1.57 +0.77 +1.97 – – – Density/g cm –3 3.43 4.1 6.07 7.19 7.21 7.8 8.7 8.9 8.9 7.1 Chemistry 94 Reprint 2025-26 Why do the transition elements exhibit higher enthalpies of ExampleExampleExampleExampleExample 4.24.24.24.24.2 atomisation? Because of large number of unpaired electrons in their atoms they SolutionSolutionSolutionSolutionSolution have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why? 4.3.3 Ionisation There is an increase in ionisation enthalpy along each series of the Enthalpies transition elements from left to right due to an increase in nuclear charge which accompanies the filling of the inner d orbitals. Table 4.2 gives the values of the first three ionisation enthalpies of the first series of transition elements. These values show that the successive enthalpies of these elements do not increase as steeply as in the case of non-transition elements. The variation in ionisation enthalpy along a series of transition elements is much less in comparison to the variation along a period of non-transition elements. The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series. The irregular trend in the first ionisation enthalpy of the metals of 3d series, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. You have learnt that when d-block elements form ions, ns electrons are lost before (n – 1) d electrons. As we move along the period in 3d series, we see that nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner subshell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly. Thus, ionization energies increase only slightly along the 3d series. The doubly or more highly charged ions have dn configurations with no 4s electrons. A general trend of increasing values of second ionisation enthalpy is expected as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction. However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn2+ and Fe3+ respectively. In both the cases, ions have d5 configuration. Similar breaks occur at corresponding elements in the later transition series. The interpretation of variation in ionisation enthalpy for an electronic configuration dn is as follows: The three terms responsible for the value of ionisation enthalpy are attraction of each electron towards nucleus, repulsion between the 95 The d- and f- Block Elements Reprint 2025-26 electrons and the exchange energy. Exchange energy is responsible for the stabilisation of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hunds rule). The loss of exchange energy increases the stability. As the stability increases, the ionisation becomes more difficult. There is no loss of exchange energy at d6 configuration. Mn+ has 3d54s1 configuration and configuration of Cr+ is d5, therefore, ionisation enthalpy of Mn+ is lower than Cr+. In the same way, Fe2+ has d6 configuration and Mn2+ has 3d5 configuration. Hence, ionisation enthalpy of Fe2+ is lower than the Mn2+. In other words, we can say that the third ionisation enthalpy of Fe is lower than that of Mn. The lowest common oxidation state of these metals is +2. To form the M 2+ ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthalpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M + ions have the d 5 and d 10 configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of one 4s electron which results in the formation of stable d 10 configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d 5 (Mn 2+) and d 10 (Zn 2+) ions. In general, the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Although ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, this problem is very complex and not amenable to ready generalisation. 4.3.4 Oxidation One of the notable features of a transition elements is the great variety States of oxidation states these may show in their compounds. Table 4.3 lists the common oxidation states of the first row transition elements. Table 4.3: Oxidation States of the first row Transition Metal (the most common ones are in bold types) Sc Ti V Cr Mn Fe Co Ni Cu Zn +2 +2 +2 +2 +2 +2 +2 +1 +2 +3 +3 +3 +3 +3 +3 +3 +3 +2 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +6 +7 Chemistry 96 Reprint 2025-26 The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus, early in the series scandium(II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II). At the other end, the only oxidation state of zinc is +2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese (Ti IVO2, VVO2 +, Cr V1O42–, MnVIIO4–) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are FeII,III, Co II,III, NiII, CuI,II, Zn II. The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity, e.g., V II, V III, VIV, VV. This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit of two. An interesting feature in the variability of oxidation states of the d– block elements is noticed among the groups (groups 4 through 10). Although in the p–block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not. Low oxidation states are found when a complex compound has ligands capable of p-acceptor character in addition to the s-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero. Name a transition element which does not exhibit variable ExampleExampleExampleExampleExample 4.34.34.34.34.3 oxidation states. Scandium (Z = 21) does not exhibit variable oxidation states. SolutionSolutionSolutionSolutionSolution IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? 97 The d- and f- Block Elements Reprint 2025-26 4.3.5 Trends in the Table 4.4 contains the thermochemical parameters related to the M2+/M transformation of the solid metal atoms to M2+ ions in solution and their V Standard standard electrode potentials. The observed values of E and those Electrode calculated using the data of Table 4.4 are compared in Fig. 4.4. Potentials The unique behaviour of Cu, having a positive EV, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration V enthalpy. The general trend towards less negative E values across the Fig. 4.4: Observed and calculated values for the standard electrode potentials (M2+ ® M°) of the elements Ti to Zn series is related to the general increase in the sum of the first and second V ionisation enthalpies. It is interesting to note that the value of E for Mn, Ni and Zn are more negative than expected from the trend. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration? ExampleExampleExampleExampleExample 4.44.44.44.44.4 Cr 2+ is reducing as its configuration changes from d 4 to d 3, the latter SolutionSolutionSolutionSolutionSolution having a half-filled t2g level (see Unit 5). On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.4 The E o(M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high DaH o and low DhydH o) Chemistry 98 Reprint 2025-26 Table 4.4: Thermochemical data (kJ mol-1) for the first row Transition Elements and the Standard Electrode Potentials for the Reduction of MII to M. Element (M) DaH o (M) DiH1o D1H2o DhydH o(M2+) Eo/V Ti 469 656 1309 -1866 -1.63 V 515 650 1414 -1895 -1.18 Cr 398 653 1592 -1925 -0.90 Mn 279 717 1509 -1862 -1.18 Fe 418 762 1561 -1998 -0.44 Co 427 758 1644 -2079 -0.28 Ni 431 736 1752 -2121 -0.25 Cu 339 745 1958 -2121 0.34 Zn 130 906 1734 -2059 -0.76 The stability of the half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are related to their E o values, whereas E o for Ni is related to the highest negative DhydH o. 4.3.6 Trends in An examination of the E o(M3+/M2+) values (Table 4.2) shows the varying the M3+/M2+ trends. The low value for Sc reflects the stability of Sc3+ which has a Standard noble gas configuration. The highest value for Zn is due to the removal Electrode of an electron from the stable d 10 configuration of Zn 2+. The Potentials comparatively high value for Mn shows that Mn 2+(d5) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe 3+ (d5). The comparatively low value for V is related to the stability of V 2+ (half-filled t2g level, Unit 5). 4.3.7 Trends in Table 4.5 shows the stable halides of the 3d series of transition metals. Stability of The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 Higher and CrF6. The +7 state for Mn is not represented in simple halides but Oxidation MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 States and CoF3. The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6. Although V +5 is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides is their instability in the low oxidation states e.g., VX2 (X = CI, Br or I) Table 4.5: Formulas of Halides of 3d Metals Oxidation Number + 6 CrF6 + 5 VF5 CrF5 + 4 TiX4 VXI4 CrX4 MnF4 + 3 TiX3 VX3 CrX3 MnF3 FeXI3 CoF3 + 2 TiX2III VX2 CrX2 MnX2 FeX2 CoX2 NiX2 CuX2II ZnX2 + 1 CuXIII Key: X = F ® I; XI = F ® Br; XII = F, CI; XIII = CI ® I 99 The d- and f- Block Elements Reprint 2025-26 and the same applies to CuX. On the other hand, all Cu II halides are known except the iodide. In this case, Cu 2+ oxidises I – to I2: 2Cu 2   4I   Cu2 I2 s  I2 However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation. 2Cu + ® Cu 2+ + Cu The stability of Cu 2+ (aq) rather than Cu+(aq) is due to the much more negative DhydH o of Cu 2+ (aq) than Cu +, which more than compensates for the second ionisation enthalpy of Cu. The ability of oxygen to stabilise the highest oxidation state is demonstrated in the oxides. The highest oxidation number in the oxides (Table 4.6) coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond Group 7, no higher oxides of Fe above Fe2O3, are known, although ferrates (VI)(FeO4)2–, are formed in alkaline media but they readily decompose to Fe2O3 and O2. Besides the oxides, oxocations stabilise V v as VO2 +, V IV as VO2+ and Ti IV as TiO 2+. The ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. Thus the highest Mn fluoride is MnF4 whereas the highest oxide is Mn2O7. The ability of oxygen to form multiple bonds to metals explains its superiority. In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O’s including a Mn–O–Mn bridge. The tetrahedral [MO4]n- ions are known for V V, Cr Vl, Mn V, Mn Vl and Mn VII. Table 4.6: Oxides of 3d Metals Oxidation Groups Number 3 4 5 6 7 8 9 10 11 12 + 7 Mn2O7 + 6 CrO3 + 5 V2O5 + 4 TiO2 V2O4 CrO2 MnO2 + 3 Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Mn3O4* Fe3O4 * Co3O4* + 2 TiO VO (CrO) MnO FeO CoO NiO CuO ZnO + 1 Cu2O * mixed oxides How would you account for the increasing oxidising power in the ExampleExampleExampleExampleExample 4.54.54.54.54.5 series VO2+ < Cr2O7 2– < MnO4 – ? This is due to the increasing stability of the lower species to which they SolutionSolutionSolutionSolutionSolution are reduced. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements? Chemistry 100 Reprint 2025-26 4.3.8 Chemical Transition metals vary widely in their chemical reactivity. Many of Reactivity them are sufficiently electropositive to dissolve in mineral acids, although and Eo a few are ‘noble’—that is, they are unaffected by single acids. Values The metals of the first series with the exception of copper are relatively more reactive and are oxidised by 1M H +, though the actual rate at which these metals react with oxidising agents like hydrogen ion (H +) is sometimes slow. For example, titanium and vanadium, in practice, are passive to dilute non oxidising acids at room temperature. The E o values for M2+/M (Table 4.2) indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative E o values is related to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the E o values for Mn, Ni and Zn are more negative than expected from the general trend. Whereas the stabilities of half-filled d subshell (d5) in Mn2+ and completely filled d subshell (d10) in zinc are related to their E e values; for nickel, Eo value is related to the highest negative enthalpy of hydration. An examination of the E o values for the redox couple M 3+/M2+ (Table 4.2) shows that Mn 3+ and Co 3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti 2+, V 2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g., 2 Cr 2+(aq) + 2 H+(aq) ® 2 Cr 3+(aq) + H2(g) ExampleExampleExampleExampleExample 4.64.64.64.64.6 For the first row transition metals the Eo values are: E o V Cr Mn Fe Co Ni Cu (M2+/M) –1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34 Explain the irregularity in the above values. SolutionSolutionSolutionSolutionSolution The E o (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (  i H1  i H 2 ) and also the sublimation enthalpies which are relatively much less for manganese and vanadium. ExampleExampleExampleExampleExample 4.74.74.74.74.7 Why is the E o value for the Mn3+/Mn 2+ couple much more positive than that for Cr 3+/Cr2+ or Fe 3+/Fe 2+? Explain. SolutionSolutionSolutionSolutionSolution Much larger third ionisation energy of Mn (where the required change is d5 to d4) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? 4.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ? 4.3.9 Magnetic When a magnetic field is applied to substances, mainly two types of Properties magnetic behaviour are observed: diamagnetism and paramagnetism. Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. Substances which are 101 The d- and f- Block Elements Reprint 2025-26 attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic. Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e.,  n  n  2  where n is the number of unpaired electrons and µ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM). The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion. The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 4.7. The experimental data are mainly for hydrated ions in solution or in the solid state. Table 4.7: Calculated and Observed Magnetic Moments (BM) Ion Configuration Unpaired Magnetic moment electron(s) Calculated Observed Sc3+ 3d0 0 0 0 Ti 3+ 3d1 1 1.73 1.75 Tl2+ 3d2 2 2.84 2.76 V2+ 3d3 3 3.87 3.86 Cr2+ 3d4 4 4.90 4.80 Mn2+ 3d5 5 5.92 5.96 Fe2+ 3d6 4 4.90 5.3 – 5.5 Co2+ 3d7 3 3.87 4.4 – 5.2 Ni2+ 3d8 2 2.84 2.9 – 3, 4 Cu 2+ 3d9 1 1.73 1.8 – 2.2 Zn2+ 3d10 0 0 Calculate the magnetic moment of a divalent ion in aqueous solution ExampleExampleExampleExampleExample 4.84.84.84.84.8 if its atomic number is 25. With atomic number 25, the divalent ion in aqueous solution will have SolutionSolutionSolutionSolutionSolution d5 configuration (five unpaired electrons). The magnetic moment, µ is  5  5  2   5.92BM Chemistry 102 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.8 Calculate the ‘spin only’ magnetic moment of M 2+ (aq) ion (Z = 27). 4.3.10 Formation When an electron from a lower energy d orbital is excited to a higher of Coloured energy d orbital, the energy of excitation corresponds to the frequency Ions of light absorbed (Unit 5). This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. In aqueous solutions where water molecules are the ligands, the colours of the ions observed are listed in Table 4.8. A few coloured solutions of Fig. 4.5: Colours of some of the first row d–block elements are transition metal ions in aqueous solutions. From illustrated in Fig. 4.5. left to right: V4+,V3+,Mn2+,Fe3+,Co2+,Ni2+and Cu2+ . Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions Configuration Example Colour 3d0 Sc3+ colourless 3d0 Ti 4+ colourless 3d1 Ti 3+ purple 3d1 V4+ blue 3d2 V3+ green 3d3 V2+ violet 3d3 Cr3+ violet 3d4 Mn 3+ violet 3d4 Cr2+ blue 3d5 Mn 2+ pink 3d5 Fe3+ yellow 3d6 Fe2+ green 3d63d7 Co3+Co2+ bluepink 3d8 Ni2+ green 3d9 Cu 2+ blue 3d10 Zn2+ colourless 4.3.11 Formation Complex compounds are those in which the metal ions bind a number of Complex of anions or neutral molecules giving complex species with Compounds characteristic properties. A few examples are: [Fe(CN)6] 3–, [Fe(CN)6]4–, [Cu(NH3)4] 2+ and [PtCl4] 2–. (The chemistry of complex compounds is 103 The d- and f- Block Elements Reprint 2025-26 dealt with in detail in Unit 5). The transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d orbitals for bond formation. 4.3.12 Catalytic The transition metals and their compounds are known for their catalytic Properties activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in Catalytic Hydrogenation) are some of the examples. Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions. 2 I– + S2O8 2– ® I2 + 2 SO4 2– An explanation of this catalytic action can be given as: 2 Fe 3+ + 2 I – ® 2 Fe 2+ + I2 2 Fe 2+ + S2O82– ® 2 Fe3+ + 2SO42– 4.3.13 Formation Interstitial compounds are those which are formed when small atoms of like H, C or N are trapped inside the crystal lattices of metals. They are Interstitial usually non stoichiometric and are neither typically ionic nor covalent, Compounds for example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. The formulas quoted do not, of course, correspond to any normal oxidation state of the metal. Because of the nature of their composition, these compounds are referred to as interstitial compounds. The principal physical and chemical characteristics of these compounds are as follows: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard, some borides approach diamond in hardness. (iii) They retain metallic conductivity. (iv) They are chemically inert. 4.3.14 Alloy An alloy is a blend of metals prepared by mixing the components. Formation Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals. The alloys so formed are hard and have often high melting points. The best known are ferrous alloys: chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel. Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance. Chemistry 104 Reprint 2025-26 ExampleExampleExampleExampleExample 4.94.94.94.94.9 What is meant by ‘disproportionation’ of an oxidation state? Give an example. SolutionSolutionSolutionSolutionSolution When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution. 3 Mn VIO4 2– + 4 H + ® 2 Mn VIIO –4 + Mn IVO2 + 2H2O IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.9 Explain why Cu+ ion is not stable in aqueous solutions? 4.44.44.44.44.4 SomeSomeSomeSomeSome 4.4.1 Oxides and Oxoanions of Metals ImportantImportantImportantImportantImportant These oxides are generally formed by the reaction of metals with CompoundsCompoundsCompoundsCompoundsCompounds ofofofofof oxygen at high temperatures. All the metals except scandium form TransitionTransitionTransitionTransitionTransition MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to ElementsElementsElementsElementsElements Mn2O7. Beyond group 7, no higher oxides of iron above Fe2O3 are known. Besides the oxides, the oxocations stabilise V V as VO2 +, V IV as VO 2+ and Ti IV as TiO 2+. As the oxidation number of a metal increases, ionic character decreases. In the case of Mn, Mn2O7 is a covalent green oil. Even CrO3 and V2O5 have low melting points. In these higher oxides, the acidic character is predominant. Thus, Mn2O7 gives HMnO4 and CrO3 gives H2CrO4 and H2Cr2O7. V2O5 is, however, amphoteric though mainly acidic and it gives VO4 3– as well as VO2+ salts. In vanadium there is gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO 2+ salts. Similarly, V2O5 reacts with alkalies as well as acids to give VO 34  and VO4 respectively. The well characterised CrO is basic but Cr2O3 is amphoteric. Potassium dichromate K2Cr2O7 Potassium dichromate is a very important chemical used in leather industry and as an oxidant for preparation of many azo compounds. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows: 4 FeCr2O4 + 8 Na2CO3 + 7 O2 ® 8 Na2CrO4 + 2 Fe2O3 + 8 CO2 The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised. 2Na2CrO4 + 2 H+ ® Na2Cr2O7 + 2 Na + + H2O 105 The d- and f- Block Elements Reprint 2025-26 Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2 KCl ® K2Cr2O7 + 2 NaCl Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same. 2 CrO4 2– + 2H + ® Cr2O7 2– + H2O Cr2O7 2– + 2 OH- ® 2 CrO4 2– + H2O The structures of chromate ion, CrO4 2– and the dichromate ion, Cr2O7 2– are shown below. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr–O–Cr bond angle of 126°. Sodium and potassium dichromates are strong oxidising agents; the sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis. In acidic solution, its oxidising action can be represented as follows: Cr2O7 2– + 14H + + 6e – ® 2Cr 3+ + 7H2O (E o = 1.33V) Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions are noted below: 6 I– ® 3I2 + 6 e – ; 3 Sn 2+ ® 3Sn 4+ + 6 e – 3 H2S ® 6H+ + 3S + 6e – ; 6 Fe 2+ ® 6Fe3+ + 6 e– The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g., Cr2O7 2– + 14 H+ + 6 Fe2+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O Potassium permanganate KMnO4 Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. 2MnO2 + 4KOH + O2 ® 2K2MnO4 + 2H2O 3MnO4 2– + 4H+ ® 2MnO4 – + MnO2 + 2H2O Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (Vl). F used with KOH, oxidised Electrolytic oxidation in MnO 2 →with air or KNO 3 MnO 24 − ; MnO 24  alkaline solution MnO 4 manganate ion manganate permanganate ion Chemistry 106 Reprint 2025-26 In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate. 2Mn2+ + 5S2O8 2– + 8H2O ® 2MnO4 – + 10SO42– + 16H + Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of KClO4. The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K. 2KMnO4 ® K2MnO4 + MnO2 + O2 It has two physical properties of considerable interest: its intense colour and its diamagnetism along with temperature-dependent weak paramagnetism. These can be explained by the use of molecular orbital theory which is beyond the present scope. The manganate and permanganate ions are tetrahedral; the p- bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese. The green manganate is paramagnetic because of one unpaired electron but the permanganate is diamagnetic due to the absence of unpaired electron. Acidified permanganate solution oxidises oxalates to carbon dioxide, iron(II) to iron(III), nitrites to nitrates and iodides to free iodine. The half-reactions of reductants are: COO – 5 10CO2 + 10e – COO – 5 Fe2+ ® 5 Fe3+ + 5e– 5NO2 – + 5H2O ® 5NO3 – + 10H+ + l0e– 10I– ® 5I2 + 10e– The full reaction can be written by adding the half-reaction for KMnO4 to the half-reaction of the reducing agent, balancing wherever necessary. If we represent the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt by half-reactions, MnO4 – + e– ® MnO4 2– (E o = + 0.56 V) MnO4 – + 4H+ + 3e– ® MnO2 + 2H2O (E o = + 1.69 V) MnO4 – + 8H+ + 5e– ® Mn2+ + 4H2O (E o = + 1.52 V) We can very well see that the hydrogen ion concentration of the solution plays an important part in influencing the reaction. Although many reactions can be understood by consideration of redox potential, kinetics of the reaction is also an important factor. Permanganate at [H+] = 1 should oxidise water but in practice the reaction is extremely slow unless either manganese(ll) ions are present or the temperature is raised. A few important oxidising reactions of KMnO4 are given below: 1. In acid solutions: (a) Iodine is liberated from potassium iodide : 10I – + 2MnO4 – + 16H + ® 2Mn2+ + 8H2O + 5I2 (b) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe 2+ + MnO4 – + 8H+ ® Mn2+ + 4H2O + 5Fe 3+ 107 The d- and f- Block Elements Reprint 2025-26 (c) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O4 2– + 2MnO4 – + 16H + ——> 2Mn 2+ + 8H2O + 10CO2 (d) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H + + S2– 5S 2– + 2MnO – 4 + 16H + ——> 2Mn2+ + 8H2O + 5S (e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid: 5SO3 2– + 2MnO4 – + 6H + ——> 2Mn 2+ + 3H2O + 5SO42– (f) Nitrite is oxidised to nitrate: 5NO2– + 2MnO4– + 6H + ——> 2Mn 2+ + 5NO3 – + 3H2O 2. In neutral or faintly alkaline solutions: (a) A notable reaction is the oxidation of iodide to iodate: 2MnO4 – + H2O + I– ——> 2MnO2 + 2OH – + IO3 – (b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4 – + 3S2O3 2– + H2O ——> 8MnO2 + 6SO4 2– + 2OH – (c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation: 2MnO4 – + 3Mn 2+ + 2H2O ——> 5MnO2 + 4H+ Note: Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine. UsesUsesUses:UsesUses Besides its use in analytical chemistry, potassium permanganate is used as a favourite oxidant in preparative organic chemistry. Its uses for the bleaching of wool, cotton, silk and other textile fibres and for the decolourisation of oils are also dependent on its strong oxidising power. THE INNER TRANSITION ELEMENTS ( f-BLOCK) The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium). Because lanthanum closely resembles the lanthanoids, it is usually included in any discussion of the lanthanoids for which the general symbol Ln is often used. Similarly, a discussion of the actinoids includes actinium besides the fourteen elements constituting the series. The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series. They have only one stable oxidation state and their chemistry provides an excellent opportunity to examine the effect of small changes in size and nuclear charge along a series of otherwise similar elements. The chemistry of the actinoids is, on the other hand, much more complicated. The complication arises partly owing to the occurrence of a wide range of oxidation states in these elements and partly because their radioactivity creates special problems in their study; the two series will be considered separately here. 4.54.54.54.54.5 TheTheTheTheThe The names, symbols, electronic configurations of atomic and some LanthanoidsLanthanoidsLanthanoidsLanthanoidsLanthanoids ionic states and atomic and ionic radii of lanthanum and lanthanoids (for which the general symbol Ln is used) are given in Table 4.9. Chemistry 108 Reprint 2025-26 4.5.1 Electronic It may be noted that atoms of these elements have electronic Configurations configuration with 6s 2 common but with variable occupancy of 4f level (Table 4.9). However, the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form 4f n (n = 1 to 14 with increasing atomic number). 4.5.2 Atomic and The overall decrease in atomic and ionic radii from lanthanum to Ionic Sizes lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. It has far reaching Sm 2+ consequences in the chemistry of the third 110 2+ transition series of the elements. The decrease Eu in atomic radii (derived from the structures of La3+ metals) is not quite regular as it is regular in 3+ M3+ ions (Fig. 4.6). This contraction is, of Ce course, similar to that observed in an ordinary Pr3+ transition series and is attributed to the same 100 Nd3+ cause, the imperfect shielding of one electron Pm 3+ by another in the same sub-shell. However, the Sm3+ shielding of one 4 f electron by another is less Eu3+ than one d electron by another with the increase Gd3+ Tm 2+radii/pm 2+ in nuclear charge along the series. There is Yb Ce 4+ Tb 3+ fairly regular decrease in the sizes with 3+ DyIonic Pr4+ 3+ increasing atomic number. 90 Ho Er 3+ The cumulative effect of the contraction of Tm3+ the lanthanoid series, known as lanthanoid Yb3+ 3+ contraction, causes the radii of the members 4+ Lu Tb of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr 57 59 61 63 65 67 69 71 (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their Atomic number occurrence together in nature and for the Fig. 4.6: Trends in ionic radii of lanthanoids difficulty faced in their separation. 4.5.3 Oxidation In the lanthanoids, La(II) and Ln(III) compounds are predominant States species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained. This irregularity (as in ionisation enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Thus, the formation of Ce IV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. The E o value for Ce 4+/ Ce 3+ is + 1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent. Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2. Eu2+ is formed by losing the two s electrons and its f 7 configuration accounts for the formation of this ion. However, Eu 2+ is a strong reducing agent changing to the common +3 state. Similarly Yb 2+ which has f 14 configuration is a reductant. Tb IV has half-filled f-orbitals and is an oxidant. The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states. 109 The d- and f- Block Elements Reprint 2025-26 Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids Electronic configurations* Radii/pm Atomic Name Symbol Ln Ln2+ Ln3+ Ln4+ Ln Ln3+ Number 57 Lanthanum La 5d16s2 5d1 4f 0 187 106 58 Cerium Ce 4f15d16s2 4f 2 4f 1 4f 0 183 103 59 Praseodymium Pr 4f 36s2 4f 3 4f 2 4f 1 182 101 60 Neodymium Nd 4f 46s2 4f 4 4f 3 4f 2 181 99 61 Promethium Pm 4f 56s2 4f 5 4f 4 181 98 62 Samarium Sm 4f 66s2 4f 6 4f 5 180 96 63 Europium Eu 4f 76s2 4f 7 4f 6 199 95 64 Gadolinium Gd 4f 75d16s2 4f 75d 1 4f 7 180 94 65 Terbium Tb 4f 96s2 4f 9 4f 8 4f 7 178 92 66 Dysprosium Dy 4f 106s2 4f 10 4f 9 4f 8 177 91 67 Holmium Ho 4f 116s2 4f 11 4f 10 176 89 68 Erbium Er 4f 126s2 4f 12 4f 11 175 88 69 Thulium Tm 4f 136s2 4f 13 4f 12 174 87 70 Ytterbium Yb 4f 146s2 4f 14 4f 13 173 86 71 Lutetium Lu 4f 145d16s2 4f 145d1 4f 14 – – – * Only electrons outside [Xe] core are indicated 4.5.4 General All the lanthanoids are silvery white soft metals and tarnish rapidly in air. Characteristics The hardness increases with increasing atomic number, samarium being steel hard. Their melting points range between 1000 to 1200 K but samarium melts at 1623 K. They have typical metallic structure and are good conductors of heat and electricity. Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm. Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of f electrons. Neither La 3+ nor Lu3+ ion shows any colour but the rest do so. However, absorption bands are narrow, probably because of the excitation within f level. The lanthanoid ions other than the f 0 type (La 3+ and Ce4+) and the f 14 type (Yb2+ and Lu3+) are all paramagnetic. The first ionisation enthalpies of the lanthanoids are around 600 kJ mol –1, the second about 1200 kJ mol–1 comparable with those of calcium. A detailed discussion of the variation of the third ionisation enthalpies indicates that the exchange enthalpy considerations (as in 3d orbitals of the first transition series), appear to impart a certain degree of stability to empty, half-filled and completely filled orbitals f level. This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium. In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium. Values for E o for the half-reaction: Ln 3+(aq) + 3e – ® Ln(s) Chemistry 110 Reprint 2025-26 Ln2 O 3 H2 are in the range of –2.2 to –2.4 V except for Eu for which the value is – 2.0 V. This is, of course, a small acids variation. The metals combine with burns in with hydrogen when gently heated in the O2 gas. The carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated heated with S Ln with halogens with carbon. They liberate hydrogen Ln 2 S3 LnX 3 from dilute acids and burn in halogens N with with toandform hydroxideshalides. They formM(OH)3.oxides M2O3The C K H2 O hydroxides are definite compounds, not heated just hydrated oxides. They are basic with 2773 like alkaline earth metal oxides and Ln N LnC2 Ln(OH)3 + H2 hydroxides. Their general reactions are depicted in Fig. 4.7. Fig 4.7: Chemical reactions of the lanthanoids. The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces. 4.64.64.64.64.6 TheTheTheTheThe ActinoidsActinoidsActinoidsActinoidsActinoids The actinoids include the fourteen elements from Th to Lr. The names, symbols and some properties of these elements are given in Table 4.10. Table 4.10: Some Properties of Actinium and Actinoids Electronic conifigurations* Radii/pm Atomic Name Symbol M M3+ M4+ M3+ M4+ Number 89 Actinium Ac 6d 17s 2 5f 0 111 90 Thorium Th 6d 27s 2 5f 1 5f 0 99 91 Protactinium Pa 5f 26d 17s 2 5f 2 5f 1 96 92 Uranium U 5f 36d 17s 2 5f 3 5f 2 103 93 93 Neptunium Np 5f 46d 17s 2 5f 4 5f 3 101 92 94 Plutonium Pu 5f 67s 2 5f 5 5f 4 100 90 95 Americium Am 5f 77s 2 5f 6 5f 5 99 89 96 Curium Cm 5f 76d 17s 2 5f 7 5f 6 99 88 97 Berkelium Bk 5f 97s 2 5f 8 5f 7 98 87 98 Californium Cf 5f 107s 2 5f 9 5f 8 98 86 99 Einstenium Es 5f 117s 2 5f 10 5f 9 – – 100 Fermium Fm 5f 127s 2 5f 11 5f 10 – – 101 Mendelevium Md 5f 137s 2 5f 12 5f 11 – – 102 Nobelium No 5f 147s 2 5f 13 5f 12 – – 103 Lawrencium Lr 5f 146d 17s 2 5f 14 5f 13 – – 111 The d- and f- Block Elements Reprint 2025-26 The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z =103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult. 4.6.1 Electronic All the actinoids are believed to have the electronic configuration of 7s2 Configurations and variable occupancy of the 5f and 6d subshells. The fourteen electrons are formally added to 5f, though not in thorium (Z = 90) but from Pa onwards the 5f orbitals are complete at element 103. The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the f 0, f 7 and f 14 occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 76d17s2. Although the 5f orbitals resemble the 4f orbitals in their angular part of the wave-function, they are not as buried as 4f orbitals and hence 5f electrons can participate in bonding to a far greater extent. 4.6.2 Ionic Sizes The general trend in lanthanoids is observable in the actinoids as well. There is a gradual decrease in the size of atoms or M3+ ions across the series. This may be referred to as the actinoid contraction (like lanthanoid contraction). The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons. 4.6.3 Oxidation There is a greater range of oxidation states, which is in part attributed to States the fact that the 5f, 6d and 7s levels are of comparable energies. The known oxidation states of actinoids are listed in Table 4.11. The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements (Table 4.11). The actinoids resemble the lanthanoids in having more compounds in +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse. Because the distribution of oxidation states among the actinoids is so uneven and so different for the former and later elements, it is unsatisfactory to review their chemistry in terms of oxidation states. Table 4.11: Oxidation States of Actinium and Actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 4.6.4 General The actinoid metals are all silvery in appearance but display Characteristics a variety of structures. The structural variability is obtained and Comparison due to irregularities in metallic radii which are far greater with Lanthanoids than in lanthanoids. Chemistry 112 Reprint 2025-26 The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives a mixture of oxide and hydride and combination with most non metals takes place at moderate temperatures. Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no action. The magnetic properties of the actinoids are more complex than those of the lanthanoids. Although the variation in the magnetic susceptibility of the actinoids with the number of unpaired 5 f electrons is roughly parallel to the corresponding results for the lanthanoids, the latter have higher values. It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids. A comparison of the actinoids with the lanthanoids, with respect to different characteristics as discussed above, reveals that behaviour similar to that of the lanthanoids is not evident until the second half of the actinoid series. However, even the early actinoids resemble the lanthanoids in showing close similarities with each other and in gradual variation in properties which do not entail change in oxidation state. The lanthanoid and actinoid contractions, have extended effects on the sizes, and therefore, the properties of the elements succeeding them in their respective periods. The lanthanoid contraction is more important because the chemistry of elements succeeding the actinoids are much less known at the present time. ExampleExampleExampleExampleExample 4.104.104.104.104.10 Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. SolutionSolutionSolutionSolutionSolution Cerium (Z = 58) IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why? 4.74.74.74.74.7 SomeSomeSomeSomeSome Iron and steels are the most important construction materials. Their ApplicationsApplicationsApplicationsApplicationsApplications production is based on the reduction of iron oxides, the removal of impurities and the addition of carbon and alloying metals such as Cr, Mn ofofofofof d-d-d-d-d- andandandandand and Ni. Some compounds are manufactured for special purposes such as f-Blockf-Blockf-Blockf-Blockf-Block TiO for the pigment industry and MnO2 for use in dry battery cells. The ElementsElementsElementsElementsElements battery industry also requires Zn and Ni/Cd. The elements of Group 11 are still worthy of being called the coinage metals, although Ag and Au 113 The d- and f- Block Elements Reprint 2025-26 are restricted to collection items and the contemporary UK ‘copper’ coins are copper-coated steel. The ‘silver’ UK coins are a Cu/Ni alloy. Many of the metals and/or their compounds are essential catalysts in the chemical industry. V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid. TiCl4 with A1(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene). Iron catalysts are used in the Haber process for the production of ammonia from N2/H2 mixtures. Nickel catalysts enable the hydrogenation of fats to proceed. In the Wacker process the oxidation of ethyne to ethanal is catalysed by PdCl2. Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene. The photographic industry relies on the special light-sensitive properties of AgBr. SummarySummarySummarySummarySummary The d-block consisting of Groups 3-12 occupies the large middle section of the periodic table. In these elements the inner d orbitals are progressively filled. The f-block is placed outside at the bottom of the periodic table and in the elements of this block, 4f and 5f orbitals are progressively filled. Corresponding to the filling of 3d, 4d and 5d orbitals, three series of transition elements are well recognised. All the transition elements exhibit typical metallic properties such as –high tensile strength, ductility, malleability, thermal and electrical conductivity and metallic character. Their melting and boiling points are high which are attributed to the involvement of (n –1) d electrons resulting into strong interatomic bonding. In many of these properties, the maxima occur at about the middle of each series which indicates that one unpaired electron per d orbital is particularly a favourable configuration for strong interatomic interaction. Successive ionisation enthalpies do not increase as steeply as in the main group elements with increasing atomic number. Hence, the loss of variable number of electrons from (n –1) d orbitals is not energetically unfavourable. The involvement of (n–1) d electrons in the behaviour of transition elements impart certain distinct characteristics to these elements. Thus, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties and tendency for the formation of coloured ions, interstitial compounds and complexes. The transition elements vary widely in their chemical behaviour. Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are ‘noble’. Of the first series, with the exception of copper, all the metals are relatively reactive. The transition metals react with a number of non-metals like oxygen, nitrogen, sulphur and halogens to form binary compounds. The first series transition metal oxides are generally formed from the reaction of metals with oxygen at high temperatures. These oxides dissolve in acids and bases to form oxometallic salts. Potassium dichromate and potassium permanganate are common examples. Potassium dichromate is prepared from the chromite ore by fusion with alkali in presence of air and acidifying the extract. Pyrolusite ore (MnO2) is used for the preparation of potassium permanganate. Both the dichromate and the permanganate ions are strong oxidising agents. The two series of inner transition elements, lanthanoids and actinoids constitute the f-block of the periodic table. With the successive filling of the inner orbitals, 4f, there is a gradual decrease in the atomic and ionic sizes of these metals along the series (lanthanoid contraction). This has far reaching consequences in the chemistry of the elements succeeding them. Lanthanum and all the lanthanoids are rather soft white metals. They react easily with water to give solutions giving +3 ions. The principal oxidation state is +3, although +4 and +2 oxidation states are also exhibited by some Chemistry 114 Reprint 2025-26 occasionally. The chemistry of the actinoids is more complex in view of their ability to exist in different oxidation states. Furthermore, many of the actinoid elements are radioactive which make the study of these elements rather difficult. There are many useful applications of the d- and f-block elements and their compounds, notable among them being in varieties of steels, catalysts, complexes, organic syntheses, etc. Exercises 4.1 Write down the electronic configuration of: (i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+ (ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+ 4.2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? 4.3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? 4.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples. 4.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d 3, 3d 5, 3d 8 and 3d 4? 4.6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. 4.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction? 4.8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements? 4.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?

3.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

3.9 A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled? 85 Chemical Kinetics Reprint 2025-26

3.5 ELECTRONIC CONFIGURATIONS transition series of elements. This starts from OF ELEMENTS AND THE PERIODIC scandium (Z = 21) which has the electronic TABLE configuration 3d14s2. The 3d orbitals are filled In the preceding unit we have learnt that an at zinc (Z=30) with electronic configuration electron in an atom is characterised by a set 3d104s2. The fourth period ends at krypton of four quantum numbers, and the principal with the filling up of the 4p orbitals. Altogether quantum number (n ) defines the main energy we have 18 elements in this fourth period. The level known as shell. We have also studied fifth period (n = 5) beginning with rubidium about the filling of electrons into different is similar to the fourth period and contains subshells, also referred to as orbitals (s, p, the 4d transition series starting at yttrium (Z = 39). This period ends at xenon with thed, f ) in an atom. The distribution of electrons filling up of the 5p orbitals. The sixth periodinto orbitals of an atom is called its electronic configuration. An element’s location in the (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals, inPeriodic Table reflects the quantum numbers the order — filling up of the 4f orbitals beginsof the last orbital filled. In this section we with cerium (Z = 58) and ends at lutetiumwill observe a direct connection between the (Z = 71) to give the 4f-inner transition serieselectronic configurations of the elements and which is called the lanthanoid series. Thethe long form of the Periodic Table. seventh period (n = 7) is similar to the sixth (a) Electronic Configurations in Periods period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes mostThe period indicates the value of n for the of the man-made radioactive elements. Thisoutermost or valence shell. In other words, period will end at the element with atomicsuccessive period in the Periodic Table is number 118 which would belong to the nobleassociated with the filling of the next higher gas family. Filling up of the 5f orbitals afterprincipal energy level (n = 1, n = 2, etc.). It can 82 chemistry actinium (Z = 89) gives the 5f-inner transition a theoretical foundation for the periodic series known as the actinoid series. The 4f- classification. The elements in a vertical column and 5f-inner transition series of elements of the Periodic Table constitute a group or are placed separately in the Periodic Table family and exhibit similar chemical behaviour. to maintain its structure and to preserve the This similarity arises because these elements principle of classification by keeping elements have the same number and same distribution with similar properties in a single column. of electrons in their outermost orbitals. We can classify the elements into four blocks viz., Problem 3.2 s-block, p-block, d-block and f-block How would you justify the presence depending on the type of atomic orbitals that of 18 elements in the 5th period of the are being filled with electrons. This is illustrated Periodic Table? in Fig. 3.3. We notice two exceptions to this Solution categorisation. Strictly, helium belongs to the s-block but its positioning in the p-block When n = 5, l = 0, 1, 2, 3. The order along with other group 18 elements is in which the energy of the available justified because it has a completely filled orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals valence shell (1s2) and as a result, exhibits available are 9. The maximum number properties characteristic of other noble gases. of electrons that can be accommodated The other exception is hydrogen. It has only is 18; and therefore 18 elements are one s-electron and hence can be placed in there in the 5th period. group 1 (alkali metals). It can also gain an electron to achieve a noble gas (b) Groupwise Electronic Configurations arrangement and hence it can behave similar to a group 17 (halogen family)Elements in the same vertical column or elements. Because it is a special case, wegroup have similar valence shell electronic shall place hydrogen separately at the top ofconfigurations, the same number of electrons the Periodic Table as shown in Fig. 3.2 andin the outer orbitals, and similar properties. Fig. 3.3. We will briefly discuss the salientFor example, the Group 1 elements (alkali metals) all have ns1 valence shell electronic features of the four types of elements marked in configuration as shown below. the Periodic Table. More about these elements Atomic number Symbol Electronic configuration 3 Li 1s22s1 (or) [He]2s1 11 Na 1s22s22p63s1 (or) [Ne]3s1 19 K 1s22s22p63s23p64s1 (or) [Ar]4s1 37 Rb 1s22s22p63s23p63d104s24p65s1 (or) [Kr]5s1 55 Cs 1s22s22p63s23p63d104s24p64d105s25p66s1 (or) [Xe]6s1 87 Fr [Rn]7s1 Thus it can be seen that the properties of will be discussed later. During the description an element have periodic dependence upon of their features certain terminology has been its atomic number and not on relative atomic used which has been classified in section 3.7. mass. 3.6.1 The s-Block Elements3.6 ELECTRONIC CONFIGURATIONS A N D T Y P E S O F E L E M E N T S : The elements of Group 1 (alkali metals) and s-, p-, d-, f- BLOCKS Group 2 (alkaline earth metals) which have The aufbau (build up) principle and the ns1 and ns2 outermost electronic configuration electronic configuration of atoms provide belong to the s-Block Elements. They are all Classification of Elements and Periodicity in Properties 83 Og Ts Mc that Nh ). METALS ( orbitalsinto the on elements of METALLOIDS based and ) Tabledivision broad ( Periodicthe theis in shown NON-METALS Also), elements of filled. types being Theare( 3.3 Fig. 84 chemistry reactive metals with low ionization enthalpies. valence (oxidation states), paramagnetism and They lose the outermost electron(s) readily to oftenly used as catalysts. However, Zn, Cd and form 1+ ion (in the case of alkali metals) or 2+ Hg which have the electronic configuration, ion (in the case of alkaline earth metals). The (n-1) d10ns2 do not show most of the properties metallic character and the reactivity increase of transition elements. In a way, transition as we go down the group. Because of high metals form a bridge between the chemically reactivity they are never found pure in nature. active metals of s-block elements and the The compounds of the s-block elements, with less active elements of Groups 13 and 14 and thus take their familiar name “Transitionthe exception of those of lithium and beryllium Elements”.are predominantly ionic. 3.6.4 The f-Block Elements3.6.2 The p-Block Elements (Inner-Transition Elements) The p-Block Elements comprise those The two rows of elements at the bottom ofbelonging to Group 13 to 18 and these the Periodic Table, called the Lanthanoids,together with the s-Block Elements are Ce(Z = 58) – Lu(Z = 71) and Actinoids,called the Representative Elements or Main Th(Z = 90) – Lr (Z = 103) are characterised by Group Elements. The outermost electronic the outer electronic configuration (n-2)f1-14 configuration varies from ns2np1 to ns2np6 (n-1)d0–1ns2. The last electron added to each in each period. At the end of each period is element is filled in f- orbital. These two series a noble gas element with a closed valence of elements are hence called the Inner- shell ns2np6 configuration. All the orbitals Transition Elements (f-Block Elements). in the valence shell of the noble gases are They are all metals. Within each series, the completely filled by electrons and it is very properties of the elements are quite similar. difficult to alter this stable arrangement by The chemistry of the early actinoids is the addition or removal of electrons. The more complicated than the corresponding lanthanoids, due to the large number ofnoble gases thus exhibit very low chemical oxidation states possible for these actinoidreactivity. Preceding the noble gas family elements. Actinoid elements are radioactive.are two chemically important groups of non- Many of the actinoid elements have been mademetals. They are the halogens (Group 17) and only in nanogram quantities or even less bythe chalcogens (Group 16). These two groups nuclear reactions and their chemistry is not of elements have highly negative electron fully studied. The elements after uranium are gain enthalpies and readily add one or two called Transuranium Elements. electrons respectively to attain the stable noble gas configuration. The non-metallic Problem 3.3 character increases as we move from left to The elements Z = 117 and 120 have not yetright across a period and metallic character been discovered. In which family/group increases as we go down the group. would you place these elements and also give the electronic configuration in3.6.3 The d-Block Elements (Transition each case. Elements) SolutionThese are the elements of Group 3 to 12 in the centre of the Periodic Table. These are We see from Fig. 3.2, that element characterised by the filling of inner d orbitals with Z = 117, would belong to the halogen family (Group 17) and theby electrons and are therefore referred to as electronic configuration would be [Rn]d-Block Elements. These elements have 5f146d107s27p5. The element with Z = 120,the general outer electronic configuration will be placed in Group 2 (alkaline earth (n-1)d1-10ns0-2 except for Pd where its electronic metals), and will have the electronic configuration is 4d105s0.. They are all metals. configuration [Uuo]8s2. They mostly form coloured ions, exhibit variable Classification of Elements and Periodicity in Properties 85 3.6.5 Metals, Non-metals and Metalloids SolutionIn addition to displaying the classification Metallic character increases down aof elements into s-, p-, d-, and f-blocks, group and decreases along a period asFig. 3.3 shows another broad classification we move from left to right. Hence the of elements based on their properties. The order of increasing metallic character elements can be divided into Metals and is: P < Si < Be < Mg < Na. Non-Metals. Metals comprise more than 78% of all known elements and appear on 3.7 PERIODIC TRENDS IN PROPERTIES the left side of the Periodic Table. Metals are OF ELEMENTS usually solids at room temperature [mercury There are many observable patterns in theis an exception; gallium and caesium also physical and chemical properties of elements have very low melting points (303K and as we descend in a group or move across a 302K, respectively)]. Metals usually have high period in the Periodic Table. For example, melting and boiling points. They are good within a period, chemical reactivity tends to conductors of heat and electricity. They are be high in Group 1 metals, lower in elements malleable (can be flattened into thin sheets by towards the middle of the table, and increases hammering) and ductile (can be drawn into to a maximum in the Group 17 non-metals. wires). In contrast, non-metals are located at Likewise within a group of representative the top right hand side of the Periodic Table. metals (say alkali metals) reactivity increases In fact, in a horizontal row, the property of on moving down the group, whereas within a elements change from metallic on the left to group of non-metals (say halogens), reactivity non-metallic on the right. Non-metals are decreases down the group. But why do the usually solids or gases at room temperature properties of elements follow these trends? with low melting and boiling points (boron And how can we explain periodicity? To and carbon are exceptions). They are poor answer these questions, we must look into the conductors of heat and electricity. Most non- theories of atomic structure and properties metallic solids are brittle and are neither of the atom. In this section we shall discuss malleable nor ductile. The elements become the periodic trends in certain physical and more metallic as we go down a group; the chemical properties and try to explain them non-metallic character increases as one goes in terms of number of electrons and energy from left to right across the Periodic Table. levels. The change from metallic to non-metallic 3.7.1 Trends in Physical Propertiescharacter is not abrupt as shown by the thick There are numerous physical properties ofzig-zag line in Fig. 3.3. The elements (e.g., elements such as melting and boiling points,silicon, germanium, arsenic, antimony and heats of fusion and vaporization, energytellurium) bordering this line and running of atomization, etc. which show periodicdiagonally across the Periodic Table show variations. However, we shall discuss theproperties that are characteristic of both periodic trends with respect to atomic andmetals and non-metals. These elements are ionic radii, ionization enthalpy, electron gaincalled Semi-metals or Metalloids. enthalpy and electronegativity. Problem 3.4 (a) Atomic Radius Considering the atomic number and You can very well imagine that finding the position in the periodic table, arrange size of an atom is a lot more complicated than the following elements in the increasing measuring the radius of a ball. Do you know order of metallic character : Si, Be, Mg, why? Firstly, because the size of an atom Na, P. (~ 1.2 Å i.e., 1.2 × 10–10 m in radius) is very 86 chemistry small. Secondly, since the electron cloud The atomic radii of a few elements are listed surrounding the atom does not have a sharp in Table 3.6. Two trends are obvious. We can boundary, the determination of the atomic explain these trends in terms of nuclear charge size cannot be precise. In other words, there and energy level. The atomic size generally is no practical way by which the size of an decreases across a period as illustrated in individual atom can be measured. However, Fig. 3.4(a) for the elements of the second an estimate of the atomic size can be made by period. It is because within the period the knowing the distance between the atoms in outer electrons are in the same valence shell the combined state. One practical approach to and the effective nuclear charge increases estimate the size of an atom of a non-metallic as the atomic number increases resulting in element is to measure the distance between the increased attraction of electrons to the two atoms when they are bound together nucleus. Within a family or vertical column by a single bond in a covalent molecule and of the periodic table, the atomic radius from this value, the “Covalent Radius” of the increases regularly with atomic number as element can be calculated. For example, the illustrated in Fig. 3.4(b). For alkali metals bond distance in the chlorine molecule (Cl2) and halogens, as we descend the groups, is 198 pm and half this distance (99 pm), is the principal quantum number (n) increases taken as the atomic radius of chlorine. For and the valence electrons are farther frommetals, we define the term “Metallic Radius” the nucleus. This happens because the innerwhich is taken as half the internuclear energy levels are filled with electrons, whichdistance separating the metal cores in the serve to shield the outer electrons from themetallic crystal. For example, the distance pull of the nucleus. Consequently the size ofbetween two adjacent copper atoms in solid the atom increases as reflected in the atomiccopper is 256 pm; hence the metallic radius radii.of copper is assigned a value of 128 pm. For simplicity, in this book, we use the term Note that the atomic radii of noble gases Atomic Radius to refer to both covalent or are not considered here. Being monoatomic, metallic radius depending on whether the their (non-bonded radii) values are very element is a non-metal or a metal. Atomic large. In fact radii of noble gases should be radii can be measured by X-ray or other compared not with the covalent radii but with spectroscopic methods. the van der Waals radii of other elements. Table 3.6(a) Atomic Radii/pm Across the Periods Atom (Period II) Li Be B C N O F Atomic radius 152 111 88 77 74 66 64 Atom (Period III) Na Mg Al Si P S Cl Atomic radius 186 160 143 117 110 104 99 Table 3.6(b) Atomic Radii/pm Down a Family Atom Atomic Atom Atomic (Group I) Radius (Group 17) Radius Li 152 F 64 Na 186 Cl 99 K 231 Br 114 Rb 244 I 133 Cs 262 At 140 Classification of Elements and Periodicity in Properties 87 Fig. 3.4 (a) Variation of atomic radius with atomic Fig. 3.4 (b) Variation of atomic radius with number across the second period atomic number for alkali metals and halogens (b) Ionic Radius cation with the greater positive charge will have a smaller radius because of the greaterThe removal of an electron from an atom attraction of the electrons to the nucleus.results in the formation of a cation, whereas Anion with the greater negative charge willgain of an electron leads to an anion. The have the larger radius. In this case, the netionic radii can be estimated by measuring repulsion of the electrons will outweigh thethe distances between cations and anions nuclear charge and the ion will expand in size.in ionic crystals. In general, the ionic radii of elements exhibit the same trend as the Problem 3.5atomic radii. A cation is smaller than its parent atom because it has fewer electrons Which of the following species will have while its nuclear charge remains the same. the largest and the smallest size? Mg, Mg2+, Al, Al3+.The size of an anion will be larger than that of the parent atom because the addition of one Solution or more electrons would result in increased Atomic radii decrease across a period. repulsion among the electrons and a decrease Cations are smaller than their parent in effective nuclear charge. For example, the atoms. Among isoelectronic species, ionic radius of fluoride ion (F–) is 136 pm the one with the larger positive nuclear whereas the atomic radius of fluorine is only charge will have a smaller radius. 64 pm. On the other hand, the atomic radius Hence the largest species is Mg; the of sodium is 186 pm compared to the ionic smallest one is Al3+. radius of 95 pm for Na+. When we find some atoms and ions which (c) Ionization Enthalpy contain the same number of electrons, we call A quantitative measure of the tendency of them isoelectronic species*. For example, an element to lose electron is given by its O2–, F–, Na+ and Mg2+ have the same number Ionization Enthalpy. It represents the of electrons (10). Their radii would be different energy required to remove an electron from an because of their different nuclear charges. The isolated gaseous atom (X) in its ground state. * Two or more species with same number of atoms, same number of valence electrons and same structure, regardless of the nature of elements involved. 88 chemistry In other words, the first ionization enthalpy for an element X is the enthalpy change (∆i H) for the reaction depicted in equation 3.1. X(g) → X+(g) + e– (3.1) The ionization enthalpy is expressed in units of kJ mol–1. We can define the second ionization enthalpy as the energy required to remove the second most loosely bound electron; it is the energy required to carry out the reaction shown in equation 3.2. X+(g) → X2+(g) + e– (3.2) Energy is always required to remove Fig. 3.5 Variation of first ionization enthalpieselectrons from an atom and hence ionization (∆iH) with atomic number for elementsenthalpies are always positive. The second with Z = 1 to 60ionization enthalpy will be higher than the first ionization enthalpy because it is more can be correlated with their high reactivity. difficult to remove an electron from a positively In addition, you will notice two trends the charged ion than from a neutral atom. In the first ionization enthalpy generally increases same way the third ionization enthalpy will be as we go across a period and decreases higher than the second and so on. The term as we descend in a group. These trends “ionization enthalpy”, if not qualified, is taken are illustrated in Figs. 3.6(a) and 3.6(b) as the first ionization enthalpy. respectively for the elements of the second The first ionization enthalpies of elements period and the first group of the periodic having atomic numbers up to 60 are plotted table. You will appreciate that the ionization in Fig. 3.5. The periodicity of the graph is enthalpy and atomic radius are closely related quite striking. You will find maxima at the properties. To understand these trends, we noble gases which have closed electron shells have to consider two factors : (i) the attraction and very stable electron configurations. On of electrons towards the nucleus, and (ii) the the other hand, minima occur at the alkali repulsion of electrons from each other. The metals and their low ionization enthalpies effective nuclear charge experienced by a 3.6 (a) 3.6 (b) Fig. 3.6(a) First ionization enthalpies (∆iH) of elements of the second period as a function of atomic number (Z) and Fig. 3.6(b) ∆iH of alkali metals as a function of Z. Classification of Elements and Periodicity in Properties 89 valence electron in an atom will be less than the 2s electrons of beryllium. Therefore, it is the actual charge on the nucleus because of easier to remove the 2p-electron from boron “shielding” or “screening” of the valence compared to the removal of a 2s- electron from electron from the nucleus by the intervening beryllium. Thus, boron has a smaller first core electrons. For example, the 2s electron ionization enthalpy than beryllium. Another in lithium is shielded from the nucleus by “anomaly” is the smaller first ionization the inner core of 1s electrons. As a result, the enthalpy of oxygen compared to nitrogen. This valence electron experiences a net positive arises because in the nitrogen atom, three charge which is less than the actual charge 2p-electrons reside in different atomic orbitals of +3. In general, shielding is effective when (Hund’s rule) whereas in the oxygen atom, the orbitals in the inner shells are completely two of the four 2p-electrons must occupy the filled. This situation occurs in the case of same 2p-orbital resulting in an increased alkali metals which have single outermost electron-electron repulsion. Consequently, ns-electron preceded by a noble gas electronic it is easier to remove the fourth 2p-electron configuration. from oxygen than it is, to remove one of the When we move from lithium to fluorine three 2p-electrons from nitrogen. across the second period, successive electrons are added to orbitals in the same principal Problem 3.6 quantum level and the shielding of the nuclear The first ionization enthalpy (∆i H ) values charge by the inner core of electrons does of the third period elements, Na, Mg and not increase very much to compensate for Si are respectively 496, 737 and 786 kJ the increased attraction of the electron to the mol–1. Predict whether the first ∆i H value nucleus. Thus, across a period, increasing for Al will be more close to 575 or 760 kJ nuclear charge outweighs the shielding. mol–1 ? Justify your answer. Consequently, the outermost electrons are Solution held more and more tightly and the ionization It will be more close to 575 kJ mol–1.enthalpy increases across a period. As we go The value for Al should be lower thandown a group, the outermost electron being that of Mg because of effective shielding increasingly farther from the nucleus, there is of 3p electrons from the nucleus by an increased shielding of the nuclear charge 3s-electrons. by the electrons in the inner levels. In this case, increase in shielding outweighs the (d) Electron Gain Enthalpy increasing nuclear charge and the removal of When an electron is added to a neutralthe outermost electron requires less energy gaseous atom (X) to convert it into a negativedown a group. ion, the enthalpy change accompanying the From Fig. 3.6(a), you will also notice that process is defined as the Electron Gain the first ionization enthalpy of boron (Z = 5) Enthalpy (∆egH). Electron gain enthalpyis slightly less than that of beryllium (Z = 4) provides a measure of the ease with which even though the former has a greater nuclear an atom adds an electron to form anion as charge. When we consider the same principal represented by equation 3.3. quantum level, an s-electron is attracted to the X(g) + e– → X –(g) (3.3)nucleus more than a p-electron. In beryllium, the electron removed during the ionization is Depending on the element, the process an s-electron whereas the electron removed of adding an electron to the atom can be during ionization of boron is a p-electron. The either endothermic or exothermic. For many penetration of a 2s-electron to the nucleus is elements energy is released when an electron more than that of a 2p-electron; hence the 2p is added to the atom and the electron gain electron of boron is more shielded from the enthalpy is negative. For example, group nucleus by the inner core of electrons than 17 elements (the halogens) have very high 90 chemistry Table 3.7 Electron Gain Enthalpies* / (kJ mol–1) of Some Main Group Elements Group 1 ∆egH Group 16 ∆egH Group 17 ∆egH Group 0 ∆egH H – 73 He + 48 Li – 60 O – 141 F – 328 Ne + 116 Na – 53 S – 200 Cl – 349 Ar + 96 K – 48 Se – 195 Br – 325 Kr + 96 Rb – 47 Te – 190 I – 295 Xe + 77 Cs – 46 Po – 174 At – 270 Rn + 68 negative electron gain enthalpies because they can attain stable noble gas electronic Problem 3.7 configurations by picking up an electron. Which of the following will have the most On the other hand, noble gases have large negative electron gain enthalpy and positive electron gain enthalpies because the which the least negative? electron has to enter the next higher principal P, S, Cl, F. quantum level leading to a very unstable Explain your answer. electronic configuration. It may be noted that Solution electron gain enthalpies have large negative Electron gain enthalpy generallyvalues toward the upper right of the periodic becomes more negative across atable preceding the noble gases. period as we move from left to right. The variation in electron gain enthalpies of Within a group, electron gain enthalpy elements is less systematic than for ionization becomes less negative down a group. enthalpies. As a general rule, electron gain However, adding an electron to the enthalpy becomes more negative with increase 2p-orbital leads to greater repulsion in the atomic number across a period. The than adding an electron to the larger effective nuclear charge increases from left to 3p-orbital. Hence the element with right across a period and consequently it will most negative electron gain enthalpy is be easier to add an electron to a smaller atom chlorine; the one with the least negative since the added electron on an average would electron gain enthalpy is phosphorus. be closer to the positively charged nucleus. We should also expect electron gain enthalpy to (e) Electronegativity become less negative as we go down a group A qualitative measure of the ability of an atombecause the size of the atom increases and in a chemical compound to attract sharedthe added electron would be farther from the electrons to itself is called electronegativity.nucleus. This is generally the case (Table Unlike ionization enthalpy and electron gain3.7). However, electron gain enthalpy of O or enthalpy, it is not a measureable quantity.F is less negative than that of the succeeding However, a number of numerical scales ofelement. This is because when an electron is added to O or F, the added electron goes to electronegativity of elements viz., Pauling the smaller n = 2 quantum level and suffers scale, Mulliken-Jaffe scale, Allred-Rochow significant repulsion from the other electrons scale have been developed. The one which present in this level. For the n = 3 quantum is the most widely used is the Pauling scale. level (S or Cl), the added electron occupies Linus Pauling, an American scientist, in 1922 a larger region of space and the electron- assigned arbitrarily a value of 4.0 to fluorine, electron repulsion is much less. the element considered to have the greatest * In many books, the negative of the enthalpy change for the process depicted in equation 3.3 is defined as the ELECTRON AFFINITY (Ae ) of the atom under consideration. If energy is released when an electron is added to an atom, the electron affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to add an electron to an atom, then the electron affinity of the atom is assigned a negative sign. However, electron affinity is defined as absolute zero and, therefore at any other temperature (T) heat capacities of the reactants and the products have to be taken into account in ∆egH = –Ae – 5/2 RT. Classification of Elements and Periodicity in Properties 91 ability to attract electrons. Approximate On the same account electronegativity values values for the electronegativity of a few decrease with the increase in atomic radii elements are given in Table 3.8(a) down a group. The trend is similar to that of ionization enthalpy. The electronegativity of any given element Knowing the relationship betweenis not constant; it varies depending on the electronegativity and atomic radius, canelement to which it is bound. Though it is you now visualise the relationship betweennot a measurable quantity, it does provide a electronegativity and non-metallic properties?means of predicting the nature of force that Non-metallic elements have strong tendencyholds a pair of atoms together – a relationship that you will explore later. Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decrease down a group (say from fluorine to astatine) in the periodic table. How can these trends be explained? Can the electronegativity be related to atomic radii, which tend to decrease across each period from left to right, but increase down each group ? The attraction between the outer (or valence) electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases. Fig. 3.7 The periodic trends of elements in the periodic table Table 3.8(a) Electronegativity Values (on Pauling scale) Across the Periods Atom (Period II) Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Atom (Period III) Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Table 3.8(b) Electronegativity Values (on Pauling scale) Down a Family Atom Electronegativity Atom Electronegativity (Group I) Value (Group 17) Value Li 1.0 F 4.0 Na 0.9 Cl 3.0 K 0.8 Br 2.8 Rb 0.8 I 2.5 Cs 0.7 At 2.2 92 chemistry to gain electrons. Therefore, electronegativity is with outer electronic configuration 2s22p5, directly related to that non-metallic properties shares one electron with oxygen in the OF2 of elements. It can be further extended to say molecule. Being highest electronegative that the electronegativity is inversely related element, fluorine is given oxidation state to the metallic properties of elements. Thus, –1. Since there are two fluorine atoms in the increase in electronegativities across this molecule, oxygen with outer electronic a period is accompanied by an increase configuration 2s22p4 shares two electrons in non-metallic properties (or decrease in with fluorine atoms and thereby exhibits metallic properties) of elements. Similarly, the oxidation state +2. In Na2O, oxygen being decrease in electronegativity down a group is more electronegative accepts two electrons, accompanied by a decrease in non-metallic one from each of the two sodium atoms and, properties (or increase in metallic properties) thus, shows oxidation state –2. On the other of elements. hand sodium with electronic configuration All these periodic trends are summarised 3s1 loses one electron to oxygen and is given in Figure 3.7. oxidation state +1. Thus, the oxidation state of an element in a particular compound can 3.7.2 Periodic Trends in Chemical be defined as the charge acquired by its atom Properties on the basis of electronegative consideration Most of the trends in chemical properties of from other atoms in the molecule. elements, such as diagonal relationships, inert pair effect, effects of lanthanoid contraction Problem 3.8 etc. will be dealt with along the discussion Using the Periodic Table, predict the of each group in later units. In this section formulas of compounds which might we shall study the periodicity of the valence be formed by the following pairs of state shown by elements and the anomalous elements; (a) silicon and bromine properties of the second period elements (from (b) aluminium and sulphur. lithium to fluorine). Solution (a) Periodicity of Valence or Oxidation (a) Silicon is group 14 element with States a valence of 4; bromine belongs to the halogen family with a valenceThe valence is the most characteristic property of 1. Hence the formula of theof the elements and can be understood in compound formed would be SiBr4.terms of their electronic configurations. The (b) Aluminium belongs to groupvalence of representative elements is usually 13 with a valence of 3; sulphur(though not necessarily) equal to the number belongs to group 16 elements withof electrons in the outermost orbitals and/or a valence of 2. Hence, the formulaequal to eight minus the number of outermost of the compound formed would be electrons as shown below. Al2S3. Nowadays the term oxidation state is Some periodic trends observed in thefrequently used for valence. Consider the valence of elements (hydrides and oxides)two oxygen containing compounds: OF2 and are shown in Table 3.9. Other such periodicNa2O. The order of electronegativity of the trends which occur in the chemical behaviourthree elements involved in these compounds of the elements are discussed elsewhere inis F > O > Na. Each of the atoms of fluorine, Group 1 2 13 14 15 16 17 18 Number of valence 1 2 3 4 5 6 7 8 electron alence 1 2 3 4 3,5 2,6 1,7 0,8 Classification of Elements and Periodicity in Properties 93 Table 3.9 Periodic Trends in Valence of Elements as shown by the Formulas of Their Compounds Group 1 2 13 14 15 16 17 Formula of LiH CaH2 B2H6 CH4 NH3 H2O HF hydride NaH AlH3 SiH4 PH3 H2S HCl KH GeH4 AsH3 H2Se HBr SnH4 H2Te HI Formula Li2O MgO B2O3 CO2 N2O3, N2O5 – of oxide Na2O CaO Al2O3 SiO2 P4O6, P4O10 SO3 Cl2 O7 SrO K2O Ga2O3 GeO2 As2O3, As2O5 SeO3 – BaO In2O3 SnO2 Sb2O3, Sb2O5 TeO3 – PbO2 Bi2O3 – – this book. There are many elements which the second element of the following group exhibit variable valence. This is particularly i.e., magnesium and aluminium, respectively. characteristic of transition elements and This sort of similarity is commonly referred actinoids, which we shall study later. to as diagonal relationship in the periodic properties. (b) Anomalous Properties of Second Period Elements What are the reasons for the different chemical behaviour of the first member ofThe first element of each of the groups 1 a group of elements in the s- and p-blocks(lithium) and 2 (beryllium) and groups 13-17 compared to that of the subsequent members(boron to fluorine) differs in many respects in the same group? The anomalous behaviourfrom the other members of their respective is attributed to their small size, large charge/group. For example, lithium unlike other radius ratio and high electronegativity of the alkali metals, and beryllium unlike other elements. In addition, the first member of alkaline earth metals, form compounds with group has only four valence orbitals (2s and pronounced covalent character; the other 2p) available for bonding, whereas the second members of these groups predominantly member of the groups have nine valence form ionic compounds. In fact the behaviour orbitals (3s, 3p, 3d). As a consequence of of lithium and beryllium is more similar with this, the maximum covalency of the first member of each group is 4 (e.g., boron Property Element can only form  BF4 , whereas the other members of the groups can expand their Metallic radius M/pm Li Be B valence shell to accommodate more than 152 111 88 four pairs of electrons e.g., aluminium forms). Furthermore, the first Na Mg Al  AlF6 3  186 160 143 member of p-block elements displays greater ability to form pπ – pπ multiple Ionic radius M+/pm Li Be bonds to itself (e.g., C = C, C ≡ C, 76 31 N = N, N ≡ Ν) and to other second period Na Mg elements (e.g., C = O, C = N, C ≡ N, 102 72 N = O) compared to subsequent members of the same group. 94 chemistry here it can be directly related to the metallic Problem 3.9 and non-metallic character of elements. Thus, Are the oxidation state and covalency of the metallic character of an element, which Al in [AlCl(H2O)5]2+ same ? is highest at the extremely left decreases and Solution the non-metallic character increases while moving from left to right across the period. No. The oxidation state of Al is +3 and the covalency is 6. The chemical reactivity of an element can be best shown by its reactions with oxygen and 3.7.3 Periodic Trends and Chemical halogens. Here, we shall consider the reaction Reactivity of the elements with oxygen only. Elements on two extremes of a period easily combineWe have observed the periodic trends in with oxygen to form oxides. The normal oxidecertain fundamental properties such as formed by the element on extreme left is theatomic and ionic radii, ionization enthalpy, most basic (e.g., Na2O), whereas that formedelectron gain enthalpy and valence. We know by now that the periodicity is related to by the element on extreme right is the most electronic configuration. That is, all chemical acidic (e.g., Cl2O7). Oxides of elements in the and physical properties are a manifestation of centre are amphoteric (e.g., Al2O3, As2O3) or the electronic configuration of elements. We neutral (e.g., CO, NO, N2O). Amphoteric oxides shall now try to explore relationships between behave as acidic with bases and as basic with these fundamental properties of elements with acids, whereas neutral oxides have no acidic their chemical reactivity. or basic properties. The atomic and ionic radii, as we know, Problem 3.10generally decrease in a period from left to right. As a consequence, the ionization enthalpies Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 isgenerally increase (with some exceptions as an acidic oxide.outlined in section 3.7.1(a)) and electron gain enthalpies become more negative across a Solution period. In other words, the ionization enthalpy Na2O with water forms a strong base of the extreme left element in a period is the whereas Cl2O7 forms strong acid. least and the electron gain enthalpy of the Na2O + H2O → 2NaOH element on the extreme right is the highest Cl2O7 + H2O → 2HClO4 negative (note : noble gases having completely Their basic or acidic nature can befilled shells have rather positive electron qualitatively tested with litmus paper.gain enthalpy values). This results into high chemical reactivity at the two extremes and the lowest in the centre. Thus, the maximum Among transition metals (3d series), the chemical reactivity at the extreme left (among change in atomic radii is much smaller as alkali metals) is exhibited by the loss of an compared to those of representative elements electron leading to the formation of a cation across the period. The change in atomic radii and at the extreme right (among halogens) is still smaller among inner-transition metals shown by the gain of an electron forming (4f series). The ionization enthalpies are an anion. This property can be related with intermediate between those of s- and p-blocks. the reducing and oxidizing behaviour of the As a consequence, they are less electropositive elements which you will learn later. However, than group 1 and 2 metals. Classification of Elements and Periodicity in Properties 95 In a group, the increase in atomic and increases down the group and non-metallic ionic radii with increase in atomic number character decreases. This trend can be related generally results in a gradual decrease in with their reducing and oxidizing property ionization enthalpies and a regular decrease which you will learn later. In the case of (with exception in some third period elements transition elements, however, a reverse trend as shown in section 3.7.1(d)) in electron is observed. This can be explained in terms of gain enthalpies in the case of main group atomic size and ionization enthalpy. elements. Thus, the metallic character SUMMARY In this Unit, you have studied the development of the Periodic Law and the Periodic Table. Mendeleev’s Periodic Table was based on atomic masses. Modern Periodic Table arranges the elements in the order of their atomic numbers in seven horizontal rows (periods) and eighteen vertical columns (groups or families). Atomic numbers in a period are consecutive, whereas in a group they increase in a pattern. Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have incrementally increasing number of electrons from left to right, and, therefore, have different valencies. Four types of elements can be recognized in the periodic table on the basis of their electronic configurations. These are s-block, p-block, d-block and f-block elements. Hydrogen with one electron in the 1s orbital occupies a unique position in the periodic table. Metals comprise more than seventy eight per cent of the known elements. Non-metals, which are located at the top of the periodic table, are less than twenty in number. Elements which lie at the border line between metals and non-metals (e.g., Si, Ge, As) are called metalloids or semi-metals. Metallic character increases with increasing atomic number in a group whereas decreases from left to right in a period. The physical and chemical properties of elements vary periodically with their atomic numbers. Periodic trends are observed in atomic sizes, ionization enthalpies, electron gain enthalpies, electronegativity and valence. The atomic radii decrease while going from left to right in a period and increase with atomic number in a group. Ionization enthalpies generally increase across a period and decrease down a group. Electronegativity also shows a similar trend. Electron gain enthalpies, in general, become more negative across a period and less negative down a group. There is some periodicity in valence, for example, among representative elements, the valence is either equal to the number of electrons in the outermost orbitals or eight minus this number. Chemical reactivity is highest at the two extremes of a period and is lowest in the centre. The reactivity on the left extreme of a period is because of the ease of electron loss (or low ionization enthalpy). Highly reactive elements do not occur in nature in free state; they usually occur in the combined form. Oxides formed of the elements on the left are basic and of the elements on the right are acidic in nature. Oxides of elements in the centre are amphoteric or neutral. 96 chemistry Exercises 3.1 What is the basic theme of organisation in the periodic table? 3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? 3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law? 3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. 3.5 In terms of period and group where would you locate the element with Z =114? 3.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table. 3.7 Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group? 3.8 Why do elements in the same group have similar physical and chemical properties? 3.9 What does atomic radius and ionic radius really mean to you? 3.10 How do atomic radius vary in a period and in a group? How do you explain the variation? 3.11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+ 3.12 Consider the following species : N3–, O2–, F–, Na+, Mg2+ and Al3+ (a) What is common in them? (b) Arrange them in the order of increasing ionic radii. 3.13 Explain why cation are smaller and anions larger in radii than their parent atoms? 3.14 What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy? Hint : Requirements for comparison purposes. 3.15 Energy of an electron in the ground state of the hydrogen atom is –2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1. Hint: Apply the idea of mole concept to derive the answer. 3.16 Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ∆i H than B (ii) O has lower ∆i H than N and F? Classification of Elements and Periodicity in Properties 97 3.17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? 3.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group? 3.19 The first ionization enthalpy values (in kJ mol–1) of group 13 elements are : B Al Ga In Tl 801 577 579 558 589 How would you explain this deviation from the general trend ? 3.20 Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl 3.21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer. 3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity? 3.23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds? 3.24 Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron 3.25 Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer. 3.26 What are the major differences between metals and non-metals? 3.27 Use the periodic table to answer the following questions. (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. 3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F > CI > Br > I. Explain. 3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements. 3.30 Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the periodic table. 98 chemistry 3.31 The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ∆H1 ∆H2 ∆egH I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 Which of the above elements is likely to be : (a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-metal. (e) the metal which can form a stable binary halide of the formula MX2(X=halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)? 3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine 3.33 In the modern periodic table, the period indicates the value of : (a) atomic number (b) atomic mass (c) principal quantum number (d) azimuthal quantum number. 3.34 Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration. Classification of Elements and Periodicity in Properties 99 3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z ) (c) Nuclear mass (d) Number of core electrons. 3.36 The size of isoelectronic species — F–, Ne and Na+ is affected by (a) nuclear charge (Z ) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same. 3.37 Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. 3.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is : (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B 3.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is : (a) B > C > Si > N > F (b) Si > C > B > N > F (c) F > N > C > B > Si (d) F > N > C > Si > B 3.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is : (a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl Unit 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE Scientists are constantly discovering new compounds, orderly arranging the facts about them, trying to explain with the existing knowledge, organising to modify the earlier views or evolve theories for explaining the newly After studying this Unit, you will be observed facts. able to • understand Kössel-Lewis approach to chemical bonding; • explain the octet rule and its Matter is made up of one or different type of elements. limitations, draw Lewis structures Under normal conditions no other element exists as an of simple molecules; independent atom in nature, except noble gases. However, • explain the formation of different a group of atoms is found to exist together as one species types of bonds; having characteristic properties. Such a group of atoms is called a molecule. Obviously there must be some force • describe the VSEPR theory and which holds these constituent atoms together in the predict the geometry of simple molecules. The attractive force which holds various molecules; constituents (atoms, ions, etc.) together in different • explain the valence bond chemical species is called a chemical bond. Since the approach for the formation of formation of chemical compounds takes place as a result of covalent bonds; combination of atoms of various elements in different ways, • predict the directional properties it raises many questions. Why do atoms combine? Why are of covalent bonds; only certain combinations possible? Why do some atoms combine while certain others do not? Why do molecules• explain the different types of hybridisation involving s, p and possess definite shapes? To answer such questions different d orbitals and draw shapes of theories and concepts have been put forward from time simple covalent molecules; to time. These are Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB)• describe the molecular orbital theory of homonuclear diatomic Theory and Molecular Orbital (MO) Theory. The evolution molecules; of various theories of valence and the interpretation of the nature of chemical bonds have closely been related to • explain the concept of hydrogen the developments in the understanding of the structure bond. of atom, the electronic configuration of elements and the periodic table. Every system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain stability. Reprint 2025-26 Chemical Bonding And Molecular Structure 101

3.17 During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1mg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.

3.16 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

3.21 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2 Cl 2 g  SO 2 g  Cl 2 g Experiment Time/s–1 Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.

3.3 MODERN PERIODIC LAW AND THE physical and chemical properties of elements PRESENT FORM OF THE PERIODIC and their compounds. TABLE Numerous forms of Periodic Table have We must bear in mind that when Mendeleev been devised from time to time. Some developed his Periodic Table, chemists forms emphasise chemical reactions and knew nothing about the internal structure valence, whereas others stress the electronic of atom. However, the beginning of the 20th configuration of elements. A modern version, century witnessed profound developments the so-called “long form” of the Periodic in theories about sub-atomic particles. In Table of the elements (Fig. 3.2), is the most 1913, the English physicist, Henry Moseley convenient and widely used. The horizontal observed regularities in the characteristic rows (which Mendeleev called series) are X-ray spectra of the elements. A plot of called periods and the vertical columns, (where is frequency of X-rays emitted) groups. Elements having similar outer against atomic number (Z) gave a straight electronic configurations in their atoms are arranged in vertical columns, referredline and not the plot of vs atomic mass. to as groups or families. According to theHe thereby showed that the atomic number recommendation of International Union ofis a more fundamental property of an element Pure and Applied Chemistry (IUPAC), thethan its atomic mass. Mendeleev’s Periodic groups are numbered from 1 to 18 replacingLaw was, therefore, accordingly modified. This the older notation of groups IA … VIIA, VIII,is known as the Modern Periodic Law and IB … VIIB and 0. can be stated as : There are altogether seven periods. The The physical and chemical properties period number corresponds to the highest of the elements are periodic functions principal quantum number (n) of the elements of their atomic numbers. in the period. The first period contains 2 The Periodic Law revealed important elements. The subsequent periods consists of analogies among the 94 naturally occurring 8, 8, 18, 18 and 32 elements, respectively. The elements (neptunium and plutonium like seventh period is incomplete and like the sixth actinium and protoactinium are also found period would have a theoretical maximum in pitch blende – an ore of uranium). It (on the basis of quantum numbers) of 32 stimulated renewed interest in Inorganic elements. In this form of the Periodic Table, Chemistry and has carried into the present 14 elements of both sixth and seventh periods with the creation of artificially produced (lanthanoids and actinoids, respectively) are short-lived elements. placed in separate panels at the bottom*. You may recall that the atomic number 3.4 NOMENCLATURE OF ELEMENTSis equal to the nuclear charge (i.e., number WITH ATOMIC NUMBERS > 100of protons) or the number of electrons in a neutral atom. It is then easy to visualize The naming of the new elements had been the significance of quantum numbers and traditionally the privilege of the discoverer electronic configurations in periodicity of (or discoverers) and the suggested name was elements. In fact, it is now recognized that the ratified by the IUPAC. In recent years this has Periodic Law is essentially the consequence led to some controversy. The new elements of the periodic variation in electronic with very high atomic numbers are so unstable configurations, which indeed determine the that only minute quantities, sometimes only * Glenn T. Seaborg’s work in the middle of the 20th century starting with the discovery of plutonium in 1940, followed by those of all the transuranium elements from 94 to 102 led to reconfiguration of the periodic table placing the actinoids below the lanthanoids. In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been named Seaborgium (Sg) in his honour. Classification of Elements and Periodicity in Properties 79 0 B VII This B VI electronic B outer V B state IV recommendations. elements. B ground IUPACthe III and for B 0 1984 II theand numbers B I with IB–VIIB atomic VIII, → their accordance VIII withinIA–VIIA, ← of Elements1-18 A scheme the VII ofnumbered A are VI Table numbering A old groups V Periodic the A The the IV of A replaces III form Longconfigurations.notation IIA 3.2 IA Fig. 80 chemistry a few atoms of them are obtained. Their digits which make up the atomic number and synthesis and characterisation, therefore, “ium” is added at the end. The IUPAC names require highly sophisticated costly equipment for elements with Z above 100 are shown in and laboratory. Such work is carried out with Table 3.5. competitive spirit only in some laboratories in the world. Scientists, before collecting the Table 3.4 Notation for IUPAC reliable data on the new element, at times Nomenclature of Elements get tempted to claim for its discovery. For example, both American and Soviet scientists Digit Name Abbreviation claimed credit for discovering element 104. 0 nil n The Americans named it Rutherfordium 1 un u whereas Soviets named it Kurchatovium. To 2 bi b avoid such problems, the IUPAC has made 3 tri t recommendation that until a new element’s 4 quad q discovery is proved, and its name is officially 5 pent precognised, a systematic nomenclature be 6 hex hderived directly from the atomic number of 7 sept sthe element using the numerical roots for 8 oct o0 and numbers 1-9. These are shown in 9 enn eTable 3.4. The roots are put together in order of Table 3.5 Nomenclature of Elements with Atomic Number Above 100 Atomic Name according to IUPAC IUPAC Symbol Number IUPAC nomenclature Official Name Symbol 101 Unnilunium Unu Mendelevium Md 102 Unnilbium Unb Nobelium No 103 Unniltrium Unt Lawrencium Lr 104 Unnilquadium Unq Rutherfordium Rf 105 Unnilpentium Unp Dubnium Db 106 Unnilhexium Unh Seaborgium Sg 107 Unnilseptium Uns Bohrium Bh 108 Unniloctium Uno Hassium Hs 109 Unnilennium Une Meitnerium Mt 110 Ununnillium Uun Darmstadtium Ds 111 Unununnium Uuu Rontgenium Rg 112 Ununbium Uub Copernicium Cn 113 Ununtrium Uut Nihonium Nh 114 Ununquadium Uuq Flerovium Fl 115 Ununpentium Uup Moscovium Mc 116 Ununhexium Uuh Livermorium Lv 117 Ununseptium Uus Tennessine Ts 118 Ununoctium Uuo Oganesson Og Classification of Elements and Periodicity in Properties 81 Thus, the new element first gets a be readily seen that the number of elements temporary name, with symbol consisting in each period is twice the number of atomic of three letters. Later permanent name orbitals available in the energy level that is and symbol are given by a vote of IUPAC being filled. The first period (n = 1) starts with representatives from each country. The the filling of the lowest level (1s) and therefore permanent name might reflect the country has two elements — hydrogen (ls1) and helium (or state of the country) in which the element (ls2) when the first shell (K) is completed. The was discovered, or pay tribute to a notable second period (n = 2) starts with lithium and the scientist. As of now, elements with atomic third electron enters the 2s orbital. The next numbers up to 118 have been discovered. element, beryllium has four electrons and has Official names of all elements have been the electronic configuration 1s22s2. Starting announced by IUPAC. from the next element boron, the 2p orbitals are filled with electrons when the L shell is Problem 3.1 completed at neon (2s22p6). Thus there are 8 elements in the second period. The third What would be the IUPAC name and period (n = 3) begins at sodium, and the added symbol for the element with atomic number 120? electron enters a 3s orbital. Successive filling of 3s and 3p orbitals gives rise to the third Solution period of 8 elements from sodium to argon. The From Table 3.4, the roots for 1, 2 and 0 fourth period (n = 4) starts at potassium, and are un, bi and nil, respectively. Hence, the added electrons fill up the 4s orbital. Now the symbol and the name respectively you may note that before the 4p orbital is filled, are Ubn and unbinilium. filling up of 3d orbitals becomes energetically favourable and we come across the so called 3d

3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26

1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

1.37 Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: 100 x xacetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 pacetone /mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 pchloroform /mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution. 1.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. 1.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water. 1.40 Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C. 1.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated. Answers to Some Intext Questions 1.1 C6H6 = 15.28%, CCl4 = 84.72% 1.2 0.459, 0.541 1.3 0.024 M, 0.03 M 1.4 36.946 g 1.5 1.5 mol kg–1 , 1.45 mol L–1 0.0263 1.9 23.4 mm Hg 1.10 121.67 g 1.11 5.077 g 1.12 30.96 Pa Chemistry 30 Reprint 2025-26 UnitUnitUnitUnit Unit22 Objectives ElectrochemistryElectrochemistry After studying this Unit, you will be able to · describe an electrochemical cell Chemical reactions can be used to produce electrical energy, and differentiate between galvanic conversely, electrical energy can be used to carry out chemical and electrolytic cells; reactions that do not proceed spontaneously.· apply Nernst equation for calculating the emf of galvanic cell and define standard potential of Electrochemistry is the study of production of the cell; · derive relation between standard electricity from energy released during spontaneous potential of the cell, Gibbs energy chemical reactions and the use of electrical energy of cell reaction and its equilibrium to bring about non-spontaneous chemical constant; transformations. The subject is of importance both · define resistivity (r), conductivity for theoretical and practical considerations. A large (k) and molar conductivity (✆m) of number of metals, sodium hydroxide, chlorine, ionic solutions; fluorine and many other chemicals are produced by · differentiate between ionic electrochemical methods. Batteries and fuel cells (electrolytic) and electronic convert chemical energy into electrical energy and are conductivity; · describe the method for used on a large scale in various instruments and measurement of conductivity of devices. The reactions carried out electrochemically electrolytic solutions and can be energy efficient and less polluting. Therefore, calculation of their molar study of electrochemistry is important for creating new conductivity; technologies that are ecofriendly. The transmission of · justify the variation of sensory signals through cells to brain and vice versa conductivity and molar and communication between the cells are known to conductivity of solutions with have electrochemical origin. Electrochemistry, is change in their concentration and therefore, a very vast and interdisciplinary subject. In define m (molar conductivity at this Unit, we will cover only some of its important zero concentration or infinite elementary aspects. dilution); · enunciate Kohlrausch law and learn its applications; · understand quantitative aspects of electrolysis; · describe the construction of some primary and secondary batteries and fuel cells; · explain corrosion as an electrochemical process. Reprint 2025-26 2.12.12.12.12.1 ElectrochemicalElectrochemicalElectrochemicalElectrochemicalElectrochemical We had studied the construction and functioning of Daniell cell CellsCellsCellsCellsCells (Fig. 2.1). This cell converts the chemical energy liberated during the redox reaction Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s) (2.1) to electrical energy and has an electrical potential equal to 1.1 V when concentration of Zn2+ and Cu2+ ions is unity (1 mol dm–3)*. Such a device is called a galvanic or a voltaic cell. If an external opposite potential is applied in the galvanic cell [Fig. 2.2(a)] and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V [Fig. 2.2(b)] when, the reaction stops altogether and no current flows through the cell. Any further increase in the external potential again starts the reaction but in the opposite direction [Fig. 2.2(c)]. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions. Both types of cells are Fig. 2.1: Daniell cell having electrodes of zinc and quite important and we shall study some of copper dipping in the solutions of their their salient features in the following pages. respective salts. Eext < 1.1V Eext = 1.1V (a) (b) e current cathodeanode I=0 Zn salt Cu Zn Cu -ve bridge +ve When Eext = 1.1 V (i) No flow of electrons or current. (ii) No chemical ZnSO4 CuSO4 ZnSO4 CuSO4 reaction. When Eext < 1.1 V Eext >1.1 (i) Electrons flow from Zn rod to (c) Cu rod hence current flows from Cu to Zn. – When Eext > 1.1 V (ii) Zn dissolves at anode and e (i) Electrons flow copper deposits at cathode. Cathode Current Anode from Cu to Zn +ve –ve and current flows Zn Cu from Zn to Cu. Fig. 2.2 (ii) Zinc is deposited Functioning of Daniell at the zinc cell when external electrode and voltage Eext opposing the copper dissolves at cell potential is applied. copper electrode. *Strictly speaking activity should be used instead of concentration. It is directly proportional to concentration. In dilute solutions, it is equal to concentration. You will study more about it in higher classes. Chemistry 32 Reprint 2025-26 2.22.22.22.22.2 GalvanicGalvanicGalvanicGalvanicGalvanic CellsCellsCellsCellsCells As mentioned earlier a galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. In this device the Gibbs energy of the spontaneous redox reaction is converted into electrical work which may be used for running a motor or other electrical gadgets like heater, fan, geyser, etc. Daniell cell discussed earlier is one such cell in which the following redox reaction occurs. Zn(s) + Cu2+(aq) ® Zn2+ (aq) + Cu(s) This reaction is a combination of two half reactions whose addition gives the overall cell reaction: (i) Cu2+ + 2e– ® Cu(s) (reduction half reaction) (2.2) (ii) Zn(s) ® Zn2+ + 2e– (oxidation half reaction) (2.3) These reactions occur in two different portions of the Daniell cell. The reduction half reaction occurs on the copper electrode while the oxidation half reaction occurs on the zinc electrode. These two portions of the cell are also called half-cells or redox couples. The copper electrode may be called the reduction half cell and the zinc electrode, the oxidation half-cell. We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half- cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge as shown in Fig. 2.1. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge. At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. At equilibrium, there is a separation of charges and depending on the tendencies of the two opposing reactions, the electrode may be positively or negatively charged with respect to the solution. A potential difference develops between the electrode and the electrolyte which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. According to IUPAC convention, standard reduction potentials are now called standard electrode potentials. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to the solution. Thus, there exists a potential difference between the two electrodes and as soon as the switch is in the on position the electrons flow from negative electrode to positive electrode. The direction of current flow is opposite to that of electron flow. 33 Electrochemistry Reprint 2025-26 The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half- cell on the right hand side minus the potential of the half-cell on the left hand side i.e., Ecell = Eright – Eleft This is illustrated by the following example: Cell reaction: Cu(s) + 2Ag+(aq) ¾® Cu2+(aq) + 2 Ag(s) (2.4) Half-cell reactions: Cathode (reduction): 2Ag+(aq) + 2e– ® 2Ag(s) (2.5) Anode (oxidation): Cu(s) ® Cu2+(aq) + 2e– (2.6) It can be seen that the sum of (3.5) and (3.6) leads to overall reaction (2.4) in the cell and that silver electrode acts as a cathode and copper electrode acts as an anode. The cell can be represented as: Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s) and we have Ecell = Eright – Eleft = EAg+úAg – ECu2+úCu (2.7) 2.2.1 The potential of individual half-cell cannot be measured. We can Measurement measure only the difference between the two half-cell potentials that of Electrode gives the emf of the cell. If we arbitrarily choose the potential of one Potential electrode (half-cell) then that of the other can be determined with respect to this. According to convention, a half-cell called standard hydrogen electrode (Fig.3.3) represented by Pt(s)ú H2(g)ú H+(aq), is assigned a zero potential at all temperatures corresponding to the reaction 1 H+ (aq) + e– ® H2(g) 2 The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity (Fig. 2.3). This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the Fig. 2.3: Standard Hydrogen Electrode (SHE). solution is one molar. Chemistry 34 Reprint 2025-26 At 298 K the emf of the cell, standard hydrogen electrode ççsecond half-cell constructed by taking standard hydrogen electrode as anode (reference half-cell) and the other half-cell as cathode, gives the reduction potential of the other half-cell. If the concentrations of the oxidised and the reduced forms of the species in the right hand half-cell are unity, then the cell potential is equal to standard electrode potential, Eo R of the given half-cell. Eo = EoR – Eo L As Eo L for standard hydrogen electrode is zero. Eo = Eo R – 0 = EoR The measured emf of the cell: Pt(s) ç H2(g, 1 bar) ç H + (aq, 1 M) çç Cu 2+ (aq, 1 M) ú Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction: Cu2+ (aq, 1M) + 2 e – ® Cu(s) Similarly, the measured emf of the cell: Pt(s) ç H2(g, 1 bar) ç H+ (aq, 1 M) çç Zn2+ (aq, 1M) ç Zn is -0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e– ® Zn(s) The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid it is oxidised by nitrate ion and not by hydrogen ion. The negative value of the standard electrode potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce hydrogen ions). In view of this convention, the half reaction for the Daniell cell in Fig. 2.1 can be written as: Left electrode: Zn(s) ® Zn 2+ (aq, 1 M) + 2 e – Right electrode: Cu 2+ (aq, 1 M) + 2 e – ® Cu(s) The overall reaction of the cell is the sum of above two reactions and we obtain the equation: Zn(s) + Cu 2+ (aq) ® Zn2+ (aq) + Cu(s) emf of the cell = Eocell = Eo R – Eo L = 0.34V – (– 0.76)V = 1.10 V Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in the reaction but provide their surface for oxidation or reduction reactions and for the conduction of electrons. For example, Pt is used in the following half-cells: Hydrogen electrode: Pt(s)|H2(g)| H+(aq) With half-cell reaction: H+ (aq)+ e– ® ½ H2(g) Bromine electrode: Pt(s)|Br2(aq)| Br–(aq) 35 Electrochemistry Reprint 2025-26 With half-cell reaction: ½ Br2(aq) + e– ® Br–(aq) The standard electrode potentials are very important and we can extract a lot of useful information from them. The values of standard electrode potentials for some selected half-cell reduction reactions are given in Table 2.1. If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species. It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction. Electrochemical cells are extensively used for determining the pH of solutions, solubility product, equilibrium constant and other thermodynamic properties and for potentiometric titrations. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.1 How would you determine the standard electrode potential of the system Mg2+|Mg? 2.2 Can you store copper sulphate solutions in a zinc pot? 2.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. 2.32.32.32.32.3 NernstNernstNernstNernstNernst We have assumed in the previous section that the concentration of all EquationEquationEquationEquationEquation the species involved in the electrode reaction is unity. This need not be always true. Nernst showed that for the electrode reaction: Mn+(aq) + ne–® M(s) the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by: RT o [M] E = E ln ( M n + / M ) ( M n + / M ) – nF [M n+ ] but concentration of solid M is taken as unity and we have o RT 1 E = E (2.8) ( M n + / M ) ( M n + /M ) – nF ln [M n+ ] o E ( M n + / M ) has already been defined, R is gas constant (8.314 JK–1 mol–1), F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+. Chemistry 36 Reprint 2025-26 Table 2.1: Standard Electrode Potentials at 298 K Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s. Reaction (Oxidised form + ne– ® Reduced form) E o/V ® 2F– 2.87 F2(g) + 2e– Co3+ + e– ® Co2+ 1.81 H2O2 + 2H+ + 2e– ® 2H2O 1.78 MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O 1.51 Au3+ + 3e– ® Au(s) 1.40 Cl2(g) + 2e– ® 2Cl– 1.36 Cr2O72– + 14H+ + 6e– ® 2Cr3+ + 7H2O 1.33 O2(g) + 4H+ + 4e– ® 2H2O 1.23 MnO2(s) + 4H+ + 2e– ® Mn2+ + 2H2O 1.23 Br2 + 2e– ® 2Br– 1.09 NO3– + 4H+ + 3e– ® NO(g) + 2H2O 0.97 2Hg2+ + 2e– ® Hg22+ 0.92 Ag+ + e– ® Ag(s) 0.80 agent agent Fe3+ + e– ® Fe2+ 0.77 O2(g) + 2H+ + 2e– ® H2O2 0.68 I2 + 2e– ® 2I– 0.54 oxidising reducing 0.52 of Cu+ + e– ® Cu(s) of Cu2+ + 2e– ® Cu(s) 0.34 AgCl(s) + e– ® Ag(s) + Cl– 0.22 strength AgBr(s) + e– ® Ag(s) + Br– strength 0.10 2H+ + 2e– ® H2(g) 0.00 Pb2+ + 2e– ® Pb(s) –0.13 Sn2+ + 2e– ® Sn(s) –0.14 Increasing Increasing Ni2+ + 2e– ® Ni(s) –0.25 Fe2+ + 2e– ® Fe(s) –0.44 Cr3+ + 3e– ® Cr(s) –0.74 Zn2+ + 2e– ® Zn(s) –0.76 2H2O + 2e– ® H2(g) + 2OH–(aq) –0.83 Al3+ + 3e– ® Al(s) –1.66 Mg2+ + 2e– ® Mg(s) –2.36 Na+ + e– ® Na(s) –2.71 Ca2+ + 2e– ® Ca(s) –2.87 K+ + e– ® K(s) –2.93 Li+ + e– ® Li(s) –3.05 1. A negative Eo means that the redox couple is a stronger reducing agent than the H+/H2 couple. 2. A positive Eo means that the redox couple is a weaker reducing agent than the H+/H2 couple. 37 Electrochemistry Reprint 2025-26 In Daniell cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we write For Cathode: E E o RT 1 (2.9)  Cu 2  /Cu  = (Cu 2 + /Cu ) – 2F ln   Cu 2   aq   For Anode: E E o RT 1 (2.10)  Zn 2  /Zn  = ( Zn 2 + / Zn ) – 2F ln   Zn 2   aq   E E 2  2  /Zn  The cell potential, E(cell) =  Cu /Cu  –  Zn o RT 1 E o RT 1 E = (Cu – ( Zn 2 + / Cu ) – 2 F ln 2 + / Zn ) + 2 F ln    Zn 2+ (aq)   Cu 2+ (aq) E o E o RT 1 1 – ln = (Cu 2 + / Cu ) – ( Zn 2 + / Zn ) – 2F ln    Cu 2+  aq    Zn 2+  aq   2  ] RT [ Zn o E(cell) = E ( cell ) – 2 F ln 2 + (2.11) [Cu ] It can be seen that E(cell) depends on the concentration of both Cu2+ and Zn2+ ions. It increases with increase in the concentration of Cu2+ ions and decrease in the concentration of Zn2+ ions. By converting the natural logarithm in Eq. (2.11) to the base 10 and substituting the values of R, F and T = 298 K, it reduces to 2 + ] 0 .059 [ Zn (2.12) 2 + ] E(cell) = E (ocell ) – 2 log [Cu We should use the same number of electrons (n) for both the electrodes and thus for the following cell Ni(s)ú Ni2+(aq) úú Ag+(aq)ú Ag The cell reaction is Ni(s) + 2Ag+(aq) ® Ni2+(aq) + 2Ag(s) The Nernst equation can be written as RT [Ni 2+ ] o + E(cell) = E ( cell ) – 2F ln [Ag ]2 and for a general electrochemical reaction of the type: a A + bB ne– cC + dD Nernst equation can be written as: RT E(cell) = E (ocell ) – nF 1nQ RT [C]c [D]d o (2.13) = E ( cell ) – nF ln [A] a [B]b Chemistry 38 Reprint 2025-26 Represent the cell in which the following reaction takes place ExampleExampleExampleExampleExample 2.12.12.12.12.1 Mg(s) + 2Ag+(0.0001M) ® Mg2+(0.130M) + 2Ag(s) Calculate its E(cell) if E (ocell ) = 3.17 V. The cell can be written as Mgú Mg2+(0.130M)úú Ag+(0.0001M)ú Ag SolutionSolutionSolutionSolutionSolution 2 + Mg RT o E = E ln (  cell  cell ) – 2F + 2 Ag 0 .059V 0.130 = 3.17 V – log 2 = 3.17 V – 0.21V = 2.96 V. 2 ( 0 . 0001) 2.3.1 Equilibrium If the circuit in Daniell cell (Fig. 2.1) is closed then we note that the reaction Constant Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s) (2.1) from Nernst takes place and as time passes, the concentration of Zn2+ keeps Equation on increasing while the concentration of Cu2+ keeps on decreasing. At the same time voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as: o 2.303 RT [Zn 2 + ] 2 + E(cell) = 0 = E ( cell ) – 2 F log [Cu ] o 2.303 RT [Zn 2  ] or E ( cell ) = log 2  2 F [Cu ] But at equilibrium, [ Zn 2 + ] = Kc for the reaction 2.1 [Cu2 + ] and at T = 298K the above equation can be written as o 0. 059 V o E ( cell ) = log KC = 1.1 V ( E ( cell ) = 1.1V) 2 (1.1V × 2) log KC =  37.288 0.059 V KC = 2 × 1037 at 298K. In general, o 2.303RT E ( cell ) = log KC (2.14) nF Thus, Eq. (2.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding Eo value of the cell. 39 Electrochemistry Reprint 2025-26 ExampleExampleExampleExampleExample 2.22.22.22.22.2 Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s) Eo( cell ) = 0.46 V o 0. 059 V SolutionSolutionSolutionSolutionSolution E ( cell ) = log KC = 0.46 V or 2 0 .46 V × 2 = 15.6 log KC = 0 .059 V KC = 3.92 × 1015 2.3.2 Electro- Electrical work done in one second is equal to electrical potential chemical multiplied by total charge passed. If we want to obtain maximum work Cell and from a galvanic cell then charge has to be passed reversibly. The Gibbs reversible work done by a galvanic cell is equal to decrease in its Gibbs Energy of energy and therefore, if the emf of the cell is E and nF is the amount the Reaction of charge passed and DrG is the Gibbs energy of the reaction, then DrG = – nFE(cell) (2.15) It may be remembered that E(cell) is an intensive parameter but DrG is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s) (2.1) DrG = – 2FE(cell) but when we write the reaction 2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s) DrG = – 4FE(cell) If the concentration of all the reacting species is unity, then E(cell) = E (ocell ) and we have DrGo = – nF E(cell)o (2.16) Thus, from the measurement of E (ocell ) we can obtain an important thermodynamic quantity, DrGo, standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: DrGo = –RT ln K. ExampleExampleExampleExampleExample 2.32.32.32.32.3 The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s) SolutionSolutionSolutionSolutionSolution DrGo = – nF E(cell)o n in the above equation is 2, F = 96487 C mol–1 and E o( cell ) = 1.1 V Therefore, DrGo = – 2 × 1.1V × 96487 C mol–1 = – 21227 J mol–1 = – 212.27 kJ mol–1 Chemistry 40 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. 2.5 Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+ (0.002 M) ® Ni2+ (0.160 M) + 2Ag(s) Given that Ecello = 1.05 V 2.6 The cell in which the following reaction occurs: E o = 0.236 V at 298 K. 2Fe 3 + ( aq ) + 2I − ( aq ) → 2Fe 2 + ( aq ) + I 2 ( s ) has cell Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. 2.42.42.42.42.4 ConductanceConductanceConductanceConductanceConductance It is necessary to define a few terms before we consider the subject of ofofofofof ElectrolyticElectrolyticElectrolyticElectrolyticElectrolytic conductance of electricity through electrolytic solutions. The electrical resistance is represented by the symbol ‘R’ and it is measured in ohm (W) SolutionsSolutionsSolutionsSolutionsSolutions which in terms of SI base units is equal to (kg m2)/(S3 A2). It can be measured with the help of a Wheatstone bridge with which you are familiar from your study of physics. The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross section, A. That is, l l R µ or R = r (2.17) A A The constant of proportionality, r (Greek, rho), is called resistivity (specific resistance). Its SI units are ohm metre (W m) and quite often its submultiple, ohm centimetre (W cm) is also used. IUPAC recommends the use of the term resistivity over specific resistance and hence in the rest of the book we shall use the term resistivity. Physically, the resistivity for a substance is its resistance when it is one metre long and its area of cross section is one m2. It can be seen that: 1 W m = 100 W cm or 1 W cm = 0.01 W m The inverse of resistance, R, is called conductance, G, and we have the relation: 1 A A G = = = κ (2.18) R ρ l l The SI unit of conductance is siemens, represented by the symbol ‘S’ and is equal to ohm–1 (also known as mho) or W–1. The inverse of resistivity, called conductivity (specific conductance) is represented by the symbol, k (Greek, kappa). IUPAC has recommended the use of term conductivity over specific conductance and hence we shall use the term conductivity in the rest of the book. The SI units of conductivity are S m–1 but quite often, k is expressed in S cm–1. Conductivity of a material in S m–1 is its conductance when it is 1 m long and its area of cross section is 1 m2. It may be noted that 1 S cm–1 = 100 S m–1. 41 Electrochemistry Reprint 2025-26 Table 2.2: The values of Conductivity of some Selected Materials at 298.15 K Material Conductivity/ Material Conductivity/ S m–1 S m–1 Conductors Aqueous Solutions Sodium 2.1×103 Pure water 3.5×10–5 Copper 5.9×103 0.1 M HCl 3.91 Silver 6.2×103 0.01M KCl 0.14 Gold 4.5×103 0.01M NaCl 0.12 Iron 1.0×103 0.1 M HAc 0.047 Graphite 1.2×10 0.01M HAc 0.016 Insulators Semiconductors Glass 1.0×10–16 CuO 1×10–7 Teflon 1.0×10–18 Si 1.5×10–2 Ge 2.0 It can be seen from Table 2.2 that the magnitude of conductivity varies a great deal and depends on the nature of the material. It also depends on the temperature and pressure at which the measurements are made. Materials are classified into conductors, insulators and semiconductors depending on the magnitude of their conductivity. Metals and their alloys have very large conductivity and are known as conductors. Certain non-metals like carbon-black, graphite and some organic polymers* are also electronically conducting. Substances like glass, ceramics, etc., having very low conductivity are known as insulators. Substances like silicon, doped silicon and gallium arsenide having conductivity between conductors and insulators are called semiconductors and are important electronic materials. Certain materials called superconductors by definition have zero resistivity or infinite conductivity. Earlier, only metals and their alloys at very low temperatures (0 to 15 K) were known to behave as superconductors, but nowadays a number of ceramic materials and mixed oxides are also known to show superconductivity at temperatures as high as 150 K. Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on (i) the nature and structure of the metal (ii) the number of valence electrons per atom (iii) temperature (it decreases with increase of temperature). * Electronically conducting polymers – In 1977 MacDiarmid, Heeger and Shirakawa discovered that acetylene gas can be polymerised to produce a polymer, polyacetylene when exposed to vapours of iodine acquires metallic lustre and conductivity. Since then several organic conducting polymers have been made such as polyaniline, polypyrrole and polythiophene. These organic polymers which have properties like metals, being composed wholly of elements like carbon, hydrogen and occasionally nitrogen, oxygen or sulphur, are much lighter than normal metals and can be used for making light-weight batteries. Besides, they have the mechanical properties of polymers such as flexibility so that one can make electronic devices such as transistors that can bend like a sheet of plastic. For the discovery of conducting polymers, MacDiarmid, Heeger and Shirakawa were awarded the Nobel Prize in Chemistry for the year 2000. Chemistry 42 Reprint 2025-26 As the electrons enter at one end and go out through the other end, the composition of the metallic conductor remains unchanged. The mechanism of conductance through semiconductors is more complex. We already know that even very pure water has small amounts of hydrogen and hydroxyl ions (~10–7M) which lend it very low conductivity (3.5 × 10–5 S m–1). When electrolytes are dissolved in water, they furnish their own ions in the solution hence its conductivity also increases. The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The conductivity of electrolytic (ionic) solutions depends on: (i) the nature of the electrolyte added (ii) size of the ions produced and their solvation (iii) the nature of the solvent and its viscosity (iv) concentration of the electrolyte (v) temperature (it increases with the increase of temperature). Passage of direct current through ionic solution over a prolonged period can lead to change in its composition due to electrochemical reactions (Section 2.4.1). 2.4.1 Measurement We know that accurate measurement of an unknown resistance can be of the performed on a Wheatstone bridge. However, for measuring the resistance Conductivity of an ionic solution we face two problems. Firstly, passing direct current of Ionic (DC) changes the composition of the solution. Secondly, a solution cannot Solutions be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. It is available in several designs and two simple ones are shown in Fig. 2.4. Connecting Connecting wires wires Platinized Pt Fig. 2.4 electrodes Two different types of conductivity cells. Platinized Pt electrode Platinized Pt electrode Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross section equal to ‘A’ and are separated by distance ‘l’. Therefore, solution confined between these electrodes is a column of length l and area of cross section A. The resistance of such a column of solution is then given by the equation: l l R = r = (2.17) A A 43 Electrochemistry Reprint 2025-26 The quantity l/A is called cell constant denoted by the symbol, G*. It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated if we know l and A. Measurement of l and A is not only inconvenient but also unreliable. The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations (Table 2.3) and at different temperatures. The cell constant, G*, is then given by the equation: l G* = = R k (2.18) A Table 2.3: Conductivity and Molar conductivity of KCl solutions at 298.15K Concentration/Molarity Conductivity Molar Conductivity mol L–1 mol m–3 S cm–1 S m–1 S cm2mol–1 S m2 mol–1 1.000 1000 0.1113 11.13 111.3 111.3×10–4 0.100 100.0 0.0129 1.29 129.0 129.0×10–4 0.010 10.00 0.00141 0.141 141.0 141.0×10–4 Once the cell constant is determined, we can use it for measuring the resistance or conductivity of any solution. The set up for the measurement of the resistance is shown in Fig. 2.5. It consists of two resistances R3 and R4, a variable resistance R1 and the conductivity cell having the unknown resistance R2. The Wheatstone bridge is fed by an oscillator O (a source of a.c. power in the audio frequency range 550 to 5000 cycles per second). P is a suitable detector (a headphone or other electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions: Fig. 2.5: Arrangement for measurement of R 1 R 4 resistance of a solution of an Unknown resistance R2 = (2.19) R 3 electrolyte. These days, inexpensive conductivity meters are available which can directly read the conductance or resistance of the solution in the conductivity cell. Once the cell constant and the resistance of the solution in the cell is determined, the conductivity of the solution is given by the equation: cell constant G*   (2.20) R R The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the Chemistry 44 Reprint 2025-26 ions in which they dissociate, the concentration of ions or ease with which the ions move under a potential gradient. It, therefore, becomes necessary to define a physically more meaningful quantity called molar conductivity denoted by the symbol Lm (Greek, lambda). It is related to the conductivity of the solution by the equation:  Molar conductivity = Lm = (2.21) c In the above equation, if k is expressed in S m–1 and the concentration, c in mol m–3 then the units of Lm are in S m2 mol–1. It may be noted that: 1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence  (S cm  1 ) Lm(S cm2 mol–1) =  3 1 1000 L m × molarity (mol L ) If we use S cm–1 as the units for k and mol cm–3, the units of concentration, then the units for Lm are S cm2 mol–1. It can be calculated by using the equation:  (S cm 1 ) × 1000 (cm 3 /L) Lm (S cm2 mol–1) = molarity (mol/L) Both type of units are used in literature and are related to each other by the equations: 1 S m2mol–1 = 104 S cm2mol–1 or 1 S cm2mol–1 = 10–4 S m2mol–1. Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is ExampleExampleExampleExampleExample 2.42.42.42.42.4 100 W . If the resistance of the same cell when filled with 0.02 mol L–1 KCl solution is 520 W , calculate the conductivity and molar conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 S/m. SolutionSolutionSolutionSolutionSolution The cell constant is given by the equation: Cell constant = G* = conductivity × resistance = 1.29 S/m × 100 W = 129 m–1 = 1.29 cm–1 Conductivity of 0.02 mol L–1 KCl solution = cell constant / resistance G * 129 m –1 = = = 0.248 S m–1 R 520  Concentration = 0.02 mol L–1 = 1000 × 0.02 mol m–3 = 20 mol m–3  Molar conductivity = m  c 248 × 10 –3 S m –1 = –3 = 124 × 10–4 S m2mol–1 20 mol m 1.29 cm –1 Alternatively, k = = 0.248 × 10–2 S cm–1 520  45 Electrochemistry Reprint 2025-26 and Lm = k × 1000 cm3 L–1 molarity–1 0.248×10 –2 S cm –1 ×1000 cm 3 L–1 = –1 0.02 mol L = 124 S cm2 mol–1 ExampleExampleExampleExampleExample 2.52.52.52.52.5 The electrical resistance of a column of 0.05 mol L–1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. SolutionSolutionSolutionSolutionSolution A = p r2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2 l = 50 cm = 0.5 m  l RA 5.55  10 3  0.785cm 2 R = or    = 87.135 W cm A l 50cm 1  1  Conductivity =  = =   S cm–1   87.135  = 0.01148 S cm–1  × 1000 Molar conductivity, m = cm3 L–1 c 0.01148 S cm –1 ×1000 cm 3 L–1 = –1 0.05 mol L = 229.6 S cm2 mol–1 If we want to calculate the values of different quantities in terms of ‘m’ instead of ‘cm’, RA = l 5.55 × 10 3  × 0.785×10 –4 m 2 = = 87.135 ×10–2 W m 0.5 m 1 100  =  m = 1.148 S m–1 = 87.135  1.148 S m –1 and m = = –3 = 229.6 × 10–4 S m2 mol–1. c 50 mol m 2.4.2 Variation of Both conductivity and molar conductivity change with the Conductivity concentration of the electrolyte. Conductivity always decreases with and Molar decrease in concentration both, for weak and strong electrolytes. Conductivity This can be explained by the fact that the number of ions per unit with volume that carry the current in a solution decreases on dilution. Concentration The conductivity of a solution at any given concentration is the conductance of one unit volume of solution kept between two Chemistry 46 Reprint 2025-26 platinum electrodes with unit area of cross section and at a distance of unit length. This is clear from the equation: A G = =  (both A and l are unity in their appropriate units in l m or cm) Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore, κA Λm = =κ l Since l = 1 and A = V ( volume containing 1 mole of electrolyte) Lm = k V (2.22) Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Lm can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by theFig. 2.6: Molar conductivity versus c½ for acetic acid (weak electrolyte) and potassium symbol L°m . The variation in Lm with chloride (strong electrolyte) in aqueous concentration is different (Fig. 2.6) for solutions. strong and weak electrolytes. Strong Electrolytes For strong electrolytes, Lm increases slowly with dilution and can be represented by the equation: Lm = L°m – A c ½ (2.23) It can be seen that if we plot (Fig. 2.6) Lm against c1/2, we obtain a straight line with intercept equal to L°m and slope equal to ‘–A’. The value of the constant ‘A’ for a given solvent and temperature depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in the solution. Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2 electrolytes respectively. All electrolytes of a particular type have the same value for ‘A’. 47 Electrochemistry Reprint 2025-26 ExampleExampleExampleExampleExample 2.62.62.62.62.6 The molar conductivity of KCl solutions at different concentrations at 298 K are given below: c/mol L–1 Lm/S cm2 mol–1 0.000198 148.61 0.000309 148.29 0.000521 147.81 0.000989 147.09 Show that a plot between Lm and c1/2 is a straight line. Determine the values of L°m and A for KCl. SolutionSolutionSolutionSolutionSolution Taking the square root of concentration we obtain: c1/2/(mol L–1 )1/2 Lm/S cm2mol–1 0.01407 148.61 0.01758 148.29 0.02283 147.81 0.03145 147.09 A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig. 3.7). It can be seen that it is nearly a straight line. From the intercept (c1/2 = 0), we find that L°m = 150.0 S cm2 mol–1 and A = – slope = 87.46 S cm2 mol–1/(mol/L–1)1/2. Fig. 2.7: Variation of Lm against c½. Chemistry 48 Reprint 2025-26 Kohlrausch examined L°m values for a number of strong electrolytes and observed certain regularities. He noted that the difference in L°m of the electrolytes NaX and KX for any X is nearly constant. For example at 298 K: m L°m (KCl) – L°m (NaCl) = L°m (KBr) – L° (NaBr) = L°m (KI) – L°m (NaI) ≃ 23.4 S cm2 mol–1 and similarly it was found that L°m (NaBr)– L°m (NaCl) = L°m (KBr) – L°m (KCl) ≃ 1.8 S cm2 mol–1 On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, – are limiting molar conductivity of the sodium and chlorideif l°Na+ and l°Cl ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation: l° l° L°m – (2.24) (NaCl) = Na+ + Cl In general, if an electrolyte on dissociation gives n+ cations and n– anions then its limiting molar conductivity is given by: L°m = n+ l°+ + n– l°– (2.25) Here, l°+ and l°– are the limiting molar conductivities of the cation and anion respectively. The values of l° for some cations and anions at 298 K are given in Table 2.4. Table 2.4: Limiting Molar Conductivity for some Ions in Water at 298 K Ion l0/(S cm2mol–1) Ion l 0/(S cm2 mol–1) H+ 349.6 OH– 199.1 Na+ 50.1 Cl– 76.3 K+ 73.5 Br– 78.1 Ca2+ 119.0 CH3COO– 40.9 2 Mg2+ 106.0 SO4 160.0 Weak Electrolytes Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations and hence for such electrolytes, the change in Lm with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains 1 mol of electrolyte. In such cases Lm increases steeply (Fig. 2.6) on dilution, especially near lower concentrations. Therefore, L°m cannot be obtained by extrapolation of Lm to zero concentration. At infinite dilution (i.e., concentration c ® zero) electrolyte dissociates completely (a =1), but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately. Therefore, L°m for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions (Example 2.8). At any concentration c, if a is the degree of dissociation 49 Electrochemistry Reprint 2025-26 then it can be approximated to the ratio of molar conductivity Lm at the concentration c to limiting molar conductivity, L0m . Thus we have: m  = ° (2.26) m But we know that for a weak electrolyte like acetic acid (Class XI, Unit 7), c  2 cm2 c m2 K = = =   a 1   m m  m   m  (2.27) m 2 1     m  Applications of Kohlrausch law Using Kohlrausch law of independent migration of ions, it is possible to calculate L0m for any electrolyte from the lo of individual ions. Moreover, for weak electrolytes like acetic acid it is possible to determine the value of its dissociation constant once we know the L0m and Lm at a given concentration c. ExampleExampleExampleExampleExample 2.72.72.72.72.7 Calculate L0m for CaCl2 and MgSO4 from the data given in Table 3.4. SolutionSolutionSolutionSolutionSolution We know from Kohlrausch law that – = 119.0 S cm2 mol–1 + 2(76.3) S cm2 mol–1 m  CaCl 2  = Ca 2+  2 Cl = (119.0 + 152.6) S cm2 mol–1 = 271.6 S cm2 mol–1 2+  m  MgSO 4  = Mg  SO 2–4 = 106.0 S cm2 mol–1 + 160.0 S cm2 mol–1 = 266 S cm2 mol–1 . ExampleExampleExampleExampleExample 2.82.82.82.82.8 L0m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate L0 for HAc.       +  Ac – H + Cl – Ac – Na + Cl – Na + SolutionSolutionSolutionSolutionSolution m  HAc  = H = m  HCl   m  NaAc   m  NaCl  = (425.9 + 91.0 – 126.4 ) S cm2 mol –1 = 390.5 S cm2 mol–1 . ExampleExampleExampleExampleExample 2.92.92.92.92.9 The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate its dissociation constant if L0m for acetic acid is 390.5 S cm2 mol–1.  4 . 95 10  5 Scm  1 1000cm 3 SolutionSolutionSolutionSolutionSolution m =  1  = 48.15 S cm3 mol–1 c 0 . 001028 mol L L m 48.15 Scm 2 mol 1 a =   2  1 = 0.1233 m 390.5 Scm mol c2 0 .001028molL–1  (0 .1233) 2 k = = 1.78 × 10–5 mol L–1  1   1  0 .1233 Chemistry 50 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.7 Why does the conductivity of a solution decrease with dilution? 2.8 Suggest a way to determine the L°m value of water. 2.9 The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given l0(H+) = 349.6 S cm2 mol–1 and l0 (HCOO–) = 54.6 S cm2 mol–1. 2.52.52.52.52.5 ElectrolyticElectrolyticElectrolyticElectrolyticElectrolytic In an electrolytic cell external source of voltage is used to bring about a chemical reaction. The electrochemical processes are of great importance CellsCellsCellsCellsCells andandandandand in the laboratory and the chemical industry. One of the simplest electrolytic ElectrolysisElectrolysisElectrolysisElectrolysisElectrolysis cell consists of two copper strips dipping in an aqueous solution of copper sulphate. If a DC voltage is applied to the two electrodes, then Cu 2+ ions discharge at the cathode (negatively charged) and the following reaction takes place: Cu2+(aq) + 2e– ® Cu (s) (2.28) Copper metal is deposited on the cathode. At the anode, copper is converted into Cu2+ ions by the reaction: Cu(s) ® Cu2+(s) + 2e– (2.29) Thus copper is dissolved (oxidised) at anode and deposited (reduced) at cathode. This is the basis for an industrial process in which impure copper is converted into copper of high purity. The impure copper is made an anode that dissolves on passing current and pure copper is deposited at the cathode. Many metals like Na, Mg, Al, etc. are produced on large scale by electrochemical reduction of their respective cations where no suitable chemical reducing agents are available for this purpose. Sodium and magnesium metals are produced by the electrolysis of their fused chlorides and aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. Quantitative Aspects of Electrolysis Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis: (i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). (ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation). 51 Electrochemistry Reprint 2025-26 There were no constant current sources available during Faraday’s times. The general practice was to put a coulometer (a standard electrolytic cell) for determining the quantity of electricity passed from the amount of metal (generally silver or copper) deposited or consumed. However, coulometers are now obsolete and we now have constant current (I) sources available and the quantity of electricity Q, passed is given by Q = It Q is in coloumbs when I is in ampere and t is in second. The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction: Ag +(aq) + e– ® Ag(s) (2.30) One mole of the electron is required for the reduction of one mole of silver ions. We know that charge on one electron is equal to 1.6021 × 10–19C. Therefore, the charge on one mole of electrons is equal to: NA × 1.6021 × 10–19 C = 6.02 × 1023 mol–1 × 1.6021 × 10–19 C = 96487 C mol–1 This quantity of electricity is called Faraday and is represented by the symbol F. For approximate calculations we use 1F ≃ 96500 C mol–1. For the electrode reactions: Mg2+(l) + 2e– ¾® Mg(s) (2.31) Al3+(l) + 3e– ¾® Al(s) (2.32) It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively. The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds. In commercial production of metals, current as high as 50,000 amperes are used that amounts to about 0.518 F per second. ExampleExampleExampleExampleExample 2.102.102.102.102.10 A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? SolutionSolutionSolutionSolutionSolution t = 600 s charge = current × time = 1.5 A × 600 s = 900 C According to the reaction: Cu2+(aq) + 2e– = Cu(s) We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu. For 900 C, the mass of Cu deposited = (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0.2938 g. 2.5.1 Products of Products of electrolysis depend on the nature of material being Electrolysis electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert Chemistry 52 Reprint 2025-26 electrodes.The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur. For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl2 gas. Here we have only one cation (Na+) which is reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is oxidised at the anode (Cl– ® ½Cl2 + e– ). During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2. In this case besides Na+ and Cl– ions we also have H+ and OH– ions along with the solvent molecules, H2O. At the cathode there is competition between the following reduction reactions: Na+ (aq) + e– ® Na (s) E (ocell ) = – 2.71 V H+ (aq) + e– ® ½ H2 (g) E (ocell ) = 0.00 V The reaction with higher value of Eo is preferred and therefore, the reaction at the cathode during electrolysis is: H+ (aq) + e– ® ½ H2 (g) (2.33) but H+ (aq) is produced by the dissociation of H2O, i.e., H2O (l ) ® H+ (aq) + OH– (aq) (2.34) Therefore, the net reaction at the cathode may be written as the sum of (2.33) and (2.34) and we have H2O (l ) + e– ® ½H2(g) + OH– (2.35) At the anode the following oxidation reactions are possible: Cl– (aq) ® ½ Cl2 (g) + e– E (ocell ) = 1.36 V (2.36) 2H2O (l ) ® O2 (g) + 4H+(aq) + 4e– E (ocell ) = 1.23 V (2.37) The reaction at anode with lower value of E o is preferred and therefore, water should get oxidised in preference to Cl– (aq). However, on account of overpotential of oxygen, reaction (2.36) is preferred. Thus, the net reactions may be summarised as: NaCl (aq) H 2 O → Na+ (aq) + Cl– (aq) Cathode: H2O(l ) + e– ® ½ H2(g) + OH– (aq) Anode: Cl– (aq) ® ½ Cl2(g) + e– Net reaction: NaCl(aq) + H2O(l) ® Na+(aq) + OH–(aq) + ½H2(g) + ½Cl2(g) The standard electrode potentials are replaced by electrode potentials given by Nernst equation (Eq. 2.8) to take into account the concentration effects. During the electrolysis of sulphuric acid, the following processes are possible at the anode: 2H2O(l) ® O2(g) + 4H+(aq) + 4e– E (ocell ) = +1.23 V (2.38) 53 Electrochemistry Reprint 2025-26 2SO42– (aq) ® S2O8 2– (aq) + 2e– E (ocell ) = 1.96 V (2.39) For dilute sulphuric acid, reaction (2.38) is preferred but at higher concentrations of H2SO4, reaction (2.39) is preferred. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? 2.11 Suggest a list of metals that are extracted electrolytically. 2.12 Consider the reaction: Cr2O7 2– + 14H+ + 6e– ® 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O7 2–? 2.62.62.62.62.6 BatteriesBatteriesBatteriesBatteriesBatteries Any battery (actually it may have one or more than one cell connected in series) or cell that we use as a source of electrical energy is basically a galvanic cell where the chemical energy of the redox reaction is converted into electrical energy. However, for a battery to be of practical use it should be reasonably light, compact and its voltage should not vary appreciably during its use. There are mainly two types of batteries. 2.6.1 Primary In the primary batteries, the reaction occurs only once and after use Batteries over a period of time battery becomes dead and cannot be reused again. The most familiar example of this type is the dry cell (known as Leclanche cell after its discoverer) which is used commonly in our transistors and clocks. The cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon (Fig.2.8). The space between the electrodes is filled by a moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2). The electrode reactions are complex, but they can be written approximately as follows : Anode: Zn(s) ¾® Zn2+ + 2e– Cathode: MnO2+ NH4 ++ e–¾® MnO(OH) + NH3 In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state. Ammonia produced in the reaction forms a complex with Zn2+ to give [Zn (NH3)4]2+. The cell has a potential of nearly 1.5 V. Mercury cell, (Fig. 2.9) suitable for low current devices like hearing aids, watches, etc. consists of zinc – mercury amalgam as anode and a paste of HgO and carbon as the Fig. 2.8: A commercial dry cell cathode. The electrolyte is a paste of KOH and ZnO. The consists of a graphite electrode reactions for the cell are given below: (carbon) cathode in a Anode: Zn(Hg) + 2OH– ¾® ZnO(s) + H2O + 2e– zinc container; the latter Cathode: HgO + H2O + 2e– ¾® Hg(l) + 2OH– acts as the anode. Chemistry 54 Reprint 2025-26 The overall reaction is represented by Zn(Hg) + HgO(s) ¾® ZnO(s) + Hg(l) The cell potential is approximately 1.35 V and remains constant during its Fig. 2.9 life as the overall reaction does not Commonly used involve any ion in solution whose mercury cell. The concentration can change during its life reducing agent is time. zinc and the oxidising agent is mercury (II) oxide. 2.6.2 Secondary A secondary cell after use can be recharged by passing current Batteries through it in the opposite direction so that it can be used again. A good secondary cell can undergo a large number of discharging and charging cycles. The most important secondary cell is the lead storage battery (Fig. 2.10) commonly used in automobiles and invertors. It consists of a lead anode and a grid of lead packed with lead dioxide (PbO2 ) as cathode. A 38% solution of sulphuric acid is used as an electrolyte. The cell reactions when the battery is in use are given below: Anode: Pb(s) + SO42–(aq) ® PbSO4(s) + 2e– Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e– ® PbSO4 (s) + 2H2O (l) i.e., overall cell reaction consisting of cathode and anode reactions is: Pb(s) + PbO2(s) + 2H2SO4(aq) ® 2PbSO4(s) + 2H2O(l) On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively. Fig. 2.10: The Lead storage battery. 55 Electrochemistry Reprint 2025-26 Another important secondary cell is the nickel-cadmium cell (Fig. 2.11) which has longer life than the lead storage cell but Fig. 2.11 more expensive to manufacture. A rechargeable We shall not go into details of nickel-cadmium cell working of the cell and the Positive plate in a jelly roll electrode reactions during arrangement and Separator charging and discharging. separated by a layer Negative plate The overall reaction during soaked in moist discharge is: sodium or potassium hydroxide. Cd (s) + 2Ni(OH)3 (s) ® CdO (s) + 2Ni(OH)2 (s) + H2O (l ) 2.72.72.72.72.7 FuelFuelFuelFuelFuel CellsCellsCellsCellsCells Production of electricity by thermal plants is not a very efficient method and is a major source of pollution. In such plants, the chemical energy (heat of combustion) of fossil fuels (coal, gas or oil) is first used for converting water into high pressure steam. This is then used to run a turbine to produce electricity. We know that a galvanic cell directly converts chemical energy into electricity and is highly efficient. It is now possible to make such cells in which reactants are fed continuously to the electrodes and products are removed continuously from the electrolyte compartment. Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells. One of the most successful fuel cells uses the reaction of hydrogen with oxygen to form water (Fig. 2.12). The cell was used for providing electrical power in the Apollo space programme. The water vapours produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode Fig. 2.12: Fuel cell using H2 and O2 produces electricity. reactions. The electrode reactions are given below: Cathode: O2(g) + 2H2O(l) + 4e–¾® 4OH–(aq) Anode: 2H2 (g) + 4OH–(aq) ¾® 4H2O(l) + 4e– Overall reaction being: 2H2(g) + O2(g) ¾® 2H2O(l ) The cell runs continuously as long as the reactants are supplied. Fuel cells produce electricity with an efficiency of about 70 % compared Chemistry 56 Reprint 2025-26 to thermal plants whose efficiency is about 40%. There has been tremendous progress in the development of new electrode materials, better catalysts and electrolytes for increasing the efficiency of fuel cells. These have been used in automobiles on an experimental basis. Fuel cells are pollution free and in view of their future importance, a variety of fuel cells have been fabricated and tried. 2.82.82.82.82.8 CorrosionCorrosionCorrosionCorrosionCorrosion Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion. It causes enormous damage to buildings, bridges, ships and to all objects made of metals especially that of iron. We lose crores of rupees every year on account of corrosion. In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex but it may be considered Oxidation: Fe (s)® Fe2+ (aq) +2e– essentially as an electrochemical Reduction: O2 (g) + 4H+(aq) +4e– ® 2H2O(l) phenomenon. At a particular spot Atomospheric (Fig. 2.13) of an object made of iron,oxidation: 2Fe2+(aq) + 2H2O(l) + ½O2(g) ® Fe2O3(s) + 4H+(aq) oxidation takes place and that spot Fig. 2.13: Corrosion of iron in atmosphere behaves as anode and we can write the reaction E o Anode: 2 Fe (s) ¾® 2 Fe2+ + 4 e– (Fe 2+ /Fe) = – 0.44 V Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+ (which is believed to be available from H2CO3 formed due to dissolution of carbon dioxide from air into water. Hydrogen ion in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction E o =1.23 V Cathode: O2(g) + 4 H+(aq) + 4 e– ¾® 2 H2O (l) H + | O 2 | H 2 O The overall reaction being: 2Fe(s) + O2(g) + 4H+(aq) ¾® 2Fe2 +(aq) + 2 H2O (l) E o(cell) =1.67 V The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3. x H2O) and with further production of hydrogen ions. Prevention of corrosion is of prime importance. It not only saves money but also helps in preventing accidents such as a bridge collapse or failure of a key component due to corrosion. One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done by covering the surface with paint or by some chemicals (e.g. bisphenol). Another simple method is to cover the surface by other metals (Sn, Zn, etc.) that are inert or react to save the object. An electrochemical method is to provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which corrodes itself but saves the object. 57 Electrochemistry Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. 2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells. 2.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell. TheTheTheTheThe HydrogenHydrogenHydrogenHydrogenHydrogen EconomyEconomyEconomyEconomyEconomy At present the main source of energy that is driving our economy is fossil fuels such as coal, oil and gas. As more people on the planet aspire to improve their standard of living, their energy requirement will increase. In fact, the per capita consumption of energy used is a measure of development. Of course, it is assumed that energy is used for productive purpose and not merely wasted. We are already aware that carbon dioxide produced by the combustion of fossil fuels is resulting in the ‘Greenhouse Effect’. This is leading to a rise in the temperature of the Earth’s surface, causing polar ice to melt and ocean levels to rise. This will flood low-lying areas along the coast and some island nations such as Maldives face total submergence. In order to avoid such a catastrope, we need to limit our use of carbonaceous fuels. Hydrogen provides an ideal alternative as its combustion results in water only. Hydrogen production must come from splitting water using solar energy. Therefore, hydrogen can be used as a renewable and non polluting source of energy. This is the vision of the Hydrogen Economy. Both the production of hydrogen by electrolysis of water and hydrogen combustion in a fuel cell will be important in the future. And both these technologies are based on electrochemical principles. SummarySummarySummarySummarySummary An electrochemical cell consists of two metallic electrodes dipping in electrolytic solution(s). Thus an important component of the electrochemical cell is the ionic conductor or electrolyte. Electrochemical cells are of two types. In galvanic cell, the chemical energy of a spontaneous redox reaction is converted into electrical work, whereas in an electrolytic cell, electrical energy is used to carry out a non- spontaneous redox reaction. The standard electrode potential for any electrode dipping in an appropriate solution is defined with respect to standard electrode potential of hydrogen electrode taken as zero. The standard potential of the cell can be obtained by taking the difference of the standard potentials of cathode and anode ( E (ocell ) = Eocathode – Eoanode). The standard potential of the cells are related to standard Gibbs energy (DrGo = –nF E (ocell ) ) and equilibrium constant (DrGo = – RT ln K) of the reaction taking place in the cell. Concentration dependence of the potentials of the electrodes and the cells are given by Nernst equation. The conductivity, k, of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent and temperature. Molar conductivity, Lm, is defined by = k/c where c is the concentration. Conductivity decreases but molar conductivity increases with decrease in concentration. It increases slowly with decrease in concentration for strong electrolytes while the increase is very steep for weak electrolytes in very dilute solutions. Kohlrausch found that molar conductivity at infinite dilution, for an electrolyte is sum of the contribution of the Chemistry 58 Reprint 2025-26 molar conductivity of the ions in which it dissociates. It is known as law of independent migration of ions and has many applications. Ions conduct electricity through the solution but oxidation and reduction of the ions take place at the electrodes in an electrochemical cell. Batteries and fuel cells are very useful forms of galvanic cell. Corrosion of metals is essentially an electrochemical phenomenon. Electrochemical principles are relevant to the Hydrogen Economy. ExercisesExercisesExercisesExercisesExercises

1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.

1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

1.8 Mole concept and Molar Masses molecules Atoms and molecules are extremely small 1 mol of sodium chloride = 6.022 ×1023 formula in size and their numbers in even a small units of sodium chloride amount of any substance is really very large. Having defined the mole, it is easier toTo handle such large numbers, a unit of convenient magnitude is required. know the mass of one mole of a substance Just as we denote one dozen for 12 items, or the constituent entities. The mass of one score for 20 items, gross for 144 items, we mole of a substance in grams is called its use the idea of mole to count entities at the molar mass. The molar mass in grams is microscopic level (i.e., atoms, molecules, numerically equal to atomic/molecular/ particles, electrons, ions, etc). formula mass in u. In SI system, mole (symbol, mol) was Molar mass of water = 18.02 g mol–1introduced as seventh base quantity for the amount of a substance. Molar mass of sodium chloride = 58.5 g mol–1 The mole, symbol mol, is the SI unit of 1.9 Percentage Compositionamount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. So far, we were dealing with the number of This number is the fixed numerical value of entities present in a given sample. But many the Avogadro constant, NA, when expressed a time, information regarding the percentage in the unit mol–1 and is called the Avogadro of a particular element present in a compound number. The amount of substance, symbol is required. Suppose, an unknown or new n, of a system is a measure of the number of compound is given to you, the first questionspecified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to: 12 g / mol 12 C 1 .992648  10 23 g /12 Catom  6 .0221367  1023 atoms/mol Fig. 1.11 One mole of various substances Reprint 2025-26 Some Basic Concepts of Chemistry 19 you would ask is: what is its formula or what 1.9.1 Empirical Formula for Molecular are its constituents and in what ratio are they Formula present in the given compound? For known An empirical formula represents the simplest compounds also, such information provides a whole number ratio of various atoms present check whether the given sample contains the in a compound, whereas, the molecular same percentage of elements as present in a formula shows the exact number of different pure sample. In other words, one can check types of atoms present in a molecule of a the purity of a given sample by analysing this compound. data. If the mass per cent of various elements Let us understand it by taking the example present in a compound is known, its empirical of water (H2O). Since water contains hydrogen formula can be determined. Molecular formula and oxygen, the percentage composition of can further be obtained if the molar mass is both these elements can be calculated as known. The following example illustrates follows: this sequence. Mass % of an element = mass of that element in the compound × 100 molar mass of the compound Problem 1.2 A compound contains 4.07% hydrogen,Molar mass of water = 18.02 g 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are itsMass % of hydrogen = empirical and molecular formulas? = 11.18 Solution 16.00Mass % of oxygen = × 100 Step 1. Conversion of mass per cent 18.02 to grams = 88.79 Since we are having mass per cent, it is Let us take one more example. What is the convenient to use 100 g of the compound percentage of carbon, hydrogen and oxygen as the starting material. Thus, in the in ethanol? 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon andMolecular formula of ethanol is: C2H5OH 71.65g chlorine are present.Molar mass of ethanol is: Step 2. Convert into number moles of(2×12.01 + 6×1.008 + 16.00) g = 46.068 g each element Mass per cent of carbon Divide the masses obtained above by 24.02g = × 100 = 52.14% respective atomic masses of various 46.068g elements. This gives the number of Mass per cent of hydrogen moles of constituent elements in the compound 6.048g = × 100 = 13.13% 46.068g 4.07 g Moles of hydrogen = = 4.04 1.008gMass per cent of oxygen 16.00g = × 100 = 34.73% 24.27 g Moles of carbon = = 2 .021 46.068g 12 .01 g After understanding the calculation of 71.65g per cent of mass, let us now see what Moles of chlorine = = 2 .021 35 .453 ginformation can be obtained from the per cent composition data. Reprint 2025-26 20 chemistry equation of a given reaction. Let us consider Step 3. Divide each of the mole values the combustion of methane. A balanced obtained above by the smallest number equation for this reaction is as given below: amongst them CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) Since 2.021 is smallest value, division Here, methane and dioxygen are called by it gives a ratio of 2:1:1 for H:C:Cl. reactants and carbon dioxide and water are In case the ratios are not whole numbers, called products. Note that all the reactants then they may be converted into whole and the products are gases in the above number by multiplying by the suitable reaction and this has been indicated by coefficient. letter (g) in the brackets next to its formula. Step 4. Write down the empirical Similarly, in case of solids and liquids, (s) and formula by mentioning the numbers (l) are written respectively. after writing the symbols of respective The coefficients 2 for O2 and H2O are elements called stoichiometric coefficients. Similarly CH2Cl is, thus, the empirical formula the coefficient for CH4 and CO2 is one in each of the above compound. case. They represent the number of molecules (and moles as well) taking part in the reaction Step 5. Writing molecular formula or formed in the reaction. (a) Determine empirical formula mass by Thus, according to the above chemical adding the atomic masses of various reaction, atoms present in the empirical formula. For CH2Cl, empirical formula mass is • One mole of CH4(g) reacts with two moles 12.01 + (2 ×1.008) + 35.453 of O2(g) to give one mole of CO2(g) and = 49.48 g two moles of H2O(g) (b) Divide Molar mass by empirical • One molecule of CH4(g) reacts with formula mass 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g) • 22.7 L of CH4(g) reacts with 45.4 L of O2 = 2 = (n) (g) to give 22.7 L of CO2 (g) and 45.4 L of (c) Multiply empirical formula by n H2O(g) obtained above to get the molecular • 16 g of CH4 (g) reacts with 2×32 g of O2 formula (g) to give 44 g of CO2 (g) and 2×18 g of Empirical formula = CH2Cl, n = 2. Hence H2O (g). molecular formula is C2H4Cl2. From these relationships, the given data can be interconverted as follows: 1.10 Stoichiometry and mass Stoichiometric Calculations The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, Mass = Density element) and metron (meaning, measure). Volume Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the 1.10.1 Limiting Reagent reactants and the products involved in a Many a time, reactions are carried out with chemical reaction. Before understanding how the amounts of reactants that are different to calculate the amounts of reactants required than the amounts as required by a balanced or the products produced in a chemical chemical reaction. In such situations, one reaction, let us study what information reactant is in more amount than the amount is available from the balanced chemical required by balanced chemical reaction. The Reprint 2025-26 Some Basic Concepts of Chemistry 21 reactant which is present in the least amount important to understand as how the amount gets consumed after sometime and after that of substance is expressed when it is present in further reaction does not take place whatever the solution. The concentration of a solution be the amount of the other reactant. Hence, or the amount of substance present in its the reactant, which gets consumed first, given volume can be expressed in any of the limits the amount of product formed and is, following ways. therefore, called the limiting reagent. 1. Mass per cent or weight per cent (w/w %) In performing stoichiometric calculations, 2. Mole fraction this aspect is also to be kept in mind. 3. Molarity 1.10.2 Reactions in Solutions 4. Molality A majority of reactions in the laboratories Let us now study each one of them in are carried out in solutions. Therefore, it is detail. Balancing a chemical equation According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides 4 Fe(s) + 3O2(g) → 2Fe2O3(s) (a) balanced equation 2 Mg(s) + O2(g) → 2MgO(s) (b) balanced equation P4(s) + O2 (g) → P4O10(s) (c) unbalanced equation Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. P4(s) + 5O2(g) → P4O10(s) balanced equation Now, let us take combustion of propane, C3H8. This equation can be balanced in steps. Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products. C3H8(g) + O2(g) → CO2 (g) + H2O(l) unbalanced equation Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. C3H8 (g) + O2 (g) → 3CO2 (g) + H2O (l) Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side. C3H8 (g) +O2 (g) → 3CO2 (g)+4H2O (l) Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. C3H8 (g) +5O2 (g) → 3CO2 (g) + 4H2O (l) Step 5 Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. Reprint 2025-26 22 chemistry Problem 1.3 the limiting reagent in the production of NH3 in this situation. Calculate the amount of water (g) produced by the combustion of 16 g Solution of methane. A balanced equation for the above Solution reaction is written as follows : The balanced equation for the combustion of methane is : CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) Calculation of moles : (i) 16 g of CH4 corresponds to one mole. Number of moles of N2 (ii) From the above equation, 1 mol of 1000 g N 2 1 mol N 2 × 2 × CH4 (g) gives 2 mol of H2O (g). = 50 .0 kg N 1 kg N 2 28 .0 g N 2 2 mol of water (H2O) = 2×(2+16) = 17.86×102 mol = 2×18 = 36 g Number of moles of H2 H 2 O ⇒18g 1000 g H 2 1 mol H 2 = 1 1 mol H2O = 18 g H2O × = 10 .00 kg H 2 × 1mol H 2 O 1 kg H 2 2 .016 g H 2 18g H 2 O = 4.96 × 103 mol Hence, 2 mol H2O× 1mol H 2 O According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the = 2×18 g H2O = 36 g H2O reaction. Hence, for 17.86×102 mol of Problem 1.4 N2, the moles of H2 (g) required would be How many moles of methane are 2 3 mol H 2 g 2  required to produce 22g CO2 (g) after 17 .86  10 mol N 1 mol N 2 g combustion? = 5.36 × 103 mol H2 Solution But we have only 4.96×103 mol H2. According to the chemical equation, Hence, dihydrogen is the limiting CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) reagent in this case. So, NH3(g) would be formed only from that amount of 44g CO2 (g) is obtained from 16 g CH4 (g). available dihydrogen i.e., 4.96 × 103 mol [∴1 mol CO2(g) is obtained from 1 mol Since 3 mol H2(g) gives 2 mol NH3(g) of CH4(g)] Number of moles of CO2 (g) 2 mol NH 3 g 2 g 4.96 ×103 mol H2 (g) × 3 mol H 2 g = 22 g CO2 (g) ×1 molCO 44 gCO 2 g = 3.30 × 103 mol NH3 (g) = 0.5 mol CO2 (g) 3.30 × 103 mol NH3 (g) is obtained. Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol If they are to be converted to grams, it of CH4 (g) would be required to produce is done as follows : 22 g CO2 (g). 1 mol NH3 (g) = 17.0 g NH3 (g) Problem 1.5 .0 g NH 3 g 50.0 kg of N2 (g) and 10.0 kg of H2 (g) 3.30 ×103 mol NH3 (g) ×17 3 g are mixed to produce NH3 (g). Calculate 1 mol NH the amount of NH3 (g) formed. Identify Reprint 2025-26 Some Basic Concepts of Chemistry 23 3. Molarity = 3.30×103×17 g NH3 (g) It is the most widely used unit and is denoted = 56.1×103 g NH3 by M. It is defined as the number of moles of = 56.1 kg NH3 the solute in 1 litre of the solution. Thus, No. of moles of solute 1. Mass per cent Molarity (M) = Volume of solution in litres It is obtained by using the following relation: Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it. 1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, Problem 1.6 we require 0.2 moles of NaOH dissolved in 1 litre solution. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate Hence, for making 0.2M solution from 1M the mass per cent of the solute. solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH Solution and dilute the solution with water to 1 litre. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution then, 0.2 mol is present in 1000 mL × 0 .2 mol solution 1 mol 2. Mole Fraction = 200 mL solutionIt is the ratio of number of moles of a particular component to the total number Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number In fact for such calculations, a general of moles are nA and nB, respectively, then the formula, M1×V1 = M2 × V2 where M and V are molarity and volume, respectively, can be used.mole fractions of A and B are given as: In this case, M1 is equal to 0.2M; V1 = 1000 mL and, M2 = 1.0M; V2 is to be calculated. Substituting the values in the formula: 0.2 M × 1000 mL = 1.0 M × V2 Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration. Reprint 2025-26 24 chemistry Problem 1.7 Problem 1.8 Calculate the molarity of NaOH in the The density of 3 M solution of NaCl is solution prepared by dissolving its 4 g 1.25 g mL–1. Calculate the molality of in enough water to form 250 mL of the the solution. solution. Solution Solution M = 3 mol L–1 Since molarity (M) Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 g mL–1) Mass of water in solution = 1250 –75.5 = 1074.5 g No. of moles of solute Molality = Mass of solvent in kg 3 mol = = 2.79 m Note that molarity of a solution depends 1 .0745 kg upon temperature because volume of a Often in a chemistry laboratory, a solution is temperature dependent. solution of a desired concentration is prepared by diluting a solution of known 4. Molality higher concentration. The solution of higher concentration is also known as It is defined as the number of moles of solute stock solution. Note that the molality present in 1 kg of solvent. It is denoted by m. of a solution does not change with temperature since mass remains No. of moles of solute Thus, Molality (m) = unaffected with temperature. Mass of solvent in kg Summary Chemistry, as we understand it today is not a very old discipline. People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life. The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures. When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English Reprint 2025-26 Some Basic Concepts of Chemistry 25 and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units). Since measurements involve recording of data, which are always associated with a certain amount of u is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers in scientific notation is used. The u figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another. The combination of different atoms is governed by basic laws of chemical combination — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass. The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities. Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality. exerciseS 1.1 Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4 1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 1.4 Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. 1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1. Reprint 2025-26 26 chemistry 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. 1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? 1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 1.9 Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659 1.10 In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. 1.11 What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? 1.12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? 1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. 1.14 What is the SI unit of mass? How is it defined? 1.15 Match the following prefixes with their multiples: Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 1.16 What do you mean by significant figures? 1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample. 1.18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012 1.19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 Reprint 2025-26 Some Basic Concepts of Chemistry 27 (iv) 126,000 (v) 500.0 (vi) 2.0034 1.20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3 1.22 If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns. 1.23 In a reaction A + B2  AB2 Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g)  2NH3 (g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? 1.25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? 1.27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg Reprint 2025-26 28 chemistry 1.28 Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). 1.30 What will be the mass of one 12C atom in g? 1.31 How many significant figures should be present in the answer of the following calculations? 0.02856 × 298.15 × 0.112 (i) (ii) 5 × 5.364 0 .5785 (iii) 0.0125 + 0.7864 + 0.0215 1.32 Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol–1 0.337% 38Ar 37.96272 g mol–1 0.063% 40Ar 39.9624 g mol–1 99.600% 1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide? Reprint 2025-26 Unit 2 structure of atom The rich diversity of chemical behaviour of different Objectives elementsstructure canof atomsbe tracedof theseto theelements.differences in the internal After studying this unit you will be able to • know about the discovery of The existence of atoms has been proposed since the time electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) who their characteristics; were of the view that atoms are the fundamental building • describe Thomson, Rutherford blocks of matter. According to them, the continued and Bohr atomic models; subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ • understand the important features has been derived from the Greek word ‘a-tomio’ which of the quantum mechanical model means ‘uncut-able’ or ‘non-divisible’. These earlier ideas of atom; were mere speculations and there was no way to test • u n d e r s t a n d n a t u r e o f them experimentally. These ideas remained dormant for electromagnetic radiation and a very long time and were revived again by scientists in Planck’s quantum theory; the nineteenth century. • explain the photoelectric effect The atomic theory of matter was first proposed and describe features of atomic on a firm scientific basis by John Dalton, a British spectra; school teacher in 1808. His theory, called Dalton’s • state the de Broglie relation and atomic theory, regarded the atom as the ultimate Heisenberg u able to explain the law of conservation of mass, law of • define an atomic orbital in terms constant composition and law of multiple proportion of quantum numbers; very successfully. However, it failed to explain the results • state aufbau principle, Pauli of many experiments, for example, it was known that exclusion principle and Hund’s substances like glass or ebonite when rubbed with silk rule of maximum multiplicity; and or fur get electrically charged. • write the electronic configurations In this unit we start with the experimental observations of atoms. made by scientists towards the end of nineteenth and beginning of twentieth century. These established that atoms are made of sub-atomic particles, i.e., electrons, protons and neutrons — a concept very different from that of Dalton. Reprint 2025-26 30 chemistry

1.5 A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution? 1.6 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both? 1.7 A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution. 1.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution? 1.9 A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample. 1.10 What role does the molecular interaction play in a solution of alcohol and water? 1.11 Why do gases always tend to be less soluble in liquids as the temperature is raised? 1.12 State Henry’s law and mention some important applications. 1.13 The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas? 1.14 What is meant by positive and negative deviations from Raoult's law and how is the sign of DmixH related to positive and negative deviations from Raoult's law? 1.15 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute? 1.16 Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane? 1.17 The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it. 1.18 Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. 1.19 A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K. 1.20 A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K. 1.21 Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B. Chemistry 28 Reprint 2025-26 1.22 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

1.28 Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

1.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. 29 Solutions Reprint 2025-26

1.23 Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O). 1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

1.30 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

1.32 Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1. 1.33 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

1.4 Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1? 27 Solutions Reprint 2025-26

1.35 Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

8.17 Complete each synthesis by giving missing starting material, reagent or products

8.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO (iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO CHO (v) (vi) PhCOPh

8.2 Name the following compounds according to IUPAC system of nomenclature: (i) CH3CH(CH3)CH2CH2CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CHO-p

8.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then H3O + (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid 8.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal 8.8 How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid 8.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile. 8.10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 8.11 An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved. 8.12 Arrange the following compounds in increasing order of their property as indicated: (i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) 8.13 Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal 8.14 How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde. 8.15 How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to a-Hydroxyphenylacetic acid (ix) Benzoic acid to m- Nitrobenzyl alcohol 8.16 Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation Chemistry 256 Reprint 2025-26

8.3 STRUCTURAL RepresenTATIONS Similarly, CH3CH2CH2CH2CH2CH2CH2CH3 OF organic COMPOUNDs can be further condensed to CH3(CH2)6CH3. 8.3.1 Complete, Condensed and Bond-line For further simplification, organic chemists Structural Formulas use another way of representing the structures, in which only lines are used.Structures of organic compounds are In this bond-line structural representationrepresented in several ways. The Lewis of organic compounds, carbon andstructure or dot structure, dash structure, hydrogen atoms are not shown and thecondensed structure and bond line structural lines representing carbon-carbon bonds areformulas are some of the specific types. The drawn in a zig-zag fashion. The only atomsLewis structures, however, can be simplified specifically written are oxygen, chlorine,by representing the two-electron covalent nitrogen etc. The terminals denote methylbond by a dash (–). Such a structural formula (–CH3) groups (unless indicated otherwise byfocuses on the electrons involved in bond a functional group), while the line junctionsformation. A single dash represents a single denote carbon atoms bonded to appropriatebond, double dash is used for double bond number of hydrogens required to satisfy theand a triple dash represents triple bond. Lone- valency of the carbon atoms. Some of thepairs of electrons on heteroatoms (e.g., oxygen, examples are represented as follows:nitrogen, sulphur, halogens etc.) may or may (i) 3-Methyloctane can be represented innot be shown. Thus, ethane (C2H6), ethene various forms as:(C2H4), ethyne (C2H2) and methanol (CH3OH) can be represented by the following structural (a) CH3CH2CHCH2CH2CH2CH2CH3 formulas. Such structural representations are | called complete structural formulas. CH3 (b) Ethane Ethene (c) Ethyne Methanol These structural formulas can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by (ii) Various ways of representing 2-bromo indicating the number of identical groups butane are: attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus, (a) CH3CHBrCH2CH3 (b)ethane, ethene, ethyne and methanol can be written as: CH3CH3 H2C=CH2 HC≡CH CH3OH (c) Ethane Ethene Ethyne Methanol Reprint 2025-26 organic chemistry – some basic principles and techniques 259 In cyclic compounds, the bond-line formulas may be given as follows: (b) Solution Condensed formula: Cyclopropane (a) HO(CH2)3CH(CH3)CH(CH3)2 (b) HOCH(CN)2 Bond-line formula: (a) Cyclopentane (b) chlorocyclohexane Problem 8.6 Problem 8.4 Expand each of the following bond-line Expand each of the following condensed formulas to show all the atoms including formulas into their complete structural carbon and hydrogen formulas. (a) (a) CH3CH2COCH2CH3 (b) CH3CH=CH(CH2)3CH3 Solution (b) (a) (c) (b) (d) Solution Problem 8.5 For each of the following compounds, write a condensed formula and also their bond-line formula. (a) HOCH2CH2CH2CH(CH3)CH(CH3)CH3 Reprint 2025-26 260 chemistry Molecular Models Molecular models are physical devices that are used for a better visualisation and perception of three-dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. Commonly three types of molecular models are used: (1) Framework model, (2) Ball-and-stick model, and (3) Space filling model. In the framework model only the bonds connecting the atoms of a molecule and not the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms. In the ball-and-stick model, both the atoms and the bonds are shown. Balls represent atoms and the stick denotes a bond. Compounds containing C=C (e.g., ethene) can best be represented by using8.3.2 Three-Dimensional springs in place of sticks. These models are Representation of Organic referred to as ball-and-spring model. The Molecules space-filling model emphasises the relative The three-dimensional (3-D) structure of size of each atom based on its van der Waals organic molecules can be represented on radius. Bonds are not shown in this model. paper by using certain conventions. For It conveys the volume occupied by each atom in the molecule. In addition to these models,example, by using solid ( ) and dashed computer graphics can also be used for( ) wedge formula, the 3-D image of a molecular modelling. molecule from a two-dimensional picture can be perceived. In these formulas the solid-wedge is used to indicate a bond projecting out of the plane of paper, towards the observer. The dashed-wedge is used to depict the bond projecting out of the plane of the paper and away from the observer. Wedges are shown in such a way that the broad end of the wedge is towards the observer. The bonds Framework model Ball and stick model lying in plane of the paper are depicted by using a normal line (—). 3-D representation of methane molecule on paper has been shown in Fig. 8.1 Space filling model Fig. 8.2 Fig. 8.1 Wedge-and-dash representation of CH4 Reprint 2025-26 organic chemistry – some basic principles and techniques 261

4.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions. 4.17 For M2+/M and M3+/M 2+ systems the E o values for some metals are as follows: Cr2+/Cr -0.9V Cr3/Cr2+ -0.4 V Mn 2+/Mn -1.2V Mn3+/Mn2+ +1.5 V Fe2+/Fe -0.4V Fe3+/Fe2+ +0.8 V Use this data to comment upon: (i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal. 4.18 Predict which of the following will be coloured in aqueous solution? Ti 3+, V3+, Cu+, Sc3+, Mn 2+, Fe3+ and Co 2+. Give reasons for each. 4.19 Compare the stability of +2 oxidation state for the elements of the first transition series. 4.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (iii) oxidation state (ii) atomic and ionic sizes and (iv) chemical reactivity. 4.21 How would you account for the following: (i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising. (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d1 configuration is very unstable in ions. 4.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution. 4.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? 4.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? 4.25 Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) The highest oxidation state is exhibited in oxoanions of a metal. 4.26 Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore. 4.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. 4.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104. 4.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. 4.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Chemistry 116 Reprint 2025-26 4.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. 4.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. 4.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity. 4.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. 4.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes. 4.36 Write down the number of 3d electrons in each of the following ions: Ti 2+, V 2+, Cr3+, Mn 2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). 4.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. 4.38 What can be inferred from the magnetic moment values of the following complex species ? Example Magnetic Moment (BM) K4[Mn(CN)6) 2.2 [Fe(H2O)6]2+ 5.3 K2[MnCl4] 5.9 Answers to Some Intext Questions 4.1 Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element. 4.2 In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. 4.3 Manganese (Z = 25), as its atom has the maximum number of unpaired electrons. 4.5 Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d 0, d 5, d 10 are exceptionally stable). 4.6 Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. 4.7 Cr 2+ is stronger reducing agent than Fe 2+ Reason: d 4  d 3 occurs in case of Cr 2+ to Cr 3+ But d 6  d 5 occurs in case of Fe2+ to Fe 3+ In a medium (like water) d 3 is more stable as compared to d 5 (see CFSE) 4.9 Cu + in aqueous solution underoes disproportionation, i.e., 2Cu +(aq) ® Cu 2+(aq) + Cu(s) The E0 value for this is favourable.

4.11 Explain giving reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomisation of the transition metals are high. (iii) The transition metals generally form coloured compounds. (iv) Transition metals and their many compounds act as good catalyst.

4.13 How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

4.5 Valence Bond Theory of the valence bond theory is based on the knowledge of atomic orbitals, electronicAs we know that Lewis approach helps in configurations of elements (Units 2), thewriting the structure of molecules but it overlap criteria of atomic orbitals, thefails to explain the formation of chemical hybridization of atomic orbitals and thebond. It also does not give any reason for the principles of variation and superposition. Adifference in bond dissociation enthalpies and rigorous treatment of the VB theory in termsbond lengths in molecules like H2 (435.8 kJ of these aspects is beyond the scope of this mol-1, 74 pm) and F2 (155 kJ mol-1, 144 pm), book. Therefore, for the sake of convenience,although in both the cases a single covalent valence bond theory has been discussed in bond is formed by the sharing of an electron terms of qualitative and non-mathematical pair between the respective atoms. It also treatment only. To start with, let us consider gives no idea about the shapes of polyatomic the formation of hydrogen molecule which is molecules. the simplest of all molecules. Similarly the VSEPR theory gives the Consider two hydrogen atoms A and B geometry of simple molecules but theoretically, approaching each other having nuclei NA it does not explain them and also it has limited and NB and electrons present in them are applications. To overcome these limitations represented by eA and eB. When the two atoms the two important theories based on quantum are at large distance from each other, there mechanical principles are introduced. These is no interaction between them. As these two are valence bond (VB) theory and molecular atoms approach each other, new attractive orbital (MO) theory. and repulsive forces begin to operate. Valence bond theory was introduced Attractive forces arise between: by Heitler and London (1927) and developed (i) nucleus of one atom and its own electron further by Pauling and others. A discussion that is NA – eA and NB– eB. Reprint 2025-26 118 chemistry (ii) nucleus of one atom and electron of together to form a stable molecule having the other atom i.e., NA– eB, NB– eA. bond length of 74 pm. Similarly repulsive forces arise between Since the energy gets released when the bond is formed between two hydrogen atoms,(i) electrons of two atoms like eA – eB, the hydrogen molecule is more stable than (ii) nuclei of two atoms NA – NB. that of isolated hydrogen atoms. The energy Attractive forces tend to bring the two so released is called as bond enthalpy, which atoms close to each other whereas repulsive is corresponding to minimum in the curve forces tend to push them apart (Fig. 4.7). depicted in Fig. 4.8. Conversely, 435.8 kJ of energy is required to dissociate one mole of H2 molecule. H2(g) + 435.8 kJ mol–1 → H(g) + H(g) Fig. 4.8 The potential energy curve for the formation of H2 molecule as a function of internuclear distance of the H atoms. The minimum in the curve corresponds to the most stable state of H2. 4.5.1 Orbital Overlap Concept In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that their atomic orbitals undergo partial interpenetration. ThisFig. 4.7 Forces of attraction and repulsion partial merging of atomic orbitals is called during the formation of H2 molecule overlapping of atomic orbitals which results in Experimentally it has been found that the pairing of electrons. The extent of overlap the magnitude of new attractive force is decides the strength of a covalent bond. Inmore than the new repulsive forces. As a general, greater the overlap the stronger is theresult, two atoms approach each other and bond formed between two atoms. Therefore,potential energy decreases. Ultimately a stage is reached where the net force of attraction according to orbital overlap concept, the balances the force of repulsion and system formation of a covalent bond between two acquires minimum energy. At this stage atoms results by pairing of electrons present two hydrogen atoms are said to be bonded in the valence shell having opposite spins. Reprint 2025-26 Chemical Bonding And Molecular Structure 119 4.5.2 Directional Properties of Bonds As we have already seen, the covalent bond is formed by overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two H atoms. In case of polyatomic molecules like CH4, NH3 and H2O, the geometry of the molecules is also important in addition to the bond formation. For example why is it so that CH4 molecule has tetrahedral shape and HCH bond angles are 109.5°? Why is the shape of NH3 molecule pyramidal ? The valence bond theory explains the shape, the formation and directional properties of bonds in polyatomic molecules like CH4, NH3 and H2O, etc. in terms of overlap and hybridisation of atomic orbitals. 4.5.3 Overlapping of Atomic Orbitals When orbitals of two atoms come close to form bond, their overlap may be positive, negative or zero depending upon the sign (phase) and direction of orientation of amplitude of orbital wave function in space (Fig. 4.9). Positive and negative sign on boundary surface diagrams in the Fig. 4.9 show the sign (phase) of orbital wave function and are not related to charge. Fig.4.9 Positive, negative and zero overlaps ofOrbitals forming bond should have same sign s and p atomic orbitals(phase) and orientation in space. This is called positive overlap. Various overlaps of s and p hydrogen. The four atomic orbitals of carbon, orbitals are depicted in Fig. 4.9. each with an unpaired electron can overlap with the 1s orbitals of the four H atoms which The criterion of overlap, as the main factor are also singly occupied. This will result in the for the formation of covalent bonds applies formation of four C-H bonds. It will, however, uniformly to the homonuclear/heteronuclear be observed that while the three p orbitals of diatomic molecules and polyatomic molecules. carbon are at 90° to one another, the HCH We know that the shapes of CH4, NH3, and angle for these will also be 90°. That is three H2O molecules are tetrahedral, pyramidal C-H bonds will be oriented at 90° to one and bent respectively. It would be therefore another. The 2s orbital of carbon and the 1s interesting to use VB theory to find out if these orbital of H are spherically symmetrical and geometrical shapes can be explained in terms they can overlap in any direction. Therefore of the orbital overlaps. the direction of the fourth C-H bond cannot Let us first consider the CH4 (methane) be ascertained. This description does not fit molecule. The electronic configuration of in with the tetrahedral HCH angles of 109.5°. carbon in its ground state is [He]2s2 2p2 which Clearly, it follows that simple atomic orbital in the excited state becomes [He] 2s1 2px1 2py1 overlap does not account for the directional 2pz1. The energy required for this excitation is characteristics of bonds in CH4. Using similar compensated by the release of energy due to procedure and arguments, it can be seen that in overlap between the orbitals of carbon and the the case of NH3 and H2O molecules, the HNH Reprint 2025-26 120 chemistry and HOH angles should be 90°. This is in above and below the plane of the disagreement with the actual bond angles of participating atoms. 107° and 104.5° in the NH3 and H2O molecules respectively. 4.5.4 Types of Overlapping and Nature of Covalent Bonds The covalent bond may be classified into two types depending upon the types of overlapping: (i) Sigma(σ) bond, and (ii) pi(π) bond (i) Sigma(σ) bond : This type of covalent 4.5.5 Strength of Sigma and pi Bonds bond is formed by the end to end (head- Basically the strength of a bond depends on) overlap of bonding orbitals along the upon the extent of overlapping. In case of internuclear axis. This is called as head sigma bond, the overlapping of orbitals takes on overlap or axial overlap. This can be place to a larger extent. Hence, it is stronger formed by any one of the following types as compared to the pi bond where the extent of combinations of atomic orbitals. of overlapping occurs to a smaller extent. • s-s overlapping : In this case, there is Further, it is important to note that in the overlap of two half filled s-orbitals along formation of multiple bonds between two the internuclear axis as shown below : atoms of a molecule, pi bond(s) is formed in addition to a sigma bond. 4.6 Hybridisation In order to explain the characteristic geometrical shapes of polyatomic molecules • s-p overlapping: This type of overlap like CH4, NH3 and H2O etc., Pauling introduced occurs between half filled s-orbitals of one the concept of hybridisation. According to him atom and half filled p-orbitals of another the atomic orbitals combine to form new set of atom. equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined as the process of intermixing of the orbitals of • p–p overlapping : This type of overlap slightly different energies so as to redistribute takes place between half filled p-orbitals their energies, resulting in the formation of of the two approaching atoms. new set of orbitals of equivalent energies and shape. For example when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp3 hybrid orbitals. Salient features of hybridisation: The main (ii) pi( ) bond : In the formation of π bond features of hybridisation are as under : the atomic orbitals overlap in such a 1. The number of hybrid orbitals is equal to way that their axes remain parallel to the number of the atomic orbitals that get each other and perpendicular to the internuclear axis. The orbitals formed hybridised. due to sidewise overlapping consists 2. The hybridised orbitals are always of two saucer type charged clouds equivalent in energy and shape. Reprint 2025-26 Chemical Bonding And Molecular Structure 121 3. The hybrid orbitals are more effective in vacant 2p orbital to account for its bivalency. forming stable bonds than the pure atomic One 2s and one 2p-orbital gets hybridised to orbitals. form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite4. These hybrid orbitals are directed in direction forming an angle of 180°. Each of space in some preferred direction to have the sp hybridised orbital overlaps with the minimum repulsion between electron 2p-orbital of chlorine axially and form two Be- pairs and thus a stable arrangement. Cl sigma bonds. This is shown in Fig. 4.10. Therefore, the type of hybridisation indicates the geometry of the molecules. Important conditions for hybridisation (i) The orbitals present in the valence shell of the atom are hybridised. (ii) The orbitals undergoing hybridisation Be should have almost equal energy. (iii) Promotion of electron is not essential condition prior to hybridisation. (iv) It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation. Fig.4.10 (a) Formation of sp hybrids from s and 4.6.1 Types of Hybridisation p orbitals; (b) Formation of the linear There are various types of hybridisation BeCl2 molecule involving s, p and d orbitals. The different (II) sp2 hybridisation : In this hybridisationtypes of hybridisation are as under: there is involvement of one s and two (I) sp hybridisation: This type of hybridisation p-orbitals in order to form three equivalent involves the mixing of one s and one p orbital sp2 hybridised orbitals. For example, in resulting in the formation of two equivalent BCl3 molecule, the ground state electronicsp hybrid orbitals. The suitable orbitals for configuration of central boron atom is sp hybridisation are s and pz, if the hybrid 1s22s22p1. In the excited state, one of the 2s orbitals are to lie along the z-axis. Each sp electrons is promoted to vacant 2p orbital as hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry. This type of hybridisation is also known as diagonal hybridisation. The two sp hybrids point in the opposite direction along the z-axis with projecting positive lobes and very small negative lobes, which provides more effective overlapping resulting in the formation of stronger bonds. Example of molecule having sp hybridisation BeCl 2: The ground state electronic configuration of Be is 1s22s2. In the exited Fig.4.11 Formation of sp2 hybrids and the BCl3 state one of the 2s-electrons is promoted to molecule Reprint 2025-26 122 chemistry a result boron has three unpaired electrons. ground state is 2S 2 2 p 1x 2 p 1y 2 p 1z having threeThese three orbitals (one 2s and two 2p) unpaired electrons in the sp3 hybrid orbitalshybridise to form three sp2 hybrid orbitals. and a lone pair of electrons is present in theThe three hybrid orbitals so formed are fourth one. These three hybrid orbitals overlaporiented in a trigonal planar arrangement with 1s orbitals of hydrogen atoms to formand overlap with 2p orbitals of chlorine to three N–H sigma bonds. We know that the form three B-Cl bonds. Therefore, in BCl3 force of repulsion between a lone pair and a(Fig. 4.11), the geometry is trigonal planar bond pair is more than the force of repulsionwith ClBCl bond angle of 120°. between two bond pairs of electrons. The (III) sp 3 hybridisation: This type of molecule thus gets distorted and the bond hybridisation can be explained by taking the angle is reduced to 107° from 109.5°. The example of CH4 molecule in which there is geometry of such a molecule will be pyramidal mixing of one s-orbital and three p-orbitals as shown in Fig. 4.13. of the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp3 hybrid orbital is 109.5° as shown in Fig. 4.12. Fig.4.13 Formation of NH3 molecule In case of H2O molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp3 hybridisation forming four sp3 hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These σ four sp3 hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by σ σ hydrogen atoms while the other two by the lone pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Fig. 4.14) σ and the molecule thus acquires a V-shape or angular geometry. Fig.4.12 Formation of sp3 hybrids by the combination of s, px , py and pz atomic orbitals of carbon and the formation of CH4 molecule The structure of NH3 and H2O molecules can also be explained with the help of sp3 hybridisation. In NH3, the valence shell (outer) electronic configuration of nitrogen in the Fig.4.14 Formation of H2O molecule Reprint 2025-26 Chemical Bonding And Molecular Structure 123 4.6.2 Other Examples of sp3, sp2 and sp used for making sp2–s sigma bond with two Hybridisation hydrogen atoms. The unhybridised orbital (2px sp3 Hybridisation in C2H6 molecule: In or 2py) of one carbon atom overlaps sidewise ethane molecule both the carbon atoms with the similar orbital of the other carbon assume sp3 hybrid state. One of the four atom to form weak π bond, which consists of sp3 hybrid orbitals of carbon atom overlaps two equal electron clouds distributed above axially with similar orbitals of other atom to and below the plane of carbon and hydrogen form sp3-sp3 sigma bond while the other three atoms. hybrid orbitals of each carbon atom are used Thus, in ethene molecule, the carbon-in forming sp3–s sigma bonds with hydrogen carbon bond consists of one sp2–sp2 sigmaatoms as discussed in section 4.6.1(iii). bond and one pi (π ) bond between p orbitalsTherefore in ethane C–C bond length is 154 which are not used in the hybridisation andpm and each C–H bond length is 109 pm. are perpendicular to the plane of molecule; thesp2 Hybridisation in C2H4: In the formation bond length 134 pm. The C–H bond is sp2–sof ethene molecule, one of the sp2 hybrid sigma with bond length 108 pm. The H–C–Horbitals of carbon atom overlaps axially with bond angle is 117.6° while the H–C–C anglesp2 hybridised orbital of another carbon atom is 121°. The formation of sigma and pi bondsto form C–C sigma bond. While the other two sp2 hybrid orbitals of each carbon atom are in ethene is shown in Fig. 4.15. Fig. 4.15 Formation of sigma and pi bonds in ethene Reprint 2025-26 124 chemistry sp Hybridisation in C2H2 : In the formation 4.6.3 Hybridisation of Elements of ethyne molecule, both the carbon atoms involving d Orbitals undergo sp-hybridisation having two The elements present in the third period unhybridised orbital i.e., 2py and 2px. contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are One sp hybrid orbital of one carbon atom comparable to the energy of the 3s and 3poverlaps axially with sp hybrid orbital of the orbitals. The energy of 3d orbitals are also other carbon atom to form C–C sigma bond, comparable to those of 4s and 4p orbitals. while the other hybridised orbital of each As a consequence the hybridisation involving carbon atom overlaps axially with the half either 3s, 3p and 3d or 3d, 4s and 4p is filled s orbital of hydrogen atoms forming possible. However, since the difference in σ bonds. Each of the two unhybridised p energies of 3p and 4s orbitals is significant, no orbitals of both the carbon atoms overlaps hybridisation involving 3p, 3d and 4s orbitals sidewise to form two π bonds between the is possible. carbon atoms. So the triple bond between the The important hybridisation schemes two carbon atoms is made up of one sigma involving s, p and d orbitals are summarised and two pi bonds as shown in Fig. 4.16. below: Shape of Hybridisation Atomic molecules/ Examples type orbitals ions Square dsp2 d+s+p(2) [Ni(CN)4]2–, planar [Pt(Cl)4]2– Trigonal sp3d s+p(3)+d PF5, PCl5 bipyramidal Square sp3d2 s+p(3)+d(2) BrF5 pyramidal Octahedral sp3d2 s+p(3)+d(2) SF6, [CrF6]3– d2sp3 d(2)+s+p(3) [Co(NH3)6]3+ (i) Formation of PCl5 (sp3d hybridisation): The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below. Fig.4.16 Formation of sigma and pi bonds in sp3d hybrid orbitals filled by electron pairs ethyne donated by five Cl atoms. Reprint 2025-26 Chemical Bonding And Molecular Structure 125 Now the five orbitals (i.e., one s, three six sp3d2 hybrid orbitals overlap with singly p and one d orbitals) are available for occupied orbitals of fluorine atoms to form hybridisation to yield a set of five sp3d hybrid six S–F sigma bonds. Thus SF6 molecule has orbitals which are directed towards the five a regular octahedral geometry as shown in corners of a trigonal bipyramidal as depicted Fig. 4.18. in the Fig. 4.17. sp3d2 hybridisation Fig. 4.17 Trigonal bipyramidal geometry of PCl5 molecule It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial Fig. 4.18 Octahedral geometry of SF6 moleculebonds. The remaining two P–Cl bonds–one lying above and the other lying below the equatorial plane, make an angle of 90° with 4.7 Molecular Orbital Theory the plane. These bonds are called axial bonds. Molecular orbital (MO) theory was developed As the axial bond pairs suffer more repulsive by F. Hund and R.S. Mulliken in 1932. The interaction from the equatorial bond pairs, salient features of this theory are : therefore axial bonds have been found to (i) The electrons in a molecule are present be slightly longer and hence slightly weaker in the various molecular orbitals as the than the equatorial bonds; which makes PCl5 electrons of atoms are present in the molecule more reactive. various atomic orbitals. (ii) Formation of SF6 (sp3d2 hybridisation): (ii) The atomic orbitals of comparableIn SF6 the central sulphur atom has the energies and proper symmetry combineground state outer electronic configuration to form molecular orbitals.3s23p4. In the exited state the available six orbitals i.e., one s, three p and two d are (iii) While an electron in an atomic orbital singly occupied by electrons. These orbitals is influenced by one nucleus, in a hybridise to form six new sp3d2 hybrid molecular orbital it is influenced by orbitals, which are projected towards the six two or more nuclei depending upon the corners of a regular octahedron in SF6. These number of atoms in the molecule. Thus, Reprint 2025-26 126 chemistry an atomic orbital is monocentric while ψA and ψB. Mathematically, the formation of a molecular orbital is polycentric. molecular orbitals may be described by the linear combination of atomic orbitals that can(iv) The number of molecular orbital formed take place by addition and by subtraction of is equal to the number of combining wave functions of individual atomic orbitals atomic orbitals. When two atomic as shown below : orbitals combine, two molecular orbitals are formed. One is known as bonding ψMO = ψA + ψB molecular orbital while the other is Therefore, the two molecular orbitals called antibonding molecular orbital. σ and σ* are formed as : (v) The bonding molecular orbital has σ = ψA + ψB lower energy and hence greater stability σ* = ψA – ψB than the corresponding antibonding The molecular orbital σ formed by the molecular orbital. addition of atomic orbitals is called the bonding (vi) Just as the electron probability molecular orbital while the molecular orbital distribution around a nucleus in an σ* formed by the subtraction of atomic orbital atom is given by an atomic orbital, the is called antibonding molecular orbital as electron probability distribution around depicted in Fig. 4.19. a group of nuclei in a molecule is given by a molecular orbital. (vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule. 4.7.1 Formation of Molecular Orbitals Linear Combination of Atomic σ* = ψA – ψB Orbitals (LCAO) According to wave mechanics, the atomic ψA ψBorbitals can be expressed by wave functions (ψ ’s) which represent the amplitude of the σ = ψA + ψBelectron waves. These are obtained from the solution of Schrödinger wave equation. However, since it cannot be solved for any system containing more than one electron, molecular orbitals which are one electron wave functions for molecules are difficult Fig.4.19 Formation of bonding (σ) and antibonding (σ*) molecular orbitals by the linearto obtain directly from the solution of combination of atomic orbitals ψA andSchrödinger wave equation. To overcome this problem, an approximate method known ψB centered on two atoms A and B respectively.as linear combination of atomic orbitals (LCAO) has been adopted. Qualitatively, the formation of molecular Let us apply this method to the orbitals can be understood in terms of the homonuclear diatomic hydrogen molecule. constructive or destructive interference of the Consider the hydrogen molecule consisting electron waves of the combining atoms. In the of two atoms A and B. Each hydrogen atom formation of bonding molecular orbital, the in the ground state has one electron in 1s two electron waves of the bonding atoms orbital. The atomic orbitals of these atoms reinforce each other due to constructive may be represented by the wave functions interference while in the formation of Reprint 2025-26 Chemical Bonding And Molecular Structure 127 antibonding molecular orbital, the electron as the molecular axis. It is important to note waves cancel each other due to destructive that atomic orbitals having same or nearly interference. As a result, the electron density in the same energy will not combine if they do a bonding molecular orbital is located between not have the same symmetry. For example, the nuclei of the bonded atoms because of 2pz orbital of one atom can combine with 2pz which the repulsion between the nuclei is very orbital of the other atom but not with the less while in case of an antibonding molecular 2px or 2py orbitals because of their different orbital, most of the electron density is located symmetries. away from the space between the nuclei. 3. The combining atomic orbitals must Infact, there is a nodal plane (on which the overlap to the maximum extent. Greater electron density is zero) between the nuclei the extent of overlap, the greater will be the and hence the repulsion between the nuclei is electron-density between the nuclei of a high. Electrons placed in a bonding molecular molecular orbital. orbital tend to hold the nuclei together and 4.7.3 Types of Molecular Orbitalsstabilise a molecule. Therefore, a bonding molecular orbital always possesses lower Molecular orbitals of diatomic molecules are energy than either of the atomic orbitals that designated as σ (sigma), π (pi), δ(delta), etc. have combined to form it. In contrast, the In this nomenclature, the sigma ( ) electrons placed in the antibonding molecular molecular orbitals are symmetrical around orbital destabilise the molecule. This is the bond-axis while pi ( ) molecular orbitals because the mutual repulsion of the electrons are not symmetrical. For example, the linear in this orbital is more than the attraction combination of 1s orbitals centered on two between the electrons and the nuclei, which nuclei produces two molecular orbitals which causes a net increase in energy. are symmetrical around the bond-axis. Such It may be noted that the energy of the molecular orbitals are of the σ type and are antibonding orbital is raised above the designated as σ1s and σ*1s [Fig. 4.20(a), page energy of the parent atomic orbitals that 124]. If internuclear axis is taken to be in have combined and the energy of the bonding the z-direction, it can be seen that a linear orbital has been lowered than the parent combination of 2pz- orbitals of two atoms orbitals. The total energy of two molecular also produces two sigma molecular orbitals orbitals, however, remains the same as that designated as 2pz and *2pz. [Fig. 4.20(b)] of two original atomic orbitals. Molecular orbitals obtained from 2px and 4.7.2 Conditions for the Combination of 2py orbitals are not symmetrical around the Atomic Orbitals bond axis because of the presence of positive lobes above and negative lobes below theThe linear combination of atomic orbitals to molecular plane. Such molecular orbitals,form molecular orbitals takes place only if the are labelled as π and =π* [Fig. 4.20(c)]. Afollowing conditions are satisfied: π bonding MO has larger electron density1. The combining atomic orbitals must above and below the inter-nuclear axis. Thehave the same or nearly the same energy. π* antibonding MO has a node between theThis means that 1s orbital can combine with nuclei.another 1s orbital but not with 2s orbital because the energy of 2s orbital is appreciably 4.7.4 Energy Level Diagram for Molecular higher than that of 1s orbital. This is not true Orbitals if the atoms are very different. We have seen that 1s atomic orbitals on two 2. The combining atomic orbitals must atoms form two molecular orbitals designated have the same symmetry about the as σ1s and σ*1s. In the same manner, the 2s molecular axis. By convention z-axis is taken and 2p atomic orbitals (eight atomic orbitals Reprint 2025-26 128 chemistry Fig. 4.20 Contours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2px atomic orbitals. on two atoms) give rise to the following eight The energy levels of these molecular molecular orbitals: orbitals have been determined experimentally from spectroscopic data for homonuclearAntibonding MOs σ∗2s σ∗2pz π∗2px π∗2py diatomic molecules of second row elements Bonding MOs σ2s σ2pz π2px π2py of the periodic table. The increasing order of Reprint 2025-26 Chemical Bonding And Molecular Structure 129 energies of various molecular orbitals for O2 The rules discussed above regarding the and F2 is given below: stability of the molecule can be restated in terms of bond order as follows: A positive bond1s <  ∗1s <  2s <  ∗2s < 2pz < (π 2px=π 2py) order (i.e., Nb > Na) means a stable molecule < (π ∗2px= π∗ 2py) <  ∗2pz while a negative (i.e., Nb<Na) or zero (i.e., However, this sequence of energy levels Nb = Na) bond order means an unstable of molecular orbitals is not correct for the molecule. remaining molecules Li2, Be2, B2, C2, N2. For Nature of the bond instance, it has been observed experimentally Integral bond order values of 1, 2 or 3 that for molecules such as B2, C2, N2, etc. correspond to single, double or triple bonds the increasing order of energies of various respectively as studied in the classical molecular orbitals is concept. 1s <  ∗1s < 2s <  ∗2s < (π 2 px = π 2 py) Bond-length < 2pz < (π ∗2px =π∗2py) <  ∗2pz The bond order between two atoms in a The important characteristic feature molecule may be taken as an approximate of this order is that the energy of 2pz measure of the bond length. The bond length molecular orbital is higher than that decreases as bond order increases. of 2px and 2py molecular orbitals. Magnetic nature 4.7.5 Electronic Configuration and If all the molecular orbitals in a molecule are Molecular Behaviour doubly occupied, the substance is diamagnetic (repelled by magnetic field). However if one orThe distribution of electrons among various more molecular orbitals are singly occupied itmolecular orbitals is called the electronic is paramagnetic (attracted by magnetic field),configuration of the molecule. From the e.g., O2 molecule.electronic configuration of the molecule, it is possible to get important information about 4.8 BONDING IN SOME HOMONUCLEAR the molecule as discussed below. DIATOMIC MOLECULES Stability of Molecules: If Nb is the number In this section we shall discuss bonding in of electrons occupying bonding orbitals and some homonuclear diatomic molecules. Na the number occupying the antibonding 1. Hydrogen molecule (H2 ): It is formed byorbitals, then the combination of two hydrogen atoms. Each (i) the molecule is stable if Nb is greater hydrogen atom has one electron in 1s orbital. than Na, and Therefore, in all there are two electrons in (ii) the molecule is unstable if Nb is less hydrogen molecule which are present in σ1s than Na. molecular orbital. So electronic configuration of hydrogen molecule is In (i) more bonding orbitals are occupied and so the bonding influence is stronger and a H2 : (σ1s)2 stable molecule results. In (ii) the antibonding The bond order of H2 molecule can be influence is stronger and therefore the calculated as given below: molecule is unstable. N b  N a 2  0 Bond order Bond order =   1 2 2 Bond order (b.o.) is defined as one half the This means that the two hydrogen atoms difference between the number of electrons are bonded together by a single covalent bond. present in the bonding and the antibonding The bond dissociation energy of hydrogen orbitals i.e., molecule has been found to be 438 kJ mol–1 and bond length equal to 74 pm. Since no Bond order (b.o.) = ½ (Nb–Na) Reprint 2025-26 130 chemistry unpaired electron is present in hydrogen vapour phase. It is important to note that molecule, therefore, it is diamagnetic. double bond in C2 consists of both pi bonds 2. Helium molecule (He2 ): The electronic because of the presence of four electrons in configuration of helium atom is 1s2. Each two pi molecular orbitals. In most of the other helium atom contains 2 electrons, therefore, molecules a double bond is made up of a in He2 molecule there would be 4 electrons. sigma bond and a pi bond. In a similar fashion the bonding in N2 molecule can be discussed.These electrons will be accommodated in σ1s and σ*1s molecular orbitals leading to 5. Oxygen molecule (O2 ): The electronic electronic configuration: configuration of oxygen atom is 1s2 2s2 2p4. Each oxygen atom has 8 electrons, hence, He2 : (σ1s)2 (σ*1s)2 in O2 molecule there are 16 electrons. The electronic configuration of O2 molecule, Bond order of He2 is ½(2 – 2) = 0 therefore, is He2 molecule is therefore unstable and does not exist. O2 : (1s)2 ( ∗1s)2 ( 2s)2 ( ∗ 2s)2 (2pz)2 Similarly, it can be shown that Be2 molecule (π2px2 ≡ π2py2) (π∗2p1x ≡ π ∗2py1) (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 also does not exist. 3. Lithium molecule (Li2 ): The electronic O2 :configuration of lithium is 1s2, 2s1. There are six electrons in Li2. The electronic configuration of Li2 molecule, therefore, is From the electronic configuration of O2 molecule it is clear that ten electrons are Li2 : (σ1s)2 (σ*1s)2 (σ2s)2 present in bonding molecular orbitals and six The above configuration is also written electrons are present in antibonding molecular as KK(σ2s)2 where KK represents the closed orbitals. Its bond order, therefore, is K shell structure (σ1s)2 (σ*1s)2. From the electronic configuration of Li2 Bond order = [Nb – Na] = [10 – 6] =2 molecule it is clear that there are four electrons So in oxygen molecule, atoms are heldpresent in bonding molecular orbitals and two by a double bond. Moreover, it may be notedelectrons present in antibonding molecular that it contains two unpaired electrons inorbitals. Its bond order, therefore, is ½ (4 – π∗2px and π∗2py molecular orbitals, therefore,2) = 1. It means that Li2 molecule is stable and since it has no unpaired electrons it O2 molecule should be paramagnetic, a prediction that corresponds toshould be diamagnetic. Indeed diamagnetic experimental observation. In this way, theLi2 molecules are known to exist in the theory successfully explains the paramagneticvapour phase. nature of oxygen. 4. Carbon molecule (C2 ): The electronic Similarly, the electronic configurationsconfiguration of carbon is 1s2 2s2 2p2. There of other homonuclear diatomic molecules of [ ]are twelve electrons in C2. The electronic the second row of the periodic table can be configuration of C2 molecule, therefore, is written. In Fig. 4.21 are given the molecular C2 : (1s)2 ( ∗1s)2 ( ∗ 2s)2 (π2p2x = π2p2y) orbital occupancy and molecular properties for B2 through Ne2. The sequence of MOs and or KK (2s)2 ( ∗ 2s)2 (π2p2x = π2p2y) their electron population are shown. The bond energy, bond length, bond order, magnetic The bond order of C2 is ½ (8 – 4) = 2 properties and valence electron configurationand C2 should be diamagnetic. Diamagnetic appear below the orbital diagrams.C2 molecules have indeed been detected in Reprint 2025-26 Chemical Bonding And Molecular Structure 131 Fig. 4.21 MO occupancy and molecular properties for B2 through Ne2.

4.2 Ionic or Electrovalent Bond other factors. The crystal structure of sodium chloride, NaCl (rock salt), for example isFrom the Kössel and Lewis treatment of the shown below.formation of an ionic bond, it follows that the formation of ionic compounds would primarily depend upon: • The ease of formation of the positive and negative ions from the respective neutral atoms; • The arrangement of the positive and negative ions in the solid, that is, the lattice of the crystalline compound. The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the Rock salt structure neutral atom. In ionic solids, the sum of the electron gain M(g) → M+(g) + e– ; enthalpy and the ionization enthalpy may be Ionization enthalpy positive but still the crystal structure gets X(g) + e– → X – (g) ; stabilized due to the energy released in the Electron gain enthalpy formation of the crystal lattice. For example: the ionization enthalpy for Na+(g) formation M+(g) + X –(g) → MX(s) from Na(g) is 495.8 kJ mol–1 ; while the electron The electron gain enthalpy, ∆egH, is the gain enthalpy for the change Cl(g) + e–→ enthalpy change (Unit 3), when a gas phase Cl– (g) is, – 348.7 kJ mol–1 only. The sum of the atom in its ground state gains an electron. two, 147.1 kJ mol-1 is more than compensated The electron gain process may be exothermic for by the enthalpy of lattice formation of or endothermic. The ionization, on the other NaCl(s) (–788 kJ mol–1). Therefore, the energy hand, is always endothermic. Electron released in the processes is more than the Reprint 2025-26 Chemical Bonding And Molecular Structure 107 energy absorbed. Thus a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state. Since lattice enthalpy plays a key role in the formation of ionic compounds, it is important that we learn more about it. 4.2.1 Lattice Enthalpy The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl is 788 kJ mol–1. This means that 788 Fig. 4.1 The bond length in a covalent kJ of energy is required to separate one mole molecule AB. of solid NaCl into one mole of Na+ (g) and one R = rA + rB (R is the bond length and rA and rB are mole of Cl– (g) to an infinite distance. the covalent radii of atoms A and B respectively) This process involves both the attractive forces between ions of opposite charges in the same molecule. The van der Waals and the repulsive forces between ions of radius represents the overall size of the like charge. The solid crystal being three- atom which includes its valence shell in a dimensional; it is not possible to calculate nonbonded situation. Further, the van der lattice enthalpy directly from the interaction Waals radius is half of the distance between of forces of attraction and repulsion only. two similar atoms in separate molecules in Factors associated with the crystal geometry a solid. Covalent and van der Waals radii of have to be included. chlorine are depicted in Fig. 4.2.

4.15 Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H2S 115 The d- and f- Block Elements Reprint 2025-26

4.9 Hydrogen Bonding Hydrogen bond is represented by a dotted line (– – –) while a solid line represents theNitrogen, oxygen and fluorine are the highly covalent bond. Thus, hydrogen bond can beelectronegative elements. When they are attached to a hydrogen atom to form covalent defined as the attractive force which binds bond, the electrons of the covalent bond are hydrogen atom of one molecule with the shifted towards the more electronegative electronegative atom (F, O or N) of another atom. This partially positively charged molecule. hydrogen atom forms a bond with the other 4.9.1 Cause of Formation of Hydrogen more electronegative atom. This bond is Bond known as hydrogen bond and is weaker When hydrogen is bonded to stronglythan the covalent bond. For example, in HF electronegative element ‘X’, the electron pairmolecule, the hydrogen bond exists between shared between the two atoms moves farhydrogen atom of one molecule and fluorine away from hydrogen atom. As a result theatom of another molecule as depicted below : hydrogen atom becomes highly electropositive – – – Hδ+–Fδ– – – –Hδ+ – Fδ– – – – Hδ+ – Fδ– with respect to the other atom ‘X’. Since Here, hydrogen bond acts as a bridge between there is displacement of electrons towards two atoms which holds one atom by covalent X, the hydrogen acquires fractional positive bond and the other by hydrogen bond. charge (δ +) while ‘X’ attain fractional negative Reprint 2025-26 132 chemistry charge (δ–). This results in the formation of a H-bond in case of HF molecule, alcohol or polar molecule having electrostatic force of water molecules, etc. attraction which can be represented as: (2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between Hδ+ – Xδ– – – – Hδ+ – Xδ– – – – Hδ+ – Xδ– the two highly electronegative (F, O, N) The magnitude of H-bonding depends atoms present within the same molecule. For on the physical state of the compound. It is example, in o-nitrophenol the hydrogen is in maximum in the solid state and minimum in between the two oxygen atoms. the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds. 4.9.2 Types of H-Bonds There are two types of H-bonds (i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond (1) Intermolecular hydrogen bond : It is formed between two different molecules of the Fig. 4.22 Intramolecular hydrogen bonding in same or different compounds. For example, o-nitrophenol molecule SUMMARY Kössel’s first insight into the mechanism of formation of electropositive and electronegative ions related the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction between ions is the cause for their stability. This gives the concept of electrovalency. The first description of covalent bonding was provided by Lewis in terms of the sharing of electron pairs between atoms and he related the process to the attainment of noble gas configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence electrons of the atoms of a given element and Lewis dot structures show pictorial representations of bonding in molecules. An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation. While a single covalent bond is formed by sharing of an electron pair between two atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional pairs of electrons not involved in bonding. These are called lone-pairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity have significant effect on the properties of compounds. A number of molecules and polyatomic ions cannot be described accurately by a single Lewis structure and a number of descriptions (representations) based on the same skeletal structure are written and these taken together represent the molecule or ion. This is a very important and extremely useful concept called resonance. The contributing structures or canonical forms taken together constitute the resonance hybrid which represents the molecule or ion. Reprint 2025-26 Chemical Bonding And Molecular Structure 133 The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible. According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions being : lp-lp > lp-bp > bp-bp The valence bond (VB) approach to covalent bonding is basically concerned with the energetics of covalent bond formation about which the Lewis and VSEPR models are silent. Basically the VB theory discusses bond formation in terms of overlap of orbitals. For example the formation of the H2 molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come near to each other. At the equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent destabilization of the molecule. Because of orbital overlap the electron density between the nuclei increases which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length values are not obtained by overlap alone and other variables have to be taken into account. For explaining the characteristic shapes of polyatomic molecules Pauling introduced the concept of hybridisation of atomic orbitals. sp, sp2, sp3 hybridizations of atomic orbitals of Be, B, C, N and O are used to explain the formation and geometrical shapes of molecules like BeCl2, BCl3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in molecules like C2H2 and C2H4. The molecular orbital (MO) theory describes bonding in terms of the combination and arrangment of atomic orbitals to form molecular orbitals that are associated with the molecule as a whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than the individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei and have more energy than the individual atomic orbitals. The electronic configuration of the molecules is written by filling electrons in the molecular orbitals in the order of increasing energy levels. As in the case of atoms, the Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals. Hydrogen bond is formed when a hydrogen atom finds itself between two highly electronegative atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds have a powerful effect on the structure and properties of many compounds. EXERCISES 4.1 Explain the formation of a chemical bond. 4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br. 4.3 Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H– 4.4 Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO32−, HCOOH 4.5 Define octet rule. Write its significance and limitations. Reprint 2025-26 134 chemistry 4.6 Write the favourable factors for the formation of ionic bond. 4.7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3 4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. 4.9 How do you express the bond strength in terms of bond order ? 4.10 Define the bond length. 4.11 Explain the important aspects of resonance with reference to the CO32− ion. 4.12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same. 4.13 Write the resonance structures for SO3, NO2 and NO3−. 4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N. 4.15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment. 4.16 Write the significance/applications of dipole moment. 4.17 Define electronegativity. How does it differ from electron gain enthalpy ? 4.18 Explain with the help of suitable example polar covalent bond. 4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. 4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. 4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ? 4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar. 4.23 Which out of NH3 and NF3 has higher dipole moment and why ? 4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals. 4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl 3  Cl   AlCl 4 Reprint 2025-26 Chemical Bonding And Molecular Structure 135 4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules. 4.28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4 4.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s. 4.30 Which hybrid orbitals are used by carbon atoms in the following molecules? CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type. 4.32 Distinguish between a sigma and a pi bond. 4.33 Explain the formation of H2 molecule on the basis of valence bond theory. 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. 4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist. 4.36 Compare the relative stability of the following species and indicate their magnetic properties; (superoxide), O22− (peroxide) 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals. 4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds? 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? 4.40 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–. Reprint 2025-26 Unit 5 Thermodynamics It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown. After studying this Unit, you will be Albert Einstein able to • explain the terms : system and surroundings; • discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released as• explain internal energy, work and heat; heat during chemical reactions when a fuel like methane, • state first law of thermodynamics cooking gas or coal burns in air. The chemical energy may and express it mathematically; also be used to do mechanical work when a fuel burns • calculate energy changes as in an engine or to provide electrical energy through a work and heat contributions in galvanic cell like dry cell. Thus, various forms of energy chemical systems; are interrelated and under certain conditions, these may • explain state functions: U, H. be transformed from one form into another. The study • correlate ∆U and ∆H; of these energy transformations forms the subject matter • measure experimentally ∆U and of thermodynamics. The laws of thermodynamics deal ∆H; with energy changes of macroscopic systems involving• define standard states for ∆H; • calculate enthalpy changes for a large number of molecules rather than microscopic various types of reactions; systems containing a few molecules. Thermodynamics is • state and apply Hess’s law of not concerned about how and at what rate these energy constant heat summation; transformations are carried out, but is based on initial and • differentiate between extensive final states of a system undergoing the change. Laws of and intensive properties; thermodynamics apply only when a system is in equilibrium • define spontaneous and non- or moves from one equilibrium state to another equilibrium spontaneous processes; state. Macroscopic properties like pressure and temperature• e x p l a i n e n t r o p y a s a thermodynamic state function do not change with time for a system in equilibrium state. and apply it for spontaneity; In this unit, we would like to answer some of the important • explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in a • establish relationship between chemical reaction/process? Will it occur or not? ∆G and spontaneity, ∆G and equilibrium constant. What drives a chemical reaction/process? To what extent do the chemical reactions proceed? Reprint 2025-26 THERMODYNAMICS 137

4.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

4.12 What are interstitial compounds? Why are such compounds well known for transition metals?

7.21 Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.

7.7 Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol

7.17 Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. (ii) Bromine in CS2 with phenol. (iii) Dilute HNO3 with phenol. (iv) Treating phenol wih chloroform in presence of aqueous NaOH.

7.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

7.16 Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

7.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl –ZnCl2 (b) HBr and (c) SOCl2. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol

7.13 Show how will you synthesise: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction. (iii) pentan-1-ol using a suitable alkyl halide?

7.12 Predict the products of the following reactions: (i) CH3 − CH2 − CH2 − O – CH3 + HBr → (ii) (iii) (iv) ( CH 3 ) 3 C − OC 2 H5 HI → SummarySummarySummarySummarySummary Alcohols and phenols are classified (i) on the basis of the number of hydroxyl groups and (ii) according to the hybridisation of the carbon atom, sp 3 or sp 2 to which the –OH group is attached. Ethers are classified on the basis of groups attached to the oxygen atom. Alcohols may be prepared (1) by hydration of alkenes (i) in presence of an acid and (ii) by hydroboration-oxidation reaction (2) from carbonyl compounds by (i) catalytic reduction and (ii) the action of Grignard reagents. Phenols may be prepared by (1) substitution of (i) halogen atom in haloarenes and (ii) sulphonic acid group in aryl sulphonic acids, by –OH group (2) by hydrolysis of diazonium salts and (3) industrially from cumene. Alcohols are higher boiling than other classes of compounds, namely hydrocarbons, ethers and haloalkanes of comparable molecular masses. The ability of alcohols, phenols and ethers to form intermolecular hydrogen bonding with water makes them soluble in it. Alcohols and phenols are acidic in nature. Electron withdrawing groups in phenol increase its acidic strength and electron releasing groups decrease it. Alcohols undergo nucleophilic substitution with hydrogen halides to yield alkyl halides. Dehydration of alcohols gives alkenes. On oxidation, primary alcohols yield aldehydes with mild oxidising agents and carboxylic acids with strong oxidising agents while secondary alcohols yield ketones. Tertiary alcohols are resistant to oxidation. The presence of –OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions due to resonance effect. Reimer-Tiemann reaction of phenol yields salicylaldehyde. In presence of sodium hydroxide, phenol generates phenoxide ion which is even more reactive than phenol. Thus, in alkaline medium, phenol undergoes Kolbe’s reaction. Ethers may be prepared by (i) dehydration of alcohols and (ii) Williamson synthesis. The boiling points of ethers resemble those of alkanes while their solubility is comparable to those of alcohols having same molecular mass. The C–O bond in ethers can be cleaved by hydrogen halides. In electrophilic substitution, the alkoxy group activates the aromatic ring and directs the incoming group to ortho and para positions. 221 Alcohols, Phenols and Ethers Reprint 2025-26 Exercises 7.1 Write IUPAC names of the following compounds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) C6H5–O–C2H5 (xi) C6H5–O–C7H15(n–) (xii) 7.2 Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane –1, 3, 5-triol (iv) 2,3 – Diethylphenol (v) 1 – Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 4-Chloro-3-ethylbutan-1-ol. 7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols. 7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane? 7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact. 7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example. 7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O. 7.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason. 7.9 Give the equations of reactions for the preparation of phenol from cumene. 7.10 Write chemical reaction for the preparation of phenol from chlorobenzene. 7.11 Write the mechanism of hydration of ethene to yield ethanol. 7.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents. Chemistry 222 Reprint 2025-26

7.23 Give IUPAC names of the following ethers: 7.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers. 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction. 7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. 7.28 Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether. 7.29 Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring. 7.30 Write the mechanism of the reaction of HI with methoxymethane. 7.31 Write equations of the following reactions: (i) Friedel-Crafts reaction – alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Craft’s acetylation of anisole. 7.32 Show how would you synthesise the following alcohols from appropriate alkenes? CH3 OH (i) OH (ii) OH (iii) (iv) OH 7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Chemistry 224 Reprint 2025-26 Answers to Some Intext Questions 7.1 Primary alcohols (i), (ii), (iii) Secondary alcohols (iv) and (v) Tertiary alcohols (vi) 7.2 Allylic alcohols (ii) and (vi) 7.3 (i) 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol 7.4 OH CH2 C OCH3 7.5 (i) CH3 CH CH3 (ii) O OH (iii) CH3 CH2 CH CH2 OH CH3 7.7 (i) 1-Methylcyclohexene (ii) A Mixture of but-1-ene and but-2-ene. But-2-ene is the major product formed due to rearrangement to give secondary carbocation.

7.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

2.1 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × k/S m–1 1.237 11.85 23.15 55.53 106.74 Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0m . 2.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant? 2.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al? (ii) 1 mol of Cu2+ to Cu? (iii) 1 mol of MnO4– to Mn2+? 2.13 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2? (ii) 40.0 g of Al from molten Al2O3? 2.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2? (ii) 1 mol of FeO to Fe2O3? 2.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? 2.16 Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I–(aq) (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) (iv) Ag(s) and Fe 3+ (aq) (v) Br2 (aq) and Fe2+ (aq). 2.18 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. Answers to Some Intext Questions 2.5 E(cell) = 0.91 V 2.6 ∆ rG o = −45.54 kJ mol −1 , Kc = 9.62 ×107 2.9 0.114, 3.67 × 10–4 mol L–1 Chemistry 60 Reprint 2025-26 UnitUnitUnit Unit33Unit Objectives ChemicalChemical KineticsKinetics After studying this Unit, you will be able to · define the average and Chemical Kinetics helps us to understand how chemical reactions instantaneous rate of a reaction; occur. · express the rate of a reaction in terms of change in concentration Chemistry, by its very nature, is concerned with change. of either of the reactants or Substances with well defined properties are converted products with time; by chemical reactions into other substances with · distinguish between elementary different properties. For any chemical reaction, chemists and complex reactions; try to find out · differentiate between the (a) the feasibility of a chemical reaction which can be molecularity and order of a reaction; predicted by thermodynamics ( as you know that a · define rate constant; reaction with DG < 0, at constant temperature and pressure is feasible);· discuss the dependence of rate of reactions on concentration, (b) extent to which a reaction will proceed can be temperature and catalyst; determined from chemical equilibrium; · derive integrated rate equations (c) speed of a reaction i.e. time taken by a reaction to for the zero and first order reach equilibrium. reactions; Along with feasibility and extent, it is equally · determine the rate constants for important to know the rate and the factors controlling zeroth and first order reactions; the rate of a chemical reaction for its complete · describe collision theory. understanding. For example, which parameters determine as to how rapidly food gets spoiled? How to design a rapidly setting material for dental filling? Or what controls the rate at which fuel burns in an auto engine? All these questions can be answered by the branch of chemistry, which deals with the study of reaction rates and their mechanisms, called chemical kinetics. The word kinetics is derived from the Greek word ‘kinesis’ meaning movement. Thermodynamics tells only about the feasibility of a reaction whereas chemical kinetics tells about the rate of a reaction. For example, thermodynamic data indicate that diamond shall convert to graphite but in reality the conversion rate is so slow that the change is not perceptible at all. Therefore, most people think Reprint 2025-26 that diamond is forever. Kinetic studies not only help us to determine the speed or rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. At the macroscopic level, we are interested in amounts reacted or formed and the rates of their consumption or formation. At the molecular level, the reaction mechanisms involving orientation and energy of molecules undergoing collisions, are discussed. In this Unit, we shall be dealing with average and instantaneous rate of reaction and the factors affecting these. Some elementary ideas about the collision theory of reaction rates are also given. However, in order to understand all these, let us first learn about the reaction rate. 3.13.13.13.13.1 RateRateRateRateRate ofofofofof aaaaa Some reactions such as ionic reactions occur very fast, for example, ChemicalChemicalChemicalChemicalChemical precipitation of silver chloride occurs instantaneously by mixing of aqueous solutions of silver nitrate and sodium chloride. On the other ReactionReactionReactionReactionReaction hand, some reactions are very slow, for example, rusting of iron in the presence of air and moisture. Also there are reactions like inversion of cane sugar and hydrolysis of starch, which proceed with a moderate speed. Can you think of more examples from each category? You must be knowing that speed of an automobile is expressed in terms of change in the position or distance covered by it in a certain period of time. Similarly, the speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R ® P One mole of the reactant R produces one mole of the product P. If [R]1 and [P]1 are the concentrations of R and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then, Dt = t2 – t1 D[R] = [R]2 – [R]1 D [P] = [P]2 – [P]1 The square brackets in the above expressions are used to express molar concentration. Rate of disappearance of R Decrease in concentration of R ∆ [ R ] = = − (3.1) Time taken ∆ t Chemistry 62 Reprint 2025-26 Rate of appearance of P Increase in concentration of P ∆ [ P ] = = + (3.2) Time taken ∆t Since, D[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make the rate of the reaction a positive quantity. Equations (3.1) and (3.2) given above represent the average rate of a reaction, rav. Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 3.1). { } Fig. 3.1: Instantaneous and average rate of a reaction Units of rate of a reaction From equations (3.1) and (3.2), it is clear that units of rate are concentration time–1. For example, if concentration is in mol L–1 and time is in seconds then the units will be mol L-1s–1. However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressures, then the units of the rate equation will be atm s–1. From the concentrations of C4H9Cl (butyl chloride) at different times given ExampleExampleExampleExampleExample 3.13.13.13.13.1 below, calculate the average rate of the reaction: C4H9Cl + H2O ® C4H9OH + HCl during different intervals of time. t/s 0 50 100 150 200 300 400 700 800 [C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017 We can determine the difference in concentration over different intervals SolutionSolutionSolutionSolutionSolution of time and thus determine the average rate by dividing D[R] by Dt (Table 3.1). 63 Chemical Kinetics Reprint 2025-26 Table 3.1: Average rates of hydrolysis of butyl chloride [C4H9CI]t1 / [C4H9CI]t2 / t1/s t2/s rav × 104/mol L–1s–1 × 10 4 mol L–1 mol L–1 = – / ( t 2 − t1 ) C 4 H 9 Cl ]t 2 – [ C 4 H 9 Cl ]t1 {[ } 0.100 0.0905 0 50 1.90 0.0905 0.0820 50 100 1.70 0.0820 0.0741 100 150 1.58 0.0741 0.0671 150 200 1.40 0.0671 0.0549 200 300 1.22 0.0549 0.0439 300 400 1.10 0.0439 0.0335 400 500 1.04 0.0210 0.017 700 800 0.4 It can be seen (Table 3.1) that the average rate falls from 1.90 × 0-4 mol L-1s-1 to 0.4 × 10-4 mol L-1s-1. However, average rate cannot be used to predict the rate of a reaction at a particular instant as it would be constant for the time interval for which it is calculated. So, to express the rate at a particular moment of time we determine the instantaneous rate. It is obtained when we consider the average rate at the smallest time interval say dt ( i.e. when Dt approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by −∆ [ R ] ∆ [ P ] rav = = (3.3) ∆t ∆ t  d  R  d P As Dt ® 0 or rinst   d t d t Fig 3.2 Instantaneous rate of hydrolysis of butyl chloride(C4H9Cl) Chemistry 64 Reprint 2025-26 It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 3.1). So in problem 3.1, rinst at 600s for example, can be calculated by plotting concentration of butyl chloride as a function of time. A tangent is drawn that touches the curve at t = 600 s (Fig. 3.2). The slope of this tangent gives the instantaneous rate. So, rinst at 600 s = – mol L–1 = 5.12 × 10–5 mol L–1s–1 At t = 250 s rinst = 1.22 × 10–4 mol L–1s–1 t = 350 s rinst = 1.0 × 10–4 mol L–1s–1 t = 450 s rinst = 6.4 ×× 10–5 mol L–1s–1 Now consider a reaction Hg(l) + Cl2 (g) ® HgCl2(s) Where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as ∆ [ Hg ] ∆ [ Cl 2 ] ∆ [ HgCl 2 ] Rate of reaction = – = – = ∆t ∆t ∆ t i.e., rate of disappearance of any of the reactants is same as the rate of appearance of the products. But in the following reaction, two moles of HI decompose to produce one mole each of H2 and I2, 2HI(g) ® H2(g) + I2(g) For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients. Since rate of consumption of HI is twice the rate of formation of H2 or I2, to make them equal, the term D[HI] is divided by 2. The rate of this reaction is given by 1 ∆ [ HI ] ∆ [ H 2 ] ∆ [ I 2 ] Rate of reaction = − = = 2 ∆ t ∆ t ∆ t Similarly, for the reaction 5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) ® 3 Br2 (aq) + 3 H2O (l) − − + 1 ∆ [ Br BrO 3 1 ∆ [ H ] 1 ∆ [ Br2 ] 1 ∆ [ H 2 O ] ] ∆  Rate = − = − = − = = 5 ∆ t ∆ t 6 ∆t 3 ∆ t 3 ∆t For a gaseous reaction at constant temperature, concentration is directly proportional to the partial pressure of a species and hence, rate can also be expressed as rate of change in partial pressure of the reactant or the product. 65 Chemical Kinetics Reprint 2025-26 ExampleExampleExampleExampleExample 3.23.23.23.23.2 The decomposition of N2O5 in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 mol L–1 and after 184 minutes, it is reduced to 2.08 mol L–1. The reaction takes place according to the equation 2 N2O5 (g) ® 4 NO2 (g) + O2 (g) Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period? 1  ( 2.08 − 2 .33 ) mol L−1  1 ∆ [ N 2 O5 ] SolutionSolutionSolutionSolutionSolution Average Rate = − = − 184 min 2  ∆t  2   = 6.79 × 10–4 mol L–1/min = (6.79 × 10–4 mol L–1 min–1) × (60 min/1h) = 4.07 × 10–2 mol L–1/h = 6.79 × 10–4 mol L–1 × 1min/60s = 1.13 × 10–5 mol L–1s–1 It may be remembered that 1 ∆ [ NO 2 ] Rate = 4  ∆t  ∆ [ NO 2 ] = 6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1 ∆t IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.1 For the reaction R ® P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. 3.2 In a reaction, 2A ® Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval? 3.23.23.23.23.2 FactorsFactorsFactorsFactorsFactors InfluencingInfluencingInfluencingInfluencingInfluencing Rate of reaction depends upon the experimental conditions such RateRateRateRateRate ofofofofof aaaaa ReactionReactionReactionReactionReaction as concentration of reactants (pressure in case of gases), temperature and catalyst. 3.2.1 Dependence The rate of a chemical reaction at a given temperature may depend on of Rate on the concentration of one or more reactants and products. The Concentration representation of rate of reaction in terms of concentration of the reactants is known as rate law. It is also called as rate equation or rate expression. 3.2.2 Rate The results in Table 3.1 clearly show that rate of a reaction decreases with Expression the passage of time as the concentration of reactants decrease. Conversely, and Rate rates generally increase when reactant concentrations increase. So, rate of Constant a reaction depends upon the concentration of reactants. Chemistry 66 Reprint 2025-26 Consider a general reaction aA + bB ® cC + dD where a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is Rate µ [A] x [B] y (3.4) where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Above equation can also be written as Rate = k [A] x [B] y (3.4a) d [ R ] x y − = k [ A ] [ B ] (3.4b) d t This form of equation (3.4 b) is known as differential rate equation, where k is a proportionality constant called rate constant. The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression. Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. For example: 2NO(g) + O2(g) ® 2NO2 (g) We can measure the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. The following results are obtained (Table 3.2). Table 3.2: Initial rate of formation of NO2 Experiment Initial [NO]/ mol L-1 Initial [O2]/ mol L-1 Initial rate of formation of NO2/ mol L-1s-1 1. 0.30 0.30 0.096 2. 0.60 0.30 0.384 3. 0.30 0.60 0.192 4. 0.60 0.60 0.768 It is obvious, after looking at the results, that when the concentration of NO is doubled and that of O2 is kept constant then the initial rate increases by a factor of four from 0.096 to 0.384 mol L–1s–1. This indicates that the rate depends upon the square of the concentration of NO. When concentration of NO is kept constant and concentration of O2 is doubled the rate also gets doubled indicating that rate depends on concentration of O2 to the first power. Hence, the rate equation for this reaction will be Rate = k [NO] 2[O2] 67 Chemical Kinetics Reprint 2025-26 The differential form of this rate expression is given as d [ R ] 2 − = k [ NO ] [ O 2 ] d t Now, we observe that for this reaction in the rate equation derived from the experimental data, the exponents of the concentration terms are the same as their stoichiometric coefficients in the balanced chemical equation. Some other examples are given below: Reaction Experimental rate expression 1. CHCl3 + Cl2 ® CCl4 + HCl Rate = k [CHCl3 ] [Cl2]1/2 2. CH3COOC2H5 + H2O ® CH3COOH + C2H5OH Rate = k [CH3COOC2H5]1 [H2O]0 In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that: Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally. 3.2.3 Order of a In the rate equation (3.4) Reaction Rate = k [A]x [B]y x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of reaction is independent of the concentration of reactants. ExampleExampleExampleExampleExample 3.33.33.33.33.3 Calculate the overall order of a reaction which has the rate expression (a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 SolutionSolutionSolutionSolutionSolution (a) Rate = k [A]x [B]y order = x + y So order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order. A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. When a sequence of elementary reactions (called mechanism) gives us the products, the reactions are called complex reactions. Chemistry 68 Reprint 2025-26 These may be consecutive reactions (e.g., oxidation of ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reverse reactions and side reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol). Units of rate constant For a general reaction aA + bB ® cC + dD Rate = k [A]x [B]y Where x + y = n = order of the reaction Rate k = x [A] [B]y concentration 1 = × n ( where [A] = [B]) time ( concentration ) Taking SI units of concentration, mol L –1 and time, s, the units of k for different reaction order are listed in Table 3.3 Table 3.3: Units of rate constant Reaction Order Units of rate constant mol L−1 1 −1 − 1 × 0 = mol L s −1 Zero order reaction 0 s ( mol L ) −1 mol L 1 −1 × = s − 1 1 First order reaction 1 s ( mol L ) − 1 mol L 1 − 1 −1 × = mol L s −1 2 Second order reaction 2 s ( mol L ) Identify the reaction order from each of the following rate constants. ExampleExampleExampleExampleExample 3.43.43.43.43.4 (i) k = 2.3 × 10–5 L mol–1 s–1 (ii) k = 3 × 10–4 s–1 (i) The unit of second order rate constant is L mol–1 s–1, therefore SolutionSolutionSolutionSolutionSolution k = 2.3 × 10–5 L mol–1 s–1 represents a second order reaction. (ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction. 3.2.4 Molecularity Another property of a reaction called molecularity helps in of a understanding its mechanism. The number of reacting species Reaction (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. The reaction can be unimolecular when one reacting species is involved, for example, decomposition of ammonium nitrite. 69 Chemical Kinetics Reprint 2025-26 NH4NO2 ® N2 + 2H2O Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI ® H2 + I2 Trimolecular or termolecular reactions involve simultaneous collision between three reacting species, for example, 2NO + O2 ® 2NO2 The probability that more than three molecules can collide and react simultaneously is very small. Hence, reactions with the molecularity three are very rare and slow to proceed. It is, therefore, evident that complex reactions involving more than three molecules in the stoichiometric equation must take place in more than one step. KClO3 + 6FeSO4 + 3H2SO4 ® KCl + 3Fe2(SO4)3 + 3H2O This reaction which apparently seems to be of tenth order is actually a second order reaction. This shows that this reaction takes place in several steps. Which step controls the rate of the overall reaction? The question can be answered if we go through the mechanism of reaction, for example, chances to win the relay race competition by a team depend upon the slowest person in the team. Similarly, the overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium. -I 2H2O2  2H2O + O2 Alkaline medium The rate equation for this reaction is found to be d  H 2 O 2   Rate  k  H2 O 2 I dt This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps (1) H2O2 + I– ® H2O + IO– (2) H2O2 + IO– ® H2O + I– + O2 Both the steps are bimolecular elementary reactions. Species IO- is called as an intermediate since it is formed during the course of the reaction but not in the overall balanced equation. The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction. Thus, from the discussion, till now, we conclude the following: (i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non integer. (ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning. Chemistry 70 Reprint 2025-26 (iii) For complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.3 For a reaction, A + B ® Product; the rate law is given by, r = k [ A]1/2 [B]2. What is the order of the reaction? 3.4 The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ? 3.33.33.33.33.3 IntegratedIntegratedIntegratedIntegratedIntegrated We have already noted that the concentration dependence of rate is RateRateRateRateRate called differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by determination EquationsEquationsEquationsEquationsEquations of slope of the tangent at point ‘t’ in concentration vs time plot (Fig. 3.1). This makes it difficult to determine the rate law and hence the order of the reaction. In order to avoid this difficulty, we can integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions. 3.3.1 Zero Order Zero order reaction means that the rate of the reaction is proportional Reactions to zero power of the concentration of reactants. Consider the reaction, R ® P d  R  Rate =   k  R 0 d t As any quantity raised to power zero is unity d  R  Rate =   k × 1 d t d[R] = – k dt Integrating both sides [R] = – k t + I (3.5) where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation (3.5) [R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation (3.5) [R] = -kt + [R]0 (3.6) 71 Chemical Kinetics Reprint 2025-26 Comparing (3.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight [R0 ] line (Fig. 3.3) with slope = –k and intercept equal to [R]0. Further simplifying equation (3.6), we get the rateR k = -slope constant, k as of [ R ]0 − [ R ] k = (3.7) t Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme Concentration catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order 0 Time reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at Fig. 3.3: Variation in the concentration high pressure. vs time plot for a zero order 1130K reaction 2NH 3 ( g ) →Pt catalyst N 2 ( g ) +3H 2 ( g ) Rate = k [NH3]0 = k In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration. The thermal decomposition of HI on gold surface is another example of zero order reaction. 3.3.2 First Order In this class of reactions, the rate of the reaction is proportional to the Reactions first power of the concentration of the reactant R. For example, R ® P d [ R ] Rate = − = k [ R ] d t d [ R ] or = – kdt [ R ] Integrating this equation, we get ln [R] = – kt + I (3.8) Again, I is the constant of integration and its value can be determined easily. When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (3.8) can be written as ln [R]0 = –k × 0 + I ln [R]0 = I Substituting the value of I in equation (3.8) ln[R] = –kt + ln[R]0 (3.9) Chemistry 72 Reprint 2025-26 Rearranging this equation [ R ] ln = −kt [ R ]0 1 R 0 or k  ln (3.10) t R At time t1 from equation (3.8) *ln[R]1 = – kt1 + *ln[R]0 (3.11) At time t2 ln[R]2 = – kt2 + ln[R]0 (3.12) where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively. Subtracting (3.12) from (3.11) ln[R]1– ln[R]2 = – kt1 – (–kt2) [ R ]1 ln = k (t 2 − t 1 ) [ R ]2 1 [ R ]1 k = ln (t 2 − t 1 ) [ R ]2 (3.13) Equation (3.9) can also be written as [ R ] ln = −kt [ R ]0 Taking antilog of both sides [R] = [R]0 e–kt (3.14) Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]0 The first order rate equation (3.10) can also be written in the form 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 kt * log = [ R ] 2.303 If we plot a graph between log [R]0/[R] vs t, (Fig. 3.5), the slope = k/2.303 Hydrogenation of ethene is an example of first order reaction. C2H4(g) + H2 (g) ® C2H6(g) Rate = k [C2H4] All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. * Refer to Appendix-IV for ln and log (logarithms). 73 Chemical Kinetics Reprint 2025-26 /[R]) 0] Slope = k /2.303 ([R log 0 Time Fig. 3.4: A plot between ln[R] and t Fig. 3.5: Plot of log [R]0/[R] vs time for a for a first order reaction first order reaction 226 88 Ra  24 He  22286 Rn Rate = k [Ra] Decomposition of N2O5 and N2O are some more examples of first order reactions. ExampleExampleExampleExampleExample 3.53.53.53.53.5 The initial concentration of N2O5 in the following first order reaction N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1. Calculate the rate constant of the reaction at 318 K. SolutionSolutionSolutionSolutionSolution For a first order reaction  R 1 k t 2  t 1  log =  R 2 2.303 2.303  R 1 log k =  t 2  t 1   R 2 2.303 1.24  10  2 mol L1 log  2  1 =  60 min  0 min  0.20  10 mol L 2.303  1 = log 6.2 min 60 k = 0.0304 min-1 Let us consider a typical first order gas phase reaction A(g) ® B(g) + C(g) Let pi be the initial pressure of A and pt the total pressure at time ‘t’. Integrated rate equation for such a reaction can be derived as Total pressure pt = pA + pB + pC (pressure units) Chemistry 74 Reprint 2025-26 pA, pB and pC are the partial pressures of A, B and C, respectively. If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each. A(g) ® B(g) + C(g) At t = 0 pi atm 0 atm 0 atm At time t (pi–x) atm x atm x atm where, pi is the initial pressure at time t = 0. pt = (pi – x) + x + x = pi + x x = (pt - pi) where, pA = pi – x = pi – (pt – pi) = 2pi – pt  2.303   p i  k =   log  (3.16)  t  p A  2.303 p i log = t  2 p i  p t  The following data were obtained during the first order thermal ExampleExampleExampleExampleExample 3.63.63.63.63.6 decomposition of N2O5 (g) at constant volume: 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) S.No. Time/s Total Pressure/(atm) 1. 0 0.5 2. 100 0.512 Calculate the rate constant. Let the pressure of N2O5(g) decrease by 2x atm. As two moles of SolutionSolutionSolutionSolutionSolution N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g), the pressure of N2O4 (g) increases by 2x atm and that of O2 (g) increases by x atm. 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) Start t = 0 0.5 atm 0 atm 0 atm At time t (0.5 – 2x) atm 2x atm x atm pt = p N 2 O 5  p N 2 O 4  p O 2 = (0.5 – 2x) + 2x + x = 0.5 + x x = tp − 0.5 p N 2 O5 = 0.5 – 2x = 0.5 – 2 (pt – 0.5) = 1.5 – 2pt At t = 100 s; pt = 0.512 atm 75 Chemical Kinetics Reprint 2025-26 p N 2 O 5 = 1.5 – 2 × 0.512 = 0.476 atm Using equation (3.16) 2.303 p i 2.303 0.5 atm k  log  log t p A 100s 0.476 atm 2.303  4 1   0.0216  4.98  10 s 100s 3.3.3 Half-Life of The half-life of a reaction is the time in which the concentration of a a Reaction reactant is reduced to one half of its initial concentration. It is represented as t1/2. For a zero order reaction, rate constant is given by equation 3.7. [ R ]0 − [ R ] k = t 1 [ R ]0 At t = t 1/2 , [ R ] = 2 The rate constant at t1/2 becomes [ R ]0 − 1/2 [ R ]0 k = t 1/2 [ R ]0 t 1/2 = 2k It is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. For the first order reaction, 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 at t1/2 [ R ] = (3.16) 2 So, the above equation becomes 2.303 [ R ]0 k = log t 1/2 [ R ] /2 2.303 or t1/2  log 2 k 2.303 t 1/2 = × 0.301 k 0.693 t 1/2 = (3.17) k Chemistry 76 Reprint 2025-26 It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. For zero order reaction t1/2 µ [R]0. For first order reaction t1/2 is independent of [R]0. A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. ExampleExampleExampleExampleExample 3.73.73.73.73.7 Find the half-life of the reaction. Half-life for a first order reaction is SolutionSolutionSolutionSolutionSolution 0.693 t 1/2 = k 0.693 t 1/2 = –14 –1 = 1.26 × 1013s 5.5×10 s Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0 ExampleExampleExampleExampleExample 3.83.83.83.83.8 2.303  R 0 log k = SolutionSolutionSolutionSolutionSolution t  R  2.303  R 0 2.303 3 log = = log10 t  R 0  0.999  R 0 t t = 6.909/k For half-life of the reaction t1/2 = 0.693/k t 6.909 k =   10 t1/2 k 0.693 Table 3.4 summarises the mathematical features of integrated laws of zero and first order reactions. Table 3.4: Integrated Rate Laws for the Reactions of Zero and First Order Order Reaction Differential Integrated Straight Half- Units of k type rate law rate law line plot life 0 R® P d[R]/dt = -k kt = [R]0-[R] [R] vs t [R]0/2k conc time-1 or mol L–1s–1 1 R® P d[R]/dt = -k[R] [R] = [R]0e-kt ln[R] vs t ln 2/k time-1 or s–1 or kt = ln{[R]0/[R]} 77 Chemical Kinetics Reprint 2025-26 The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are given as under. H CH3COOH + C2H5OH CH3COOC2H5 + H2O  t = 0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.99 mol 0.01 mol 0.01 mol The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction. C12H22O11 + H2O →H+ C6H12O6 + C6H12O6 Cane sugar Glucose Fructose Rate = k [C12H22O11] IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.5 A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this reactant take to reduce to 3 g? 3.6 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. 3.43.43.43.43.4 TemperatureTemperatureTemperatureTemperatureTemperature Most of the chemical reactions are accelerated by increase in temperature. For example, in decomposition of N2O5, the time taken for half of the DependenceDependenceDependenceDependenceDependence ofofofofof original amount of material to decompose is 12 min at 50oC, 5 h at thethethethethe RateRateRateRateRate ofofofofof aaaaa 25oC and 10 days at 0oC. You also know that in a mixture of potassium ReactionReactionReactionReactionReaction permanganate (KMnO4) and oxalic acid (H2C2O4), potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation (3.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation. Chemistry 78 Reprint 2025-26 k = A e -Ea /RT (3.18) where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol –1). It can be understood clearly using the following simple reaction H 2 g  I 2 g  2HI g According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine Intermediate collide to form an unstable intermediate (Fig. 3.6). It exists for a very short time and then breaks up to form two Fig. 3.6: Formation of HI through molecules of hydrogen iodide. the intermediate The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea). Fig. 3.7 is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products. All the molecules in the reacting species do not have the same kinetic Fig. 3.7: Diagram showing plot of potential energy. Since it is difficult to predict the energy vs reaction coordinate behaviour of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules. According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given kinetic energy (E) vs kinetic energy (Fig. 3.8). Here, NE is the number of molecules with energy E and NT is total number of molecules. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number Fig. 3.8: Distribution curve showing energies of molecules with energies higher or among gaseous molecules lower than this value. When the 79 Chemical Kinetics Reprint 2025-26 temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 3.9) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of Ea on Fig. 3.9: Distribution curve showing temperature Maxwell Boltzmann distribution curve dependence of rate of a reaction (Fig. 3.9). Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. In the Arrhenius equation (3.18) the factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) E a ln k = – + ln A (3.19) RT The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. E a In Fig. 3.10, slope = – and intercept = ln R A. So we can calculate Ea and A using these values. At temperature T1, equation (3.19) is E a ln k1 = – RT1 + ln A (3.20) At temperature T2, equation (3.19) is E a ln k2 = – RT2 + ln A (3.21) (since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Fig. 3.10: A plot between ln k and 1/T Chemistry 80 Reprint 2025-26 Subtracting equation (3.20) from (3.21), we obtain E a E a ln k2 – ln k1 = RT1 – RT2 k 2 E a  1 1  ln = − k1 R  T1 T2  k 2 E a  1 1  log = − (3.22) k1 2.303 R  T1 T2  k 2 E a  T2 − T1  log = k1 2. 303R  T1T2  ExampleExampleExampleExampleExample 3.93.93.93.93.9 The rate constants of a reaction at 500K and 700K are 0.02s–1 and 0.07s–1 respectively. Calculate the values of Ea and A. k 2 E a  T2  T1 SolutionSolutionSolutionSolutionSolution log =   k1 2.303 R  T1T2  0.07  E a   700  500  log =      700  500 0.02  2.303  8.314 JK  1 mol  1  0.544 = Ea × 5.714 × 10-4/19.15 Ea = 0.544 × 19.15/5.714 × 10–4 = 18230.8 J Since k = Ae-Ea/RT × 500 0.02 = Ae-18230.8/8.314 A = 0.02/0.012 = 1.61 ExampleExampleExampleExampleExample 3.103.103.103.103.10 The first order rate constant for the decomposition of ethyl iodide by the reaction C2H5I(g) ® C2H4 (g) + HI(g) at 600K is 1.60 × 10–5 s–1. Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700K. SolutionSolutionSolutionSolutionSolution We know that E a  1 1     log k2 – log k1 = 2.303R  T1 T2  81 Chemical Kinetics Reprint 2025-26 E a  1 1  log k2 = log k1     2.303R  T1 T2  209000 J mol L 1  1 1  5 = log 1.60  10     1  1    2.303  8.314 J mol L K  600 K 700K  log k2 = – 4.796 + 2.599 = – 2.197 k2 = 6.36 × 10–3 s–1 3.4.1 Effect of A catalyst is a substance which increases the rate of a reaction without Catalyst itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase its rate considerably. 2KClO3 MnO2 2 KCl + 3O2 The word catalyst should not be used when the added substance reduces the rate of raction. The substance is then called inhibitor. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst. It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in Fig. 3.11. It is clear from Arrhenius equation (3.18) that lower the value of activation energy faster will be the rate of a reaction. A small amount of the catalyst can catalyse a large amount of reactants. A catalyst does Fig. 3.11: Effect of catalyst on activation energy not alter Gibbs energy, DG of a reaction. It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions. It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reactions to the same extent so that the equilibrium state remains same but is reached earlier. 3.53.53.53.53.5 CollisionCollisionCollisionCollisionCollision Though Arrhenius equation is applicable under a wide range of TheoryTheoryTheoryTheoryTheory ofofofofof circumstances, collision theory, which was developed by Max Trautz ChemicalChemicalChemicalChemicalChemical and William Lewis in 1916 -18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic ReactionsReactionsReactionsReactionsReactions theory of gases. According to this theory, the reactant molecules are Chemistry 82 Reprint 2025-26 assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Another factor which affects the rate of chemical reactions is activation energy (as we have already studied). For a bimolecular elementary reaction A + B ® Products rate of reaction can be expressed as a / RT (3.23) Rate = Z AB e − E where ZAB represents the collision frequency of reactants, A and B and e-Ea /RT represents the fraction of molecules with energies equal to or greater than Ea. Comparing (3.23) with Arrhenius equation, we can say that A is related to collision frequency. Equation (3.23) predicts the value of rate constants fairly accurately for the reactions that involve atomic species or simple molecules but for complex molecules significant deviations are observed. The reason could be that all collisions do not lead to the formation of products. The collisions in which molecules collide with sufficient kinetic energy (called threshold energy*) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collisions. For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown in Fig. 3.12. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed. Fig. 3.12: Diagram showing molecules having proper and To account for effective collisions, improper orientation another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e., − E a / RT Rate = PZ AB e Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and ignores their structural aspect. You will study details about this theory and more on other theories in your higher classes. * Threshold energy = Activation Energy + energy possessed by reacting species. 83 Chemical Kinetics Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.7 What will be the effect of temperature on rate constant ? 3.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. 3.9 The activation energy for the reaction 2 HI(g) ® H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? SummarySummarySummarySummarySummary Chemical kinetics is the study of chemical reactions with respect to reaction rates, effect of various variables, rearrangement of atoms and formation of intermediates. The rate of a reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. The order of a reaction is the sum of all such powers of concentration of terms for different reactants. Rate constant is the proportionality factor in the rate law. Rate constant and order of a reaction can be determined from rate law or its integrated rate equation. Molecularity is defined only for an elementary reaction. Its values are limited from 1 to 3 whereas order can be 0, 1, 2, 3 or even a fraction. Molecularity and order of an elementary reaction are same. Temperature dependence of rate constants is described by Arrhenius equation (k = Ae–Ea/RT). Ea corresponds to the activation energy and is given by the energy difference between activated complex and the reactant molecules, and A (Arrhenius factor or pre-exponential factor) corresponds to the collision frequency. The equation clearly shows that increase of temperature or lowering of Ea will lead to an increase in the rate of reaction and presence of a catalyst lowers the activation energy by providing an alternate path for the reaction. According to collision theory, another factor P called steric factor which refers to the orientation of molecules which collide, is important and contributes to effective collisions, thus, modifying the Arrhenius equation to k  P Z AB e  E a / RT . Chemistry 84 Reprint 2025-26 ExercisesExercisesExercisesExercisesExercises 3.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) ® N2O (g) Rate = k[NO]2 (ii) H2O2 (aq) + 3I– (aq) + 2H+ ® 2H2O (l) + 3I Rate = k[H2O2][I-] (iii) CH3CHO (g) ® CH4 (g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl (g) ® C2H4 (g) + HCl (g) Rate = k [C2H5Cl] 3.2 For the reaction: 2A + B ® A2B the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. 3.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1? 3.4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., 3/2 Rate = k ( p CH 3 OCH 3 ) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? 3.5 Mention the factors that affect the rate of a chemical reaction. 3.6 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 3.7 What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 3.8 In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L–1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.

2.2 Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V Arrange these metals in their increasing order of reducing power.

2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) ® 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) ® Fe3+(aq) + Ag(s) Calculate the DrGo and equilibrium constant of the reactions. 2.5 Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s) (ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)| Pt(s) (iii) Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s) (iv) Pt(s)|Br–(0.010 M)|Br2(l )||H+(0.030 M)| H2(g) (1 bar)|Pt(s).

2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.

2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 59 Electrochemistry Reprint 2025-26

2.6 In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) ® Zn2+(aq) + 2Ag(s) + 2OH–(aq) Determine DrGo and Eo for the reaction.

9.9 Give the structures of A, B and C in the following reactions: B NaOH  Br 2 C (i) CH 3 CH 2 I NaCN A Partial OHhydrolysis CuCN H 2 O/H  NH 3 (ii) C 6 H 5 N 2 Cl  A  B  C 4 B HNO 2 C (iii) CH 3 CH 2 Br KCN A LiAlH 0  C  2  2 O/H HCl B H C  (iv) C6 H 5 NO 2 Fe/HCl A NaNO273 K 3 A NaOBr B NaNO 2 /HCl C (v) CH 3 COOH NH 2 B C 6 H 5 OH C (vi) C6 H 5 NO 2 Fe/HCl A HNO273K 279 Amines Reprint 2025-26

9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

9.5 How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid? 9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

9.7 Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

9.8 Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-chloroaniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.

9.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C. 9.11 Complete the following reactions: (i) C6 H 5 NH 2  CHCl 3  alc.KOH  (ii) C 6 H 5 N 2 Cl  H 3 PO 2  H 2 O  (iii) C6 H 5 NH 2  H 2 SO 4  conc.  (iv) C6 H 5 N 2 Cl  C 2 H 5 OH  (v) C6 H5 NH 2  Br2  aq   O  (vi) C6 H 5 NH 2   CH 3 CO  2 4  (vii) C 6 H5 N 2 Cl ii NaNOi HBF2 /Cu, 9.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis? 9.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid. 9.14 Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines? Answers to Some Intext Questions 9.4 (i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH (ii) C6H5NH2 < C2H5NH2. < (C2H5)3N < (C2H5)2NH (iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH Chemistry 280 Reprint 2025-26 UnitUnitUnitUnit Unit1010 BiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesObjectives After studying this Unit, you will be able to “It is the harmonious and synchronous progress of chemical • explain the characteristics of reactions in body which leads to life”. biomolecules like carbohydrates, proteins and nucleic acids and hormones; A living system grows, sustains and reproduces itself. • classify carbohydrates, proteins, The most amazing thing about a living system is that it nucleic acids and vitamins on the basis of their structures; is composed of non-living atoms and molecules. The • explain the difference between pursuit of knowledge of what goes on chemically within DNA and RNA; a living system falls in the domain of biochemistry. Living • describe the role of biomolecules systems are made up of various complex biomolecules in biosystem. like carbohydrates, proteins, nucleic acids, lipids, etc. Proteins and carbohydrates are essential constituents of our food. These biomolecules interact with each other and constitute the molecular logic of life processes. In addition, some simple molecules like vitamins and mineral salts also play an important role in the functions of organisms. Structures and functions of some of these biomolecules are discussed in this Unit. 10.110.110.110.110.1 CarbohydratesCarbohydratesCarbohydratesCarbohydratesCarbohydrates Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. Some common examples of carbohydrates are cane sugar, glucose, starch, etc. Most of them have a general formula, Cx(H2O)y, and were considered as hydrates of carbon from where the name carbohydrate was derived. For example, the molecular formula of glucose (C6H12O6) fits into this general formula, C6(H2O)6. But all the compounds which fit into this formula may not be classified as carbohydrates. For example acetic acid (CH3COOH) fits into this general formula, C2(H2O)2 but is not a carbohydrate. Similarly, rhamnose, C6H12O5 is a carbohydrate but does not fit in this definition. A large number of their reactions have shown that they contain specific functional groups. Chemically, the carbohydrates may be defined as optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Some of the carbohydrates, which are sweet in taste, are also called sugars. The most common sugar, used in our homes is named as sucrose whereas the sugar present Reprint 2025-26 in milk is known as lactose. Carbohydrates are also called saccharides (Greek: sakcharon means sugar). Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have been broadly divided into following three groups. 10.1.1 (i) Monosaccharides: A carbohydrate that cannot be hydrolysed further Classification of to give simpler unit of polyhydroxy aldehyde or ketone is called a Carbohydrates monosaccharide. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc. (ii) Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. They are further classified as disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the number of monosaccharides, they provide on hydrolysis. Amongst these the most common are disaccharides. The two monosaccharide units obtained on hydrolysis of a disaccharide may be same or different. For example, one molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose whereas maltose gives two molecules of only glucose. (iii) Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Some common examples are starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste, hence they are also called non-sugars. The carbohydrates may also be classified as either reducing or non- reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars. 10.1.2 Monosaccharides are further classified on the basis of number of carbon Monosaccharides atoms and the functional group present in them. If a monosaccharide contains an aldehyde group, it is known as an aldose and if it contains a keto group, it is known as a ketose. Number of carbon atoms constituting the monosaccharide is also introduced in the name as is evident from the examples given in Table 10.1 Table 10.1: Different Types of Monosaccharides Carbon atoms General term Aldehyde Ketone 3 Triose Aldotriose Ketotriose 4 Tetrose Aldotetrose Ketotetrose 5 Pentose Aldopentose Ketopentose 6 Hexose Aldohexose Ketohexose 7 Heptose Aldoheptose Ketoheptose 10.1.2.1 Glucose Glucose occurs freely in nature as well as in the combined form. It is present in sweet fruits and honey. Ripe grapes also contain glucose in large amounts. It is prepared as follows: Preparation of 1. From sucrose (Cane sugar): If sucrose is boiled with dilute HCl or Glucose H2SO4 in alcoholic solution, glucose and fructose are obtained in equal amounts. Chemistry 282 Reprint 2025-26 H+ C12 H 22 O11 + H 2 O → C 6 H12 O 6 + C 6 H12 O 6 Sucrose Glucose Fructose 2. From starch: Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H2SO4 at 393 K under pressure. H + (C 6 H10 O 5 ) n + nH 2 O →393K; 2-3 atm nC 6 H12 O 6 Starch or cellulose Glucose Glucose is an aldohexose and is also known as dextrose. It is theStructure of monomer of many of the larger carbohydrates, namely starch, cellulose.Glucose It is probably the most abundant organic compound on earth. It was assigned the structure given below on the basis of the following evidences: CHO 1. Its molecular formula was found to be C6H12O6. 2. On prolonged heating with HI, it forms n-hexane, suggesting that all(CH OH )4 the six carbon atoms are linked in a straight chain. CH2OH Glucose 3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C = O) in glucose. 4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group. CHO COOH Br2 water (CHOH )4 (CH OH )4 CH2OH CH2OH Gluconic acid 5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms. 283 Biomolecules Reprint 2025-26 6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose. CHO COOH COOH Oxidation Oxidation (CHOH)4 (CHOH )4 (CHOH )4 CH 2OH COOH CH2 OH Saccharic Gluconic acid acid The exact spatial arrangement of different —OH groups was given by Fischer after studying many other properties. Its configuration is correctly represented as I. So gluconic acid is represented as II and saccharic acid as III. CHO COOH COOH H OH H OH H OH OH H OH H OH H H OH H OH H OH H OH H OH H OH CH2OH CH2OH COOH I II III Glucose is correctly named as D(+)-glucose. ‘D’ before the name of glucose represents the configuration whereas ‘(+)’ represents dextrorotatory nature of the molecule. It should be remembered that ‘D’ and ‘L’ have no relation with the optical activity of the compound. They are also not related to letter ‘d’ and ‘l’ (see Unit 6). The meaning of D– and L– notations is as follows. The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer of a compound with respect to configuration of some other compound, configuration of which is known. In the case of carbohydrates, this refers to their relation with a particular isomer of glyceraldehyde. Glyceraldehyde contains one asymmetric carbon atom and exists in two enantiomeric forms as shown below. (+) Isomer of glyceraldehyde has ‘D’ configuration. It means that when its structural formula is written on paper following specific conventions which you will study in higher classes, the –OH group lies on right hand side in the structure. All those compounds which can be chemically correlated to D (+) isomer of glyceraldehyde are said to have D- configuration whereas those which can be correlated to ‘L’ (–) isomer of glyceraldehyde are said to have L—configuration. In L (–) isomer –OH group is on left hand side as you can see in the structure. For assigning Chemistry 284 Reprint 2025-26 the configuration of monosaccharides, it is the lowest asymmetric carbon atom (as shown below) which is compared. As in (+) glucose, —OH on the lowest asymmetric carbon is on the right side which is comparable to (+) glyceraldehyde, so (+) glucose is assigned D-configuration. Other asymmetric carbon atoms of glucose are not considered for this comparison. Also, the structure of glucose and glyceraldehyde is written in a way that most oxidised carbon (in this case –CHO)is at the top. CHO H OH OH H CHO H OH H OH H OH CH2OH CH2OH D– (+) – Glyceraldehyde D–(+) – Glucose Cyclic The structure (I) of glucose explained most of its properties but the Structure following reactions and facts could not be explained by this structure. of Glucose 1. Despite having the aldehyde group, glucose does not give Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3. 2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group. 3. Glucose is found to exist in two different crystalline forms which are named as a and b. The a-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the b-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K. This behaviour could not be explained by the open chain structure (I) for glucose. It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. This explains the absence of —CHO group and also existence of glucose in two forms as shown below. These two cyclic forms exist in equilibrium with open chain structure. The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C1, called anomeric carbon 285 Biomolecules Reprint 2025-26 (the aldehyde carbon before cyclisation). Such isomers, i.e., a-form and b-form, are called anomers. The six membered cyclic structure of glucose is called pyranose structure (a– or b–), in analogy with pyran. Pyran is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring. The cyclic structure of glucose is more correctly represented by Haworth structure as given below. 10.1.2.2 Fructose Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide, sucrose. It is a natural monosaccharide found in fruits, honey and vegetables. In its pure form it is used as a sweetner. It is also an important ketohexose. Fructose also has the molecular formula C6H12O6 and Structure on the basis of its reactions it was found to contain a of Fructose ketonic functional group at carbon number 2 and six carbons in straight chain as in the case of glucose. It belongs to D-series and is a laevorotatory compound. It is appropriately written as D-(–)-fructose. Its open chain structure is as shown. It also exists in two cyclic forms which are obtained by the addition of —OH at C5 to the ( ) group. The ring, thus formed is a five membered ring and is named as furanose with analogy to the compound furan. Furan is a five membered cyclic compound with one oxygen and four carbon atoms. The cyclic structures of two anomers of fructose are represented by Haworth structures as given. Chemistry 286 Reprint 2025-26 10.1.3 You have already read that disaccharides on hydrolysis with dilute Disaccharides acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. In disaccharides, if the reducing groups of monosaccharides i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars, e.g., sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose. (i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. These two monosaccharides are held together by a glycosidic linkage between C1 of a-D-glucose and C2 of b-D-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar. Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and the product is named as invert sugar. (ii) Maltose: Another disaccharide, maltose is composed of two a-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C1 of second glucose in solution and it shows reducing properties so it is a reducing sugar. 287 Biomolecules Reprint 2025-26 (iii) Lactose: It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of b-D-galactose and b-D-glucose. The linkage is between C1 of galactose and C4 of glucose. Free aldehyde group may be produced at C-1 of glucose unit, hence it is also a reducing sugar. 10.1.4 Polysaccharides Polysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. These are the most commonly encountered carbohydrates in nature. They mainly act as the food storage or structural materials. (i) Starch: Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of a-glucose and consists of two components— Amylose and Amylopectin. Amylose is water soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 a-D-(+)-glucose units held together by C1– C4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80- 85% of starch. It is a branched chain polymer of a-D-glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage. Chemistry 288 Reprint 2025-26 (ii) Cellulose: Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain polysaccharide composed only of b-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit. (iii) Glycogen: The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin and is rather more highly branched. It is present in liver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi. 10.1.5 Carbohydrates are essential for life in both plants and animals. They Importance of form a major portion of our food. Honey has been used for a long time Carbohydrates as an instant source of energy by ‘Vaids’ in ayurvedic system of medicine. Carbohydrates are used as storage molecules as starch in plants and glycogen in animals. Cell wall of bacteria and plants is made up of cellulose. We build furniture, etc. from cellulose in the form 289 Biomolecules Reprint 2025-26 of wood and clothe ourselves with cellulose in the form of cotton fibre. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. Two aldopentoses viz. D-ribose and 2-deoxy-D-ribose (Section 10.5.1, Class XII) are present in nucleic acids. Carbohydrates are found in biosystem in combination with many proteins and lipids. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. 10.2 What are the expected products of hydrolysis of lactose? 10.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? 10.210.210.210.210.2 ProteinsProteinsProteinsProteinsProteins Proteins are the most abundant biomolecules of the living system. Chief sources of proteins are milk, cheese, pulses, peanuts, fish, meat, etc. They occur in every part of the body and form the fundamental basis of structure and functions of life. They are also required for growth and maintenance of body. The word protein is derived from Greek word, “proteios” which means primary or of prime importance. All proteins are polymers of a-amino acids. 10.2.1 Amino Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups. Depending upon the relative position of amino group with Acids respect to carboxyl group, the amino acids can be R CH COOH classified as a, b, g, d and so on. Only a-amino acids are obtained on hydrolysis of proteins. They NH2 may contain other functional groups also. a-amino acid All a-amino acids have trivial names, which (R = side chain) usually reflect the property of that compound or its source. Glycine is so named since it has sweet taste (in Greek glykos means sweet) and tyrosine was first obtained from cheese (in Greek, tyros means cheese.) Amino acids are generally represented by a three letter symbol, sometimes one letter symbol is also used. Structures of some commonly occurring amino acids along with their 3-letter and 1-letter symbols are given in Table 10.2. COOH Table 10.2: Natural Amino Acids H2N H R Name of the Characteristic feature Three letter One letter amino acids of side chain, R symbol code 1. Glycine H Gly G 2. Alanine – CH3 Ala A 3. Valine* (H3C)2CH- Val V 4. Leucine* (H3C)2CH-CH2- Leu L Chemistry 290 Reprint 2025-26 5. Isoleucine* H3C-CH2-CH- Ile I | CH3 6. Arginine* HN=C-NH-(CH2)3- Arg R | NH2 7. Lysine* H2N-(CH2)4- Lys K 8. Glutamic acid HOOC-CH2-CH2- Glu E 9. Aspartic acid HOOC-CH2- Asp D O || 10. Glutamine H2N-C-CH2-CH2- Gln Q O || 11. Asparagine H2N-C-CH2- Asn N 12. Threonine* H3C-CHOH- Thr T 13. Serine HO-CH2- Ser S 14. Cysteine HS-CH2- Cys C 15. Methionine* H3C-S-CH2-CH2- Met M 16. Phenylalanine* C6H5-CH2- Phe F 17. Tyrosine (p)HO-C6H4-CH2- Tyr Y –CH2 18. Tryptophan* Trp W N H 19. Histidine* His H 20. Proline Pro P * essential amino acid, a = entire structure 10.2.2 Amino acids are classified as acidic, basic or neutral depending upon Classification of the relative number of amino and carboxyl groups in their molecule. Amino Acids Equal number of amino and carboxyl groups makes it neutral; more number of amino than carboxyl groups makes it basic and more carboxyl groups as compared to amino groups makes it acidic. The amino acids, which can be synthesised in the body, are known as non- essential amino acids. On the other hand, those which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids (marked with asterisk in Table 10.2). 291 Biomolecules Reprint 2025-26 Amino acids are usually colourless, crystalline solids. These are water-soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges. In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases. Except glycine, all other naturally occurring a-amino acids are optically active, since the a-carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L-configuration. L-Aminoacids are represented by writing the –NH2 group on left hand side. 10.2.3 Structure You have already read that proteins are the polymers of a-amino acids of Proteins and they are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between –COOH group and –NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other. This results in the elimination of a water molecule and formation of a peptide bond –CO–NH–. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine. If a third amino acid combines to a dipeptide, the product is called a tripeptide. A tripeptide contains three amino acids linked by two peptide linkages. Similarly when four, five or six amino acids are linked, the respective products are known as tetrapeptide, pentapeptide or hexapeptide, respectively. When the number of such amino acids is more than ten, then the products are called polypeptides. A polypeptide with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a protein. However, the distinction between a polypeptide and a protein is not very sharp. Polypeptides with fewer amino acids are likely to be called proteins if they ordinarily have a well defined conformation of a protein such as insulin which contains 51 amino acids. Proteins can be classified into two types on the basis of their molecular shape. (a) Fibrous proteins When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibre– like structure is formed. Such proteins are generally insoluble in water. Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc. Chemistry 292 Reprint 2025-26 (b) Globular proteins This structure results when the chains of polypeptides coil around to give a spherical shape. These are usually soluble in water. Insulin and albumins are the common examples of globular proteins. Structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quaternary, each level being more complex than the previous one. (i) Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amino acids creates a different protein. (ii) Secondary structure of proteins: The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz. a-helix and b-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between and –NH– groups of the peptide bond. a-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the Fig. 10.1: a-Helix –NH group of each amino acid residue hydrogen bonded to the structure of proteins C O of an adjacent turn of the helix as shown in Fig.10.1. In b-pleated sheet structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as b-pleated sheet. (iii) Tertiary structure of proteins: The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous and globular. The main forces which stabilise the 2° and 3° structures of proteins are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction. Fig. 10.2: b-Pleated sheet structure of (iv) Quaternary structure of proteins: Some of the proteins proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other is known as quaternary structure. 293 Biomolecules Reprint 2025-26 A diagrammatic representation of all these four structures is given in Figure 10.3 where each coloured ball represents an amino acid. Fig. 10.3: Diagrammatic representation of protein structure (two sub-units of two types in quaternary structure) Fig. 10.4: Primary, secondary, tertiary and quaternary structures of haemoglobin 10.2.4 Protein found in a biological system with a unique three-dimensional Denaturation of structure and biological activity is called a native protein. When a Proteins protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of Chemistry 294 Reprint 2025-26 protein. During denaturation secondary and tertiary structures are destroyed but primary structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. 10.5 Where does the water present in the egg go after boiling the egg? 10.310.310.310.310.3 EnzymesEnzymesEnzymesEnzymesEnzymes Life is possible due to the coordination of various chemical reactions in living organisms. An example is the digestion of food, absorption of appropriate molecules and ultimately production of energy. This process involves a sequence of reactions and all these reactions occur in the body under very mild conditions. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular proteins. Enzymes are very specific for a particular reaction and for a particular substrate. They are generally named after the compound or class of compounds upon which they work. For example, the enzyme that catalyses hydrolysis of maltose into glucose is named as maltase. Maltase C12 H 22 O11  2 C 6 H12 O 6 Maltose G lucose Sometimes enzymes are also named after the reaction, where they are used. For example, the enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are named as oxidoreductase enzymes. The ending of the name of an enzyme is -ase. 10.3.1 Mechanism Enzymes are needed only in small quantities for the progress of a reaction. of Enzyme Similar to the action of chemical catalysts, enzymes are said to reduce Action the magnitude of activation energy. For example, activation energy for acid hydrolysis of sucrose is 6.22 kJ mol–1, while the activation energy is only 2.15 kJ mol–1 when hydrolysed by the enzyme, sucrase. Mechanism for the enzyme action has been discussed. 10.410.410.410.410.4 VitaminsVitaminsVitaminsVitaminsVitamins It has been observed that certain organic compounds are required in small amounts in our diet but their deficiency causes specific diseases. These compounds are called vitamins. Most of the vitamins cannot be synthesised in our body but plants can synthesise almost all of them, so they are considered as essential food factors. However, the bacteria of the gut can produce some of the vitamins required by us. All the vitamins are generally available in our diet. Different vitamins belong to various chemical classes and it is difficult to define them on the basis of structure. They are generally regarded as organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth 295 Biomolecules Reprint 2025-26 and health of the organism. Vitamins are designated by alphabets A, B, C, D, etc. Some of them are further named as sub-groups e.g. B1, B2, B6, B12, etc. Excess of vitamins is also harmful and vitamin pills should not be taken without the advice of doctor. The term “Vitamine” was coined from the word vital + amine since the earlier identified compounds had amino groups. Later work showed that most of them did not contain amino groups, so the letter ‘e’ was dropped and the term vitamin is used these days. 10.4.1 Vitamins are classified into two groups depending upon their solubility Classification of in water or fat. Vitamins (i) Fat soluble vitamins: Vitamins which are soluble in fat and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. (ii) Water soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body. Some important vitamins, their sources and diseases caused by their deficiency are listed in Table 10.3. Table 10.3: Some important Vitamins, their Sources and their Deficiency Diseases Sl. Name of Sources Deficiency diseases No. Vitamins 1. Vitamin A Fish liver oil, carrots, X e r o p h t h a l m i a butter and milk (hardening of cornea of eye) Night blindness 2. Vitamin B1 Yeast, milk, green Beri beri (loss of appe- (Thiamine) vegetables and cereals tite, retarded growth) 3. Vitamin B2 Milk, eggwhite, liver, Cheilosis (fissuring at (Riboflavin) kidney corners of mouth and lips), digestive disorders and burning sensation of the skin. 4. Vitamin B6 Yeast, milk, egg yolk, Convulsions (Pyridoxine) cereals and grams 5. Vitamin B12 Meat, fish, egg and Pernicious anaemia curd (RBC deficient in haemoglobin) 6. Vitamin C Citrus fruits, amla and Scurvy (bleeding gums) (Ascorbic acid) green leafy vegetables 7. Vitamin D Exposure to sunlight, Rickets (bone deformities fish and egg yolk in children) and osteo- malacia (soft bones and joint pain in adults) Chemistry 296 Reprint 2025-26 8. Vitamin E Vegetable oils like wheat Increased fragility of germ oil, sunflower oil, RBCs and muscular etc. weakness 9. Vitamin K Green leafy vegetables Increased blood clotting time 10.5.5.5.5.5 NucleicNucleicNucleicNucleicNucleic AcidsAcidsAcidsAcidsAcids Every generation of each and every species resembles its ancestors in many ways. How are these characteristics transmitted from one generation to the next? It has been observed that nucleus of a living cell is responsible for this transmission of inherent characters, also called heredity. The particles in nucleus of the cell, responsible for heredity, are called chromosomes which are made up of proteins and another type of biomolecules called nucleic acids. These are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Since nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides. James Dewey Watson Born in Chicago, Illinois, in 1928, Dr Watson received his Ph.D. (1950) from Indiana University in Zoology. He is best known for his discovery of the structure of DNA for which he shared with Francis Crick and Maurice Wilkins the 1962 Nobel prize in Physiology and Medicine. They proposed that DNA molecule takes the shape of a double helix, an elegantly simple structure that resembles a gently twisted ladder. The rails of the ladder are made of alternating units of phosphate and the sugar deoxyribose; the rungs are each composed of a pair of purine/ pyrimidine bases. This research laid the foundation for the emerging field of molecular biology. The complementary pairing of nucleotide bases explains how identical copies of parental DNA pass on to two daughter cells. This research launched a revolution in biology that led to modern recombinant DNA techniques. 10.5.1 Chemical Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric Composition acid and nitrogen containing heterocyclic compounds (called bases). In of Nucleic DNA molecules, the sugar moiety is b-D-2-deoxyribose whereas in Acids RNA molecule, it is b-D-ribose. 297 Biomolecules Reprint 2025-26 DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains four bases, the first three bases are same as in DNA but the fourth one is uracil (U). Cytosine (C) Thymine (T) Uracil (U) 10.5.2 Structure A unit formed by the attachment of a base to 1¢ position of sugar is of Nucleic known as nucleoside. In nucleosides, the sugar carbons are numbered Acids as 1¢, 2¢, 3¢, etc. in order to distinguish these from the bases (Fig. 10.5a). When nucleoside is linked to phosphoric acid at 5¢-position of sugar moiety, we get a nucleotide (Fig. 10.5). Fig. 10.5: Structure of (a) a nucleoside and (b) a nucleotide Nucleotides are joined together by phosphodiester linkage between 5¢ and 3¢ carbon atoms of the pentose sugar. The formation of a typical dinucleotide is shown in Fig. 10.6. Chemistry 298 Reprint 2025-26 Fig. 10.6: Formation of a dinucleotide A simplified version of nucleic acid chain is as shown below. Base Base Base Sugar Phosphate Sugar Phosphate Sugar n Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also. James Watson and Francis Crick gave a double strand helix structure for DNA (Fig. 10.7). Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine. In secondary structure of RNA single stranded helics is present which sometimes foldsback on itself. RNA molecules are of three types and they perform different functions. They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA). Fig. 10.7: Double strand helix structure for DNA 299 Biomolecules Reprint 2025-26 Har Gobind Khorana Har Gobind Khorana, was born in 1922. He obtained his M.Sc. degree from Punjab University in Lahore. He worked with Professor Vladimir Prelog, who moulded Khorana’s thought and philosophy towards science, work and effort. After a brief stay in India in 1949, Khorana went back to England and worked with Professor G.W. Kenner and Professor A.R.Todd. It was at Cambridge, U.K. that he got interested in both proteins and nucleic acids. Dr Khorana shared the Nobel Prize for Medicine and Physiology in 1968 with Marshall Nirenberg and Robert Holley for cracking the genetic code. DNA Fingerprinting It is known that every individual has unique fingerprints. These occur at the tips of the fingers and have been used for identification for a long time but these can be altered by surgery. A sequence of bases on DNA is also unique for a person and information regarding this is called DNA fingerprinting. It is same for every cell and cannot be altered by any known treatment. DNA fingerprinting is now used (i) in forensic laboratories for identification of criminals. (ii) to determine paternity of an individual. (iii) to identify the dead bodies in any accident by comparing the DNA’s of parents or children. (iv) to identify racial groups to rewrite biological evolution. 10.5.3 Biological DNA is the chemical basis of heredity and may be regarded as the reserve Functions of genetic information. DNA is exclusively responsible for maintaining of Nucleic the identity of different species of organisms over millions of years. A Acids DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells. Another important function of nucleic acids is the protein synthesis in the cell. Actually, the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA. 10.610.610.610.610.6 HormonesHormonesHormonesHormonesHormones Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly in the blood stream which transports them to the site of action. In terms of chemical nature, some of these are steroids, e.g., estrogens and androgens; some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine. Hormones have several functions in the body. They help to maintain the balance of biological activities in the body. The role of insulin in keeping the blood glucose level within the narrow limit is an example of this function. Insulin is released in response to the rapid rise in blood glucose level. On the other hand hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally low level of thyroxine leads Chemistry 300 Reprint 2025-26 to hypothyroidism which is characterised by lethargyness and obesity. Increased level of thyroxine causes hyperthyroidism. Low level of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This condition is largely being controlled by adding sodium iodide to commercial table salt (“Iodised” salt). Steroid hormones are produced by adrenal cortex and gonads (testes in males and ovaries in females). Hormones released by the adrenal cortex play very important role in the functions of the body. For example, glucocorticoids control the carbohydrate metabolism, modulate inflammatory reactions, and are involved in reactions to stress. The mineralocorticoids control the level of excretion of water and salt by the kidney. If adrenal cortex does not function properly then one of the results may be Addison’s disease characterised by hypoglycemia, weakness and increased susceptibility to stress. The disease is fatal unless it is treated by glucocorticoids and mineralocorticoids. Hormones released by gonads are responsible for development of secondary sex characters. Testosterone is the major sex hormone produced in males. It is responsible for development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It is responsible for development of secondary female characteristics and participates in the control of menstrual cycle. Progesterone is responsible for preparing the uterus for implantation of fertilised egg. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.6 Why cannot vitamin C be stored in our body? 10.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? 10.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA? SummarySummarySummarySummarySummary Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis. They are broadly classified into three groups — monosaccharides, disaccharides and polysaccharides. Glucose, the most important source of energy for mammals, is obtained by the digestion of starch. Monosaccharides are held together by glycosidic linkages to form disaccharides or polysaccharides. Proteins are the polymers of about twenty different a-amino acids which are linked by peptide bonds. Ten amino acids are called essential amino acids because they cannot be synthesised by our body, hence must be provided through diet. Proteins perform various structural and dynamic functions in the organisms. Proteins which contain only a-amino acids are called simple proteins. The secondary or tertiary structure of proteins get disturbed on change of pH or temperature and they are not able to perform their functions. This is called denaturation of proteins. Enzymes are biocatalysts which speed up the reactions in biosystems. They are very specific and selective in their action and chemically majority of enzymes are proteins. Vitamins are accessory food factors required in the diet. They are classified as fat soluble (A, D, E and K) and water soluble (B group and C). Deficiency of vitamins leads to many diseases. 301 Biomolecules Reprint 2025-26 Nucleic acids are the polymers of nucleotides which in turn consist of a base, a pentose sugar and phosphate moiety. Nucleic acids are responsible for the transfer of characters from parents to offsprings. There are two types of nucleic acids — DNA and RNA. DNA contains a five carbon sugar molecule called 2-deoxyribose whereas RNA contains ribose. Both DNA and RNA contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in RNA. The structure of DNA is a double strand whereas RNA is a single strand molecule. DNA is the chemical basis of heredity and have the coded message for proteins to be synthesised in the cell. There are three types of RNA — mRNA, rRNA and tRNA which actually carry out the protein synthesis in the cell. ExercisesExercisesExercisesExercisesExercises 10.1 What are monosaccharides? 10.2 What are reducing sugars? 10.3 Write two main functions of carbohydrates in plants. 10.4 Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose. 10.5 What do you understand by the term glycosidic linkage? 10.6 What is glycogen? How is it different from starch? 10.7 What are the hydrolysis products of (i) sucrose and (ii) lactose? 10.8 What is the basic structural difference between starch and cellulose? 10.9 What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3 10.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure. 10.11 What are essential and non-essential amino acids? Give two examples of each type. 10.12 Define the following as related to proteins (i) Peptide linkage (ii) Primary structure (iii) Denaturation. 10.13 What are the common types of secondary structure of proteins? 10.14 What type of bonding helps in stabilising the a-helix structure of proteins? 10.15 Differentiate between globular and fibrous proteins. 10.16 How do you explain the amphoteric behaviour of amino acids? 10.17 What are enzymes? 10.18 What is the effect of denaturation on the structure of proteins? 10.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood. 10.20 Why are vitamin A and vitamin C essential to us? Give their important sources. 10.21 What are nucleic acids? Mention their two important functions. 10.22 What is the difference between a nucleoside and a nucleotide? 10.23 The two strands in DNA are not identical but are complementary. Explain. 10.24 Write the important structural and functional differences between DNA and RNA. 10.25 What are the different types of RNA found in the cell? Chemistry 302 Reprint 2025-26

9.3 Alkenes Alkenes are unsaturated hydrocarbons Fig. 9.4 Orbital picture of ethene depictingcontaining at least one double bond. What σ bonds only should be the general formula of alkenes? If there is one double bond between two carbon 9.3.2 Nomenclature atoms in alkenes, they must possess two For nomenclature of alkenes in IUPAC system, hydrogen atoms less than alkanes. Hence, the longest chain of carbon atoms containing general formula for alkenes is CnH2n. Alkenes the double bond is selected. Numbering of the are also known as olefins (oil forming) since chain is done from the end which is nearer to Reprint 2025-26 Hydrocarbons 307 Fig. 9.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ Solutionof alkanes. It may be remembered that first member of alkene series is: CH2 (replacing (i) 2,8-Dimethyl-3, 6-decadiene; n by 1 in CnH2n) known as methene but has (ii) 1,3,5,7 Octatetraene; a very short life. As already mentioned, first (iii) 2-n-Propylpent-1-ene; stable member of alkene series is C2H4 known (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes Problem 9.8 are given below : Calculate number of sigma (σ) and pi (π) Structure IUPAC name bonds in the above structures (i-iv). CH3 – CH = CH2 Propene SolutionCH3 – CH2 – CH = CH2 But – l - ene σ bonds : 33, π bonds : 2CH3 – CH = CH–CH3 But-2-ene σ bonds : 17, π bonds : 4CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 23, π bond : 1CH2 = C – CH3 2-Methylprop-1-ene | σ bonds : 41, π bond : 1 CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene 9.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 9.7 (C2H4) and propene (C3H6) can have only one Write IUPAC names of the following structure but alkenes higher than propene compounds: have different structures. Alkenes possessing (i) (CH3)2CH – CH = CH – CH2 – CH C4H8 as molecular formula can be written in  the following three ways: CH3 – CH – CH | I. 1 2 3 4 C2H5 CH2 = CH – CH2 – CH3 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 3 4 | | CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 | But-2-ene CH3 (C4H8) Reprint 2025-26 308 chemistry III. 1 2 3 In (a), the two identical atoms i.e., both CH2 = C – CH3 the X or both the Y lie on the same side | of the double bond but in (b) the two X or CH3 two Y lie across the double bond or on the 2-Methylprop-1-ene opposite sides of the double bond. This (C4H8) results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in Structures I and III, and II and III are the two arrangements is different. Therefore,the examples of chain isomerism whereas they are stereoisomers. They would have thestructures I and II are position isomers. same geometry if atoms or groups around C=C bond can be rotated but rotation around Problem 9.9 C=C bond is not free. It is restricted. For Write structures and IUPAC names of understanding this concept, take two pieces different structural isomers of alkenes of strong cardboards and join them with the corresponding to C5H10. help of two nails. Hold one cardboard in your Solution one hand and try to rotate the other. Can (a) CH2 = CH – CH2 – CH2 – CH3 you really rotate the other cardboard ? The answer is no. The rotation is restricted. This Pent-1-ene illustrates that the restricted rotation of atoms (b) CH3 – CH=CH – CH2 – CH3 or groups around the doubly bonded carbon Pent-2-ene atoms gives rise to different geometries of such compounds. The stereoisomers of this (c) CH3 – C = CH – CH3 type are called geometrical isomers. The | isomer of the type (a), in which two identical CH3 atoms or groups lie on the same side of the 2-Methylbut-2-ene double bond is called cis isomer and the (d) CH3 – CH – CH = CH2 other isomer of the type (b), in which identical | atoms or groups lie on the opposite sides of CH3 the double bond is called trans isomer . Thus 3-Methylbut-1-ene cis and trans isomers have the same structure but have different configuration (arrangement (e) CH2 = C – CH2 – CH3 of atoms or groups in space). Due to different | arrangement of atoms or groups in space, CH3 these isomers differ in their properties like 2-Methylbut-1-ene melting point, boiling point, dipole moment, solubility etc. Geometrical or cis-trans isomersGeometrical isomerism: Doubly bonded of but-2-ene are represented below :carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that Reprint 2025-26 Hydrocarbons 309 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two (iii) C6H5CH = CH – CH3forms as given below from which it is clear that in the trans-but-2-ene, the two methyl (iv) CH3CH = CCl CH3 groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus Solution making the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. 9.3.4 Preparation 1. From alkynes: Alkynes on partial reduction with calculated amount of cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised (µ = 0.33D) (µ = 0) charcoal partially deactivated with poisons like sulphur compounds or quinoline give In the case of solids, it is observed that the alkenes. Partially deactivated palladisedtrans isomer has higher melting point than charcoal is known as Lindlar’s catalyst.the cis form. Alkenes thus obtained are having cis Geometrical or cis-trans isomerism geometry. However, alkynes on reductionis also shown by alkenes of the types with sodium in liquid ammonia form transXYC = CXZ and XYC = CZW alkenes. Problem 9.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (9.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (9.31) iii) CH≡ CH+H2 Pd/C CH2 =CH2 (9.32) Ethyne Ethene CH3–C≡ CH+H2 Pd/C CH3–CH =CH2 iv) Propyne Propene (9.33) Will propene thus obtained show Problem 9.11 geometrical isomerism? Think for the reason in support of your answer. Which of the following compounds will show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on (i) (CH3)2C = CH – C2H5 heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, Reprint 2025-26 310 chemistry ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is β-carbon atom. known as dehydrohalogenation i.e., removal of halogen acid. This is example of β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (9.37) 9.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. (9.34) The first three members are gases, the next fourteen are liquids and the higher ones are Nature of halogen atom and the alkyl group solids. Ethene is a colourless gas with a faint determine rate of the reaction. It is observed sweet smell. All other alkenes are colourless that for halogens, the rate is: iodine > and odourless, insoluble in water but fairly bromine > chlorine, while for alkyl groups soluble in non-polar solvents like benzene, it is : tertiary > secondary > primary. petroleum ether. They show a regular increase 3. From vicinal dihalides: Dihalides in in boiling point with increase in size i.e., every which two halogen atoms are attached – CH2 group added increases boiling point by to two adjacent carbon atoms are known 20–30 K. Like alkanes, straight chain alkenes as vicinal dihalides. Vicinal dihalides on have higher boiling point than isomeric treatment with zinc metal lose a molecule branched chain compounds. of ZnX2 to form an alkene. This reaction Chemical properties is known as dehalogenation. Alkenes are the rich source of loosely held CH2Br–CH2Br + Zn CH2=CH2+ ZnBr2 pi (π) electrons, due to which they show (9.35) addition reactions in which the electrophiles add on to the carbon-carbon double bond toCH3CHBr–CH2Br + Zn CH3CH=CH2 form the addition products. Some reagents +ZnBr2 also add by free radical mechanism. There (9.36) are cases when under special conditions, alkenes also undergo free radical substitution4. From alcohols by acidic dehydration: reactions. Oxidation and ozonolysis reactions You have read during nomenclature of are also quite prominent in alkenes. A brief different homologous series in Unit 12 description of different reactions of alkenes that alcohols are the hydroxy derivatives is given below: of alkanes. They are represented by R–OH where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add with concentrated sulphuric acid form up one molecule of dihydrogen gas in alkenes with the elimination of one water the presence of finely divided nickel, molecule. Since a water molecule is palladium or platinum to form alkanes eliminated from the alcohol molecule in (Section 9.2.2) the presence of an acid, this reaction is 2. Addition of halogens : Halogens like known as acidic dehydration of alcohols. bromine or chlorine add up to alkene to This reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under Reprint 2025-26 Hydrocarbons 311 normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation (9.42) which you will study in higher classes. Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding (9.38) molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov (9.39) rule can be better understood in terms of mechanism of the reaction. 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes Mechanism to form alkyl halides. The order of Hydrogen bromide provides an electrophile, reactivity of the hydrogen halides is H +, which attacks the double bond to form HI > HBr > HCl. Like addition of halogens carbocation as shown below : to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes (a) less stable (b) more stableAddition reactions of HBr to symmetrical primary carbocation secondary carbocationalkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates CH2=CH2+H–Br CH3–CH2–Br (9.40) because it is formed at a faster rate. (ii) The carbocation (b) is attacked by Br– ionCH3–CH=CH–CH3+HBr CH3–CH–CHCH3 to form the product as follows : Br (9.41) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two 2-Bromopropane possible products are I and II. (major product) Reprint 2025-26 312 chemistry Anti Markovnikov addition or peroxide effect or Kharash effect In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl The secondary free radical obtained in the and Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 the primary. This explains the formation of at the University of Chicago. This reaction 1-bromopropane as the major product. It may is known as peroxide or Kharash effect be noted that the peroxide effect is not observed or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due rule. to the fact that the H–Cl bond being (C6H5CO)2O2 stronger (430.5 kJ mol –1) than H–Br bond CH3 – CH=CH2+HBr CH3–CH2 (363.7 kJ mol –1), is not cleaved by the free radical, whereas the H–I bond is weaker CH2Br (296.8 kJ mol –1) and iodine free radicals 1–Bromopropane combine to form iodine molecules instead of adding to the double bond. (9.43) Mechanism : Peroxide effect proceeds via Problem 9.12 free radical chain mechanism as given below: Write IUPAC names of the products obtained by addition reactions of HBr to(i) hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution Homolysis C. 6H5+H–Br C6H3+ B. r(ii) 4. Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. Reprint 2025-26 Hydrocarbons 313 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (9.49) KMnO4/H+ CH3 – CH=CH–CH3 2CH3COOH (9.44) But-2-ene Ethanoic acid (9.50) 7. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H2O to smaller molecules. This reaction is highly useful in detecting the position of the double (9.45) bond in alkenes or other unsaturated compounds.5. Addition of water : In the presence of a few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (9.51) (9.46) 6. Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. (9.52) 8. Polymerisation: You are familiar with (9.47) polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called (9.48) polymers. This reaction is known as b) Acidic potassium permanganate or acidic polymerisation. The simple compounds potassium dichromate oxidises alkenes to from which polymers are made are called Reprint 2025-26 314 chemistry monomers. Other alkenes also undergo are named as derivatives of the corresponding polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. n(CH2 =CH2) High temp./pressureCatalyst —( CH2–CH2 )— The position of the triple bond is indicated by the first triply bonded carbon. Common Polythene and IUPAC names of a few members of alkyne (9.53) series are given in Table 9.2. High temp./pressure You have already learnt that ethyne and n(CH3 –CH=CH2) Catalyst —( CH–CH2 )—n propyne have got only one structure but there are two possible structures for butyne – CH3 (i) but-1-yne and (ii) but-2-yne. Since these Polypropene two compounds differ in their structures (9.54) due to the position of the triple bond, they Polymers are used for the manufacture of plastic are known as position isomers. In how bags, squeeze bottles, refrigerator dishes, toys, many ways, you can construct the structure pipes, radio and T.V. cabinets etc. Polypropene for the next homologue i.e., the next alkyne is used for the manufacture of milk crates, with molecular formula C5H8? Let us try to plastic buckets and other moulded articles. arrange five carbon atoms with a continuous Though these materials have now become chain and with a side chain. Following are the common, excessive use of polythene and possible structures : polypropylene is a matter of great concern for Structure IUPAC name all of us. 1 2 3 4 5 I. HC≡ C– CH2– CH2– CH3 Pent–1-yne

6.7 Relationship between to such a small degree that only a very Equilibrium Constant K, minute quantity of product is formed. Reaction Quotient Q and Gibbs Energy G Problem 6.10 The value of Kc for a reaction does not depend The value of ∆G  for the phosphorylation of on the rate of the reaction. However, as you glucose in glycolysis is 13.8 kJ/mol. Find have studied in Unit 5, it is directly related the value of Kc at 298 K. to the thermodynamics of the reaction and Solutionin particular, to the change in Gibbs energy, ∆G. If, ∆G  = 13.8 kJ/mol = 13.8 × 103J/mol • ∆G is negative, then the reaction is Also, ∆G  = – RT lnKc spontaneous and proceeds in the forward Hence, ln Kc = –13.8 × 103J/mol direction. (8.314 J mol–1K–1 × 298 K) • ∆G is positive, then reaction is considered ln Kc = – 5.569 non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the Kc = e–5.569 products of the forward reaction shall be Kc = 3.81 × 10–3 converted to the reactants. Problem 6.11• ∆G is 0, reaction has achieved equilibrium; Hydrolysis of sucrose gives, at this point, there is no longer any free energy left to drive the reaction. Sucrose + H2O Glucose + Fructose A mathematical expression of this Equilibrium constant Kc for the reaction is thermodynamic view of equilibrium can be 2 ×1013 at 300K. Calculate ∆G  at 300K. described by the following equation: Solution ∆G = ∆G + RT lnQ (6.21) ∆G  = – RT lnKcwhere, G is standard Gibbs energy. ∆G  = – 8.314J mol–1K–1× At equilibrium, when ∆G = 0 and Q = Kc, 300K × ln(2×1013) the equation (6.21) becomes, ∆G  = – 7.64 ×104 J mol–1 ∆G = ∆G + RT ln K = 0 6.8 FACTORS AFFECTING EQUILIBRIA ∆G = – RT lnK (6.22) One of the principal goals of chemical lnK = – ∆G / RT synthesis is to maximise the conversion of the Reprint 2025-26 EQUILIBRIUM 185 reactants to products while minimising the “When the concentration of any of the expenditure of energy. This implies maximum reactants or products in a reaction at yield of products at mild temperature and equilibrium is changed, the composition pressure conditions. If it does not happen, of the equilibrium mixture changes so as then the experimental conditions need to be to minimize the effect of concentration adjusted. For example, in the Haber process changes”. for the synthesis of ammonia from N2 and Let us take the reaction, H2, the choice of experimental conditions is of real economic importance. Annual world H2(g) + I2(g) 2HI(g) production of ammonia is about hundred If H2 is added to the reaction mixture million tones, primarily for use as fertilisers. at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, Equilibrium constant, Kc is independent the reaction proceeds in a direction whereinof initial concentrations. But if a system at equilibrium is subjected to a change in the H2 is consumed, i.e., more of H2 and I2 react to form HI and finally the equilibrium shiftsconcentration of one or more of the reacting in right (forward) direction (Fig.6.8). This is insubstances, then the system is no longer at accordance with the Le Chatelier’s principleequilibrium; and net reaction takes place in which implies that in case of addition of asome direction until the system returns to reactant/product, a new equilibrium willequilibrium once again. Similarly, a change be set up in which the concentration of thein temperature or pressure of the system may reactant/product should be less than what italso alter the equilibrium. In order to decide was after the addition but more than what itwhat course the reaction adopts and make was in the original mixture.a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria. We shall now be discussing factors which can influence the equilibrium. 6.8.1 Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that: • The concentration stress of an added reactant/product is relieved by net Fig. 6.8 Effect of addition of H2 on change reaction in the direction that consumes of concentration for the reactants the added substance. and products in the reaction, • The concentration stress of a removed H2(g) + I2 (g) 2HI(g) reactant/product is relieved by net reaction in the direction that replenishes The same point can be explained in terms the removed substance. of the reaction quotient, Qc, or in other words, Qc = [HI]2/ [H2][I2] Reprint 2025-26 186 chemistry Addition of hydrogen at equilibrium concentration of [Fe(SCN)]2+ decreases, the results in value of Qc being less than Kc . Thus, intensity of red colour decreases. in order to attain equilibrium again reaction Addition of aq. HgCl2 also decreases redmoves in the forward direction. Similarly, colour because Hg2+ reacts with SCN– ions to we can say that removal of a product also form stable complex ion [Hg(SCN)4]2–. Removalboosts the forward reaction and increases of free SCN– (aq) shifts the equilibrium the concentration of the products and this in equation (6.24) from right to left to has great commercial application in cases replenish SCN– ions. Addition of potassium of reactions, where the product is a gas or a thiocyanate on the other hand increases the volatile substance. In case of manufacture of colour intensity of the solution as it shift the ammonia, ammonia is liquified and removed equilibrium to right. from the reaction mixture so that reaction keeps moving in forward direction. Similarly, 6.8.2 Effect of Pressure Change in the large scale production of CaO (used A pressure change obtained by changing the as important building material) from CaCO3, volume can affect the yield of products in constant removal of CO2 from the kiln drives case of a gaseous reaction where the total the reaction to completion. It should be number of moles of gaseous reactants and remembered that continuous removal of a total number of moles of gaseous products are product maintains Qc at a value less than Kc different. In applying Le Chatelier’s principle and reaction continues to move in the forward to a heterogeneous equilibrium the effect direction. of pressure changes on solids and liquids can be ignored because the volume (and Effect of Concentration – An experiment concentration) of a solution/liquid is nearly This can be demonstrated by the following independent of pressure. reaction: Consider the reaction, Fe3+(aq)+ SCN–(aq) [Fe(SCN)]2+(aq) (6.24) CO(g) + 3H2(g) CH4(g) + H2O(g)yellow colourless deep red Here, 4 mol of gaseous reactants (CO + 3H2) become 2 mol of gaseous products (CH4 + H2O). Suppose equilibrium mixture (for above (6.25) reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to A reddish colour appears on adding two one half of its original volume. Then, totaldrops of 0.002 M potassium thiocynate solution pressure will be doubled (according to to 1 mL of 0.2 M iron(III) nitrate solution due pV = constant). The partial pressure and to the formation of [Fe(SCN)]2+. The intensity therefore, concentration of reactants and of the red colour becomes constant on products have changed and the mixture is no attaining equilibrium. This equilibrium can be longer at equilibrium. The direction in which shifted in either forward or reverse directions the reaction goes to re-establish equilibrium depending on our choice of adding a reactant can be predicted by applying the Le Chatelier’s or a product. The equilibrium can be shifted principle. Since pressure has doubled, in the opposite direction by adding reagents the equilibrium now shifts in the forward that remove Fe3+ or SCN– ions. For example, direction, a direction in which the number oxalic acid (H2C2O4), reacts with Fe3+ ions of moles of the gas or pressure decreases (we to form the stable complex ion [Fe(C2O4)3]3–, know pressure is proportional to moles of the thus decreasing the concentration of free gas). This can also be understood by using Fe3+(aq). In accordance with the Le Chatelier’s reaction quotient, Qc. Let [CO], [H2], [CH4] principle, the concentration stress of removed and [H2O] be the molar concentrations at Fe3+ is relieved by dissociation of [Fe(SCN)]2+ equilibrium for methanation reaction. When to replenish the Fe3+ ions. Because the volume of the reaction mixture is halved, the Reprint 2025-26 EQUILIBRIUM 187 partial pressure and the concentration are Production of ammonia according to the doubled. We obtain the reaction quotient by reaction, replacing each equilibrium concentration by N2(g) + 3H2(g) 2NH3(g);double its value. ∆H= – 92.38 kJ mol–1  CH 4 ( g )  H 2 O ( g ) is an exothermic process. According to Qc = 3  CO ( g )  H 2 ( g ) Le Chatelier’s principle, raising the temperature shifts the equilibrium to left As Qc < Kc , the reaction proceeds in the and decreases the equilibrium concentration forward direction. of ammonia. In other words, low temperature is favourable for high yield of ammonia, but In reaction C(s) + CO2(g) 2CO(g), when practically very low temperatures slow downpressure is increased, the reaction goes in the the reaction and thus a catalyst is used.reverse direction because the number of moles of gas increases in the forward direction. Effect of Temperature – An experiment Effect of temperature on equilibrium can6.8.3 Effect of Inert Gas Addition be demonstrated by taking NO2 gas (brown If the volume is kept constant and an inert gas in colour) which dimerises into N2O4 gas such as argon is added which does not take (colourless). part in the reaction, the equilibrium remains 2NO2(g) N2O4(g); ∆H = –57.2 kJ mol–1undisturbed. It is because the addition of an inert gas at constant volume does not NO2 gas prepared by addition of Cu change the partial pressures or the molar turnings to conc. HNO3 is collected in two 5 mL test tubes (ensuring same intensityconcentrations of the substance involved in of colour of gas in each tube) and stopperthe reaction. The reaction quotient changes sealed with araldite. Three 250 mL beakersonly if the added gas is a reactant or product 1, 2 and 3 containing freesing mixture, waterinvolved in the reaction. at room temperature and hot water (363K), 6.8.4 Effect of Temperature Change respectively, are taken (Fig. 6.9). Both the test tubes are placed in beaker 2 for 8-10 minutes.Whenever an equilibrium is disturbed by After this one is placed in beaker 1 and thea change in the concentration, pressure or other in beaker 3. The effect of temperaturevolume, the composition of the equilibrium on direction of reaction is depicted very wellmixture changes because the reaction in this experiment. At low temperatures inquotient, Qc no longer equals the equilibrium beaker 1, the forward reaction of formation ofconstant, Kc. However, when a change in temperature occurs, the value of equilibrium N2O4 is preferred, as reaction is exothermic, and thus, intensity of brown colour dueconstant, Kc is changed. to NO2 decreases. While in beaker 3, high In general, the temperature dependence temperature favours the reverse reaction of of the equilibrium constant depends on the sign of ∆H for the reaction. • The equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases. • The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases. Temperature changes affect the Fig. 6.9 Effect of temperature on equilibrium for the reaction, 2NO2 (g) N2O4 (g)equilibrium constant and rates of reactions. Reprint 2025-26 188 chemistry formation of NO2 and thus, the brown colour Similarly, in manufacture of sulphuric intensifies. acid by contact process, Effect of temperature can also be seen in 2SO2(g) + O2(g) 2SO3(g); Kc = 1.7 × 1026 an endothermic reaction, though the value of K is suggestive of reaction [Co(H2O)6]3+(aq) + 4Cl–(aq) [CoCl4]2–(aq) + going to completion, but practically the 6H2O(l) oxidation of SO2 to SO3 is very slow. Thus, pink colourless blue platinum or divanadium penta-oxide (V2O5) is used as catalyst to increase the rate of the At room temperature, the equilibrium reaction.mixture is blue due to [CoCl4]2–. When cooled Note: If a reaction has an exceedingly smallin a freesing mixture, the colour of the mixture K, a catalyst would be of little help.turns pink due to [Co(H2O)6]3+. 6.9 IONIC EQUILIBRIUM IN SOLUTION6.8.5 Effect of a Catalyst Under the effect of change of concentrationA catalyst increases the rate of the chemical on the direction of equilibrium, you havereaction by making available a new low energy pathway for the conversion of reactants to incidently come across with the following products. It increases the rate of forward equilibrium which involves ions: and reverse reactions that pass through the Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) same transition state and does not affect There are numerous equilibria that involve equilibrium. Catalyst lowers the activation ions only. In the following sections we will energy for the forward and reverse reactions study the equilibria involving ions. It is well by exactly the same amount. Catalyst does known that the aqueous solution of sugar not affect the equilibrium composition of does not conduct electricity. However, when a reaction mixture. It does not appear in common salt (sodium chloride) is added the balanced chemical equation or in the to water it conducts electricity. Also, the equilibrium constant expression. conductance of electricity increases with an Let us consider the formation of NH3 increase in concentration of common salt. from dinitrogen and dihydrogen which is Michael Faraday classified the substances highly exothermic reaction and proceeds into two categories based on their ability with decrease in total number of moles to conduct electricity. One category of formed as compared to the reactants. substances conduct electricity in their Equilibrium constant decreases with increase aqueous solutions and are called electrolytes in temperature. At low temperature rate while the other do not and are thus, referred to decreases and it takes long time to reach at as non-electrolytes. Faraday further classified equilibrium, whereas high temperatures give electrolytes into strong and weak electrolytes. satisfactory rates but poor yields. Strong electrolytes on dissolution in water German chemist, Fritz Haber discovered are ionized almost completely, while the weak that a catalyst consisting of iron catalyse electrolytes are only partially dissociated. the reaction to occur at a satisfactory rate For example, an aqueous solution of at temperatures, where the equilibrium sodium chloride is comprised entirely of concentration of NH3 is reasonably favourable. sodium ions and chloride ions, while that Since the number of moles formed in the of acetic acid mainly contains unionized reaction is less than those of reactants, the acetic acid molecules and only some acetate yield of NH3 can be improved by increasing ions and hydronium ions. This is because the pressure. there is almost 100% ionization in case Optimum conditions of temperature of sodium chloride as compared to less and pressure for the synthesis of NH3 using than 5% ionization of acetic acid which is catalyst are around 500°C and 200 atm. a weak electrolyte. It should be noted Reprint 2025-26 EQUILIBRIUM 189 that in weak electrolytes, equilibrium is exists in solid state as a cluster of positively established between ions and the unionized charged sodium ions and negatively charged molecules. This type of equilibrium involving chloride ions which are held together due to ions in aqueous solution is called ionic electrostatic interactions between oppositely equilibrium. Acids, bases and salts come charged species (Fig.6.10). The electrostatic under the category of electrolytes and may act forces between two charges are inversely as either strong or weak electrolytes. proportional to dielectric constant of the medium. Water, a universal solvent, possesses

6.6 APPLICATIONS OF EQUILIBRIUM in the denominator). This implies that a high value of K is suggestive of a high concentration CONSTANTS of products and vice-versa.Before considering the applications of We can make the following generalisationsequilibrium constants, let us summarise the concerning the composition of equilibriumimportant features of equilibrium constants mixtures:as follows: 1. Expression for equilibrium constant is • If Kc > 103, products predominate over applicable only when concentrations of reactants, i.e., if Kc is very large, the the reactants and products have attained reaction proceeds nearly to completion. constant value at equilibrium state. Consider the following examples: 2. The value of equilibrium constant is (a) The reaction of H2 with O2 at 500 K independent of initial concentrations of has a very large equilibrium constant, the reactants and products. Kc = 2.4 × 1047. 3. Equilibrium constant is temperature (b) H2(g) + Cl2(g) 2HCl(g) at 300K has dependent having one unique value for Kc = 4.0 × 1031. a particular reaction represented by a (c) H2(g) + Br2(g) 2HBr (g) at 300 K, balanced equation at a given temperature. Kc = 5.4 × 1018 4. The equilibrium constant for the reverse • If Kc < 10–3, reactants predominate over reaction is equal to the inverse of the products, i.e., if Kc is very small, the equilibrium constant for the forward reaction proceeds rarely. Consider the reaction. following examples: Reprint 2025-26 182 chemistry (a) The decomposition of H2O into H2 and If Qc = Kc, the reaction mixture is already O2 at 500 K has a very small equilibrium at equilibrium. constant, Kc = 4.1 × 10–48 Consider the gaseous reaction of H2 (b) N2(g) + O2(g) 2NO(g), with I2, at 298 K has Kc = 4.8 ×10–31. H2(g) + I2(g) 2HI(g); Kc = 57.0 at 700 K. • If Kc is in the range of 10 – 3 to 103, Suppose we have molar concentrations appreciable concentrations of both [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M. reactants and products are present. (the subscript t on the concentration symbols Consider the following examples: means that the concentrations were measured at some arbitrary time t, not necessarily at(a) For reaction of H2 with I2 to give HI, equilibrium). Kc = 57.0 at 700K. Thus, the reaction quotient, Qc at this(b) Also, gas phase decomposition of N2O4 stage of the reaction is given by, to NO2 is another reaction with a value 2 –3 Qc = [HI]t / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20) of Kc = 4.64 × 10 at 25°C which is neither too small nor too large. Hence, = 8.0 equilibrium mixtures contain appreciable Now, in this case, Qc (8.0) does not equal concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g) These generarlisations are illustrated in is not at equilibrium; that is, more H2(g) and Fig. 6.6 I2(g) will react to form more HI(g) and their concentrations will decrease till Qc = Kc. The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc. Thus, we can make the following generalisations concerning the direction of the reaction (Fig. 6.7) :Fig.6.6 Dependence of extent of reaction on Kc 6.6.2 Predicting the Direction of the Reaction The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Qc with molar Fig. 6.7 Predicting the direction of the reactionconcentrations and QP with partial pressures) is defined in the same way as the equilibrium • If Qc < Kc, net reaction goes from left to constant Kc except that the concentrations right in Qc are not necessarily equilibrium values. • If Qc > Kc, net reaction goes from right to For a general reaction: left. a A + b B c C + d D (6.19) • If Qc = Kc, no net reaction occurs. Qc = [C]c[D]d / [A]a[B]b (6.20) Problem 6.7 Then, The value of Kc for the reaction If Qc > Kc, the reaction will proceed in the 2A B + C is 2 × 10–3. At a given time, direction of reactants (reverse reaction). the composition of reaction mixture is [A] = [B] = [C] = 3 × 10–4 M. In which direction If Qc < Kc, the reaction will proceed in the the reaction will proceed?direction of the products (forward reaction). Reprint 2025-26 EQUILIBRIUM 183 Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc is found to be 9.15 bar. Calculate Kc, Kp and given by, partial pressure at equilibrium. Qc = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10–4M Qc = (3 ×10–4)(3 × 10–4) / (3 ×10–4)2 = 1 We know pV = nRT as Qc > Kc so the reaction will proceed in the Total volume (V ) = 1 L reverse direction. Molecular mass of N2O4 = 92 g 6.6.3 Calculating Equilibrium Number of moles = 13.8g/92 g = 0.15 Concentrations of the gas (n) In case of a problem in which we know the Gas constant (R) = 0.083 bar L mol–1K–1 initial concentrations but do not know any of Temperature (T ) = 400 K the equilibrium concentrations, the following pV = nRTthree steps shall be followed: Step 1. Write the balanced equation for the p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 × 400 Kreaction. Step 2. Under the balanced equation, make p = 4.98 bar a table that lists for each substance involved N2O4 2NO2 in the reaction: Initial pressure: 4.98 bar 0 (a) the initial concentration, At equilibrium: (4.98 – x) bar 2x bar (b) the change in concentration on going to Hence, equilibrium, and ptotal at equilibrium = pN2O4 + pNO2(c) the equilibrium concentration. 9.15 = (4.98 – x) + 2x In constructing the table, define x as the 9.15 = 4.98 + xconcentration (mol/L) of one of the substances that reacts on going to equilibrium, then use x = 9.15 – 4.98 = 4.17 bar the stoichiometry of the reaction to determine Partial pressures at equilibrium are, the concentrations of the other substances in terms of x. pN2O4 = 4.98 – 4.17 = 0.81bar Step 3. Substitute the equilibrium pNO2 = 2x = 22 × 4.17 = 8.34 bar concentrations into the equilibrium equation K p = p N 2O 4  p NO 2 / for the reaction and solve for x. If you are = (8.34)2/0.81 = 85.87to solve a quadratic equation choose the mathematical solution that makes chemical Kp = Kc(RT)∆n sense. 85.87 = Kc(0.083 × 400)1 Step 4. Calculate the equilibrium Kc = 2.586 = 2.6 concentrations from the calculated value of x. Problem 6.9Step 5. Check your results by substituting them into the equilibrium equation. 3.00 mol of PCl5 kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the Problem 6.8 mixture at equilibrium. Kc= 1.80 13.8g of N2O4 was placed in a 1L reaction Solution vessel at 400K and allowed to attain PCl5 PCl3 + Cl2 equilibrium Initial N2O4 (g) 2NO2 (g) concentration: 3.0 0 0 Reprint 2025-26 184 chemistry Taking antilog of both sides, we get, Let x mol per litre of PCl5 be dissociated, K = e–∆G/RT (6.23) At equilibrium: (3-x) x x Hence, using the equation (6.23), the reaction spontaneity can be interpreted in Kc = [PCl3][Cl2]/[PCl5] terms of the value of ∆G . 1.8 = x2/ (3 – x) • If ∆G  < 0, then –∆G /RT is positive, x2 + 1.8x – 5.4 = 0 and e –∆DG /RT>1, making K >1, which x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 implies a spontaneous reaction or the x = [–1.8 ± √3.24 + 21.6]/2 reaction which proceeds in the forward direction to such an extent that the x = [–1.8 ± 4.98]/2 products are present predominantly. x = [–1.8 + 4.98]/2 = 1.59 • If ∆G  > 0, then –∆G /RT is negative, and [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M e –∆G </RT 1, that is , K < 1, which implies [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction which proceeds in the forward direction

6.11 IONIZATION OF ACIDS AND BASES and the equilibrium will shift in the direction of weaker acid. Say, if HA is a stronger acidArrhenius concept of acids and bases than H3O+, then HA will donate protons andbecomes useful in case of ionization of acids not H3O+, and the solution will mainly containand bases as mostly ionizations in chemical A– and H3O+ ions. The equilibrium moves inand biological systems occur in aqueous the direction of formation of weaker acid medium. Strong acids like perchloric acid Reprint 2025-26 EQUILIBRIUM 193 and weaker base because the stronger acid H2O(l) + H2O(l) H3O+(aq) + OH–(aq) donates a proton to the stronger base. acid base conjugate conjugate It follows that as a strong acid dissociates acid base completely in water, the resulting base formed The dissociation constant is represented by, would be very weak i.e., strong acids have K = [H3O+] [OH–] / [H2O] (6.26)very weak conjugate bases. Strong acids like perchloric acid (HClO4), hydrochloric acid The concentration of water is omitted from (HCl), hydrobromic acid (HBr), hydroiodic acid the denominator as water is a pure liquid and (HI), nitric acid (HNO3) and sulphuric acid its concentration remains constant. [H2O] is (H2SO4) will give conjugate base ions ClO4–, Cl, incorporated within the equilibrium constant Br–, I–, NO3– and HSO4– , which are much weaker to give a new constant, Kw, which is called the bases than H2O. Similarly a very strong base ionic product of water. would give a very weak conjugate acid. On the Kw = [H+][OH–] (6.27) other hand, a weak acid say HA is only partially The concentration of H+ has been founddissociated in aqueous medium and thus, the out experimentally as 1.0 × 10–7 M at 298 K.solution mainly contains undissociated HA And, as dissociation of water produces equalmolecules. Typical weak acids are nitrous number of H+ and OH– ions, the concentrationacid (HNO2), hydrofluoric acid (HF) and acetic of hydroxyl ions, [OH–] = [H+] = 1.0 × 10–7 M.acid (CH3COOH). It should be noted that the Thus, the value of Kw at 298K,weak acids have very strong conjugate bases. For example, NH2–, O 2– and H– are very good Kw = [H3O+][OH–] = (1 × 10–7)2 = 1 × 10–14 M2 proton acceptors and thus, much stronger (6.28) bases than H2O. The value of Kw is temperature dependent Certain water soluble organic compounds as it is an equilibrium constant. like phenolphthalein and bromothymol blue The density of pure water is 1000 g / Lbehave as weak acids and exhibit different and its molar mass is 18.0 g /mol. From thiscolours in their acid (HIn) and conjugate base the molarity of pure water can be given as,(In– ) forms. [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.HIn(aq) + H2O(l) H3O+(aq) + In–(aq) Therefore, the ratio of dissociated water toacid conjugate conjugate that of undissociated water can be given as: indicator acid base 10–7 / (55.55) = 1.8 × 10–9 or ~ 2 in 10–9 (thus, colour A colour B equilibrium lies mainly towards undissociated Such compounds are useful as indicators water) in acid-base titrations, and finding out H+ ion We can distinguish acidic, neutral andconcentration. basic aqueous solutions by the relative values 6.11.1 The Ionization Constant of Water of the H3O+ and OH– concentrations: and its Ionic Product Acidic: [H3O+] > [OH– ]Some substances like water are unique in Neutral: [H3O+] = [OH– ]their ability of acting both as an acid and a base. We have seen this in case of water in Basic : [H3O+] < [OH–] section 6.10.2. In presence of an acid, HA it accepts a proton and acts as the base while 6.11.2 The pH Scale in the presence of a base, B– it acts as an Hydronium ion concentration in molarity is acid by donating a proton. In pure water, one more conveniently expressed on a logarithmic H2O molecule donates proton and acts as an scale known as the pH scale. The pH of a acid and another water molecules accepts a solution is defined as the negative logarithm proton and acts as a base at the same time. of hydrogen to base 10 of the activity  aH The following equilibrium exists: Reprint 2025-26 194 chemistry ion. In dilute solutions (< 0.01 M), activity when the hydrogen ion concentration, [H+] of hydrogen ion (H+) is equal in magnitude changes by a factor of 100, the value of pH to molarity represented by [H+]. It should changes by 2 units. Now you can realise why be noted that activity has no units and is the change in pH with temperature is often defined as: ignored. = [H+] / mol L–1 Measurement of pH of a solution is very essential as its value should be known From the definition of pH, the following when dealing with biological and cosmeticcan be written, applications. The pH of a solution can be pH = – log aH+ = – log {[H+] / mol L–1} found roughly with the help of pH paper that has different colour in solutions of different Thus, an acidic solution of HCl (10–2 M) will have a pH = 2. Similarly, a basic solution pH. Now-a-days pH paper is available with of NaOH having [OH–] =10–4 M and [H3O+] = four strips on it. The different strips have 10–10 M will have a pH = 10. At 25 °C, pure different colours (Fig. 6.11) at the same pH. water has a concentration of hydrogen ions, The pH in the range of 1-14 can be determined [H+] = 10–7 M. Hence, the pH of pure water is with an accuracy of ~0.5 using pH paper. given as: pH = –log(10–7) = 7 Acidic solutions possess a concentration of hydrogen ions, [H+] > 10–7 M, while basic solutions possess a concentration of hydrogen ions, [H+] < 10–7 M. thus, we can summarise that Fig.6.11 pH-paper with four strips that may haveAcidic solution has pH < 7 different colours at the same pH Basic solution has pH > 7 For greater accuracy pH meters are used.Neutral solution has pH = 7 pH meter is a device that measures the Now again, consider the equation (6.28) pH-dependent electrical potential of the testat 298 K solution within 0.001 precision. pH meters Kw = [H3O+] [OH–] = 10–14 of the size of a writing pen are now available Taking negative logarithm on both sides in the market. The pH of some very common of equation, we obtain substances are given in Table 6.5 (page 195). –log Kw = – log {[H3O+] [OH–]} Problem 6.16 = – log [H3O+] – log [OH–] –14 The concentration of hydrogen ion in a = – log 10 sample of soft drink is 3.8 × 10–3M. what pKw = pH + pOH = 14 (6.29) is its pH ? Note that although Kw may change with temperature the variations in pH with Solution temperature are so small that we often pH = – log[3.8 × 10–3] ignore it. = – {log[3.8] + log[10–3]} pKw is a very important quantity for = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42 aqueous solutions and controls the relative Therefore, the pH of the soft drink is 2.42 concentrations of hydrogen and hydroxyl and it can be inferred that it is acidic. ions as their product is a constant. It should Problem 6.17be noted that as the pH scale is logarithmic, Calculate pH of a 1.0 × 10 –8 M solution ofa change in pH by just one unit also means HCl.change in [H+] by a factor of 10. Similarly, Reprint 2025-26 EQUILIBRIUM 195 Table 6.5 The pH of Some Common Substances Name of the Fluid pH Name of the Fluid pH Saturated solution of NaOH ~15 Black Coffee 5.0 0.1 M NaOH solution 13 Tomato juice ~4.2 Lime water 10.5 Soft drinks and vinegar ~3.0 Milk of magnesia 10 Lemon juice ~2.2 Egg white, sea water 7.8 Gastric juice ~1.2 Human blood 7.4 1M HCl solution ~0 Milk 6.8 Concentrated HCl ~–1.0 Human Saliva 6.4 equilibrium constant for the above discussed Solution acid-dissociation equilibrium: 2H2O (l) H3O+ (aq) + OH–(aq) Ka = c2α2 / c(1-α) = cα2 / 1-α Kw = [OH–][H3O+] Ka is called the dissociation or ionization = 10–14 constant of acid HX. It can be represented Let, x = [OH–] = [H3O+] from H2O. The alternatively in terms of molar concentration H3O+ concentration is generated (i) from as follows, the ionization of HCl dissolved i.e., Ka = [H+][X–] / [HX] (6.30) HCl(aq) + H2O(l) H3O+ (aq) + Cl –(aq), At a given temperature T, Ka is a and (ii) from ionization of H2O. In these very measure of the strength of the acid HX i.e., dilute solutions, both sources of H3O+ must larger the value of Ka, the stronger is the be considered: acid. Ka is a dimensionless quantity with [H3O+] = 10–8 + x the understanding that the standard state Kw = (10–8 + x)(x) = 10–14 concentration of all species is 1M. or x2 + 10–8 x – 10–14 = 0 The values of the ionization constants [OH– ] = x = 9.5 × 10–8 of some selected weak acids are given in Table 6.6. So, pOH = 7.02 and pH = 6.98 Table 6.6 The Ionization Constants of Some 6.11.3 Ionization Constants of Weak Acids Selected Weak Acids (at 298K) Consider a weak acid HX that is partially Acid Ionization Constant, ionized in the aqueous solution. The Ka equilibrium can be expressed by: Hydrofluoric Acid (HF) 3.5 × 10–4 HX(aq) + H2O(l) H3O+(aq) + X–(aq) Nitrous Acid (HNO2) 4.5 × 10–4 Initial Formic Acid (HCOOH) 1.8 × 10–4 concentration (M) c 0 0 Niacin (C5H4NCOOH) 1.5 × 10–5 Let α be the extent of ionization Acetic Acid (CH3COOH) 1.74 × 10–5 Change (M) Benzoic Acid (C6H5COOH) 6.5 × 10–5 -cα +cα +cα Hypochlorous Acid (HCIO) 3.0 × 10–8 Equilibrium concentration (M) Hydrocyanic Acid (HCN) 4.9 × 10–10 c-cα cα cα Phenol (C6H5OH) 1.3 × 10–10 Here, c = initial concentration of the undissociated acid, HX at time, t = 0. α = The pH scale for the hydrogen ion extent up to which HX is ionized into ions. concentration has been so useful that besides Using these notations, we can derive the pKw, it has been extended to other species and Reprint 2025-26 196 chemistry quantities. Thus, we have: Solution pKa = –log (Ka) (6.31) The following proton transfer reactions are Knowing the ionization constant, Ka possible: of an acid and its initial concentration, c, 1) HF + H2O H3O+ + F–it is possible to calculate the equilibrium Ka = 3.2 × 10–4concentration of all species and also the 2) H2O + H2O H3O+ + OH–degree of ionization of the acid and the pH of the solution. Kw = 1.0 × 10–14 As Ka >> Kw, [1] is the principle reaction. A general step-wise approach can be adopted to evaluate the pH of the weak HF + H2O H3O+ + F– electrolyte as follows: Initial Step 1. The species present before concentration (M) dissociation are identified as Brönsted-Lowry 0.02 0 0 acids/bases. Change (M) Step 2. Balanced equations for all possible –0.02α +0.02α +0.02α reactions i.e., with a species acting both as Equilibriumacid as well as base are written. concentration (M)Step 3. The reaction with the higher Ka is identified as the primary reaction whilst the 0.02 – 0.02 α 0.02 α 0.02α other is a subsidiary reaction. Substituting equilibrium concentrations Step 4. Enlist in a tabular form the following in the equilibrium reaction for principal reaction gives:values for each of the species in the primary reaction Ka = (0.02α)2 / (0.02 – 0.02α) (a) Initial concentration, c. = 0.02 α2 / (1 –α) = 3.2 × 10–4 (b) Change in concentration on proceeding We obtain the following quadratic equation: to equilibrium in terms of α, degree of α2 + 1.6 × 10–2α – 1.6 × 10–2 = 0 ionization. The quadratic equation in α can be solved (c) Equilibrium concentration. and the two values of the roots are: α = + 0.12 and – 0.12Step 5. Substitute equilibrium concentrations into equilibrium constant equation for The negative root is not acceptable and hence,principal reaction and solve for α. Step 6. Calculate the concentration of species α = 0.12 in principal reaction. This means that the degree of ionization, α = 0.12, then equilibrium concentrationsStep 7. Calculate pH = – log[H3O+] – of other species viz., HF, F and H3O+ are The above mentioned methodology has given by: been elucidated in the following examples. [H3O+] = [F –] = cα = 0.02 × 0.12 = 2.4 × 10–3 M Problem 6.18 [HF] = c(1 – α) = 0.02 (1 – 0.12) The ionization constant of HF is = 17.6 × 10-3 M 3.2 × 10–4. Calculate the degree of dissociation of HF in its 0.02 M solution. pH = – log[H+] = –log(2.4 × 10–3) = 2.62 Calculate the concentration of all species Problem 6.19 present (H3O+, F – and HF) in the solution The pH of 0.1M monobasic acid is 4.50. and its pH. Calculate the concentration of species H+, A– Reprint 2025-26 EQUILIBRIUM 197 and HA at equilibrium. Also, determine the Percent dissociation value of Ka and pKa of the monobasic acid. = {[HOCl]dissociated / [HOCl]initial }× 100 Solution = 1.41 × 10–3 × 102/ 0.08 = 1.76 %. pH = –log(1.41 × 10–3) = 2.85. pH = – log [H+] Therefore, [H+] = 10 –pH = 10–4.50 6.11.4 Ionization of Weak Bases = 3.16 × 10–5 The ionization of base MOH can be represented by equation: [H+] = [A–] = 3.16 × 10–5 MOH(aq) M+(aq) + OH–(aq) Thus, Ka = [H+][A-] / [HA] In a weak base there is partial ionization [HA]eqlbm = 0.1 – (3.16 × 10-5)  0.1 of MOH into M+ and OH–, the case is similar to that of acid-dissociation equilibrium. The Ka = (3.16 × 10–5)2 / 0.1 = 1.0 × 10–8 equilibrium constant for base ionization pKa = – log(10–8) = 8 is called base ionization constant and is represented by Kb. It can be expressed in Alternatively, “Percent dissociation” is terms of concentration in molarity of various another useful method for measure of strength of a weak acid and is given as: species in equilibrium by the following equation: Percent dissociation Kb = [M+][OH–] / [MOH] (6.33) = [HA]dissociated/[HA]initial × 100% (6.32) Alternatively, if c = initial concentration Problem 6.20 of base and α = degree of ionization of base i.e. the extent to which the base ionizes. Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization When equilibrium is reached, the equilibrium constant of the acid is 2.5 × 10–5. Determine constant can be written as: the percent dissociation of HOCl. Kb = (cα)2 / c (1-α) = cα2 / (1-α) Solution The values of the ionization constants of some selected weak bases, Kb are given in HOCl(aq) + H2O (l) H3O+(aq) + ClO–(aq) Table 6.7. Initial concentration (M) Table 6.7 The Values of the Ionization 0.08 0 0 Constant of Some Weak Bases at Change to reach 298 K equilibrium concentration Base Kb (M) Dimethylamine, (CH3)2NH 5.4 × 10–4 – x + x +x Triethylamine, (C2H5)3N 6.45 × 10–5 equilibrium concentartion (M) Ammonia, NH3 or NH4OH 1.77 × 10–5 0.08 – x x x Quinine, (A plant product) 1.10 × 10–6 Ka = {[H3O+][ClO–] / [HOCl]} Pyridine, C5H5N 1.77 × 10–9 = x2 / (0.08 –x) Aniline, C6H5NH2 4.27 × 10–10 As x << 0.08, therefore 0.08 – x  0.08 Urea, CO (NH2)2 1.3 × 10–14 x2 / 0.08 = 2.5 × 10–5 Many organic compounds like amines x2 = 2.0 × 10–6, thus, x = 1.41 × 10–3 are weak bases. Amines are derivatives of [H+] = 1.41 × 10–3 M. ammonia in which one or more hydrogen Therefore, atoms are replaced by another group. For example, methylamine, codeine, quinine and Reprint 2025-26 198 chemistry nicotine all behave as very weak bases due to Kb = 10–4.75 = 1.77 × 10–5 Mtheir very small Kb. Ammonia produces OH– in aqueous solution: NH3 + H2O NH4+ + OH– Initial concentration (M) NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) 0.10 0.20 0 The pH scale for the hydrogen ion Change to reachconcentration has been extended to get: pKb = –log (Kb) (6.34) equilibrium (M) –x +x +x At equilibrium (M) Problem 6.21 0.10 – x 0.20 + x x The pH of 0.004M hydrazine solution is 9.7. Kb = [NH4+][OH–] / [NH3] Calculate its ionization constant Kb and pKb. = (0.20 + x)(x) / (0.1 – x) = 1.77 × 10–5 Solution As Kb is small, we can neglect x in comparison NH2NH2 + H2O NH2NH3+ + OH– to 0.1M and 0.2M. Thus, From the pH we can calculate the hydrogen [OH–] = x = 0.88 × 10–5 ion concentration. Knowing hydrogen ion Therefore, [H+] = 1.12 × 10–9 concentration and the ionic product of pH = – log[H+] = 8.95. water we can calculate the concentration of hydroxyl ions. Thus we have: 6.11.5 Relation between Ka and Kb [H+] = antilog (–pH) As seen earlier in this chapter, Ka and Kb = antilog (–9.7) = 1.67 ×10–10 represent the strength of an acid and a base, [OH–] = Kw / [H+] = 1 × 10–14 / 1.67 × 10–10 respectively. In case of a conjugate acid-base pair, they are related in a simple manner so = 5.98 × 10–5 that if one is known, the other can be deduced. The concentration of the corresponding + Considering the example of NH4 and NH3 hydrazinium ion is also the same as that we see, of hydroxyl ion. The concentration of both these ions is very small so the concentration NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) of the undissociated base can be taken Ka = [H3O+][ NH3] / [NH4+] = 5.6 × 10–10 equal to 0.004M. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) Thus, Kb =[ NH4+][ OH–] / NH3 = 1.8 × 10–5 Kb = [NH2NH3+][OH–] / [NH2NH2] Net: 2 H2O(l) H3O+(aq) + OH–(aq) = (5.98 × 10–5)2 / 0.004 = 8.96 × 10–7 Kw = [H3O+][ OH– ] = 1.0 × 10–14 M pKb = –logKb = –log(8.96 × 10–7) = 6.04. + Where, Ka represents the strength of NH4 Problem 6.22 as an acid and Kb represents the strength of Calculate the pH of the solution in which NH3 as a base. 0.2M NH4Cl and 0.1M NH3 are present. The It can be seen from the net reaction that pKb of ammonia solution is 4.75. the equilibrium constant is equal to the product of equilibrium constants Ka and Kb Solution for the reactions added. Thus, NH3 + H2O NH4+ + OH– Ka × Kb = {[H3O+][ NH3] / [NH4+ ]} × {[NH4 +] The ionization constant of NH3, [OH–] / [NH3]} Kb = antilog (–pKb) i.e. = [H3O+][OH–] = Kw = (5.6 ×10–10) × (1.8 × 10–5) = 1.0 × 10–14 M Reprint 2025-26 EQUILIBRIUM 199 This can be extended to make a + NH3 + H2O NH4 + OH–generalisation. The equilibrium constant We use equation (6.33) to calculatefor a net reaction obtained after adding hydroxyl ion concentration,two (or more) reactions equals the product [OH–] = c α = 0.05 αof the equilibrium constants for individual reactions: Kb = 0.05 α2 / (1 – α) The value of α is small, therefore the KNET = K1 × K2 × …… (6.35) quadratic equation can be simplified by Similarly, in case of a conjugate acid-base neglecting α in comparison to 1 in the pair, denominator on right hand side of the Ka × Kb = Kw (6.36) equation, Knowing one, the other can be obtained. Thus, It should be noted that a strong acid will have Kb = c α2 or α = √ (1.77 × 10–5 / 0.05) a weak conjugate base and vice-versa. = 0.018. Alternatively, the above expression [OH–] = c α = 0.05 × 0.018 = 9.4 × 10–4M. Kw = Ka × Kb, can also be obtained by [H+] = Kw / [OH–] = 10–14 / (9.4 × 10–4)considering the base-dissociation equilibrium reaction: = 1.06 × 10–11 B(aq) + H2O(l) BH+(aq) + OH–(aq) pH = –log(1.06 × 10–11) = 10.97. Kb = [BH+][OH–] / [B] Now, using the relation for conjugate As the concentration of water remains acid-base pair, constant it has been omitted from the Ka × Kb = Kw denominator and incorporated within the using the value of Kb of NH3 fromdissociation constant. Then multiplying and Table 6.7.dividing the above expression by [H+], we get: We can determine the concentration of Kb = [BH+][OH–][H+] / [B][H+] + conjugate acid NH4 ={[ OH–][H+]}{[BH+] / [B][H+]} Ka = Kw / Kb = 10–14 / 1.77 × 10–5 = Kw / Ka = 5.64 × 10–10. or Ka × Kb = Kw It may be noted that if we take negative logarithm of both sides of the equation, then 6.11.6 Di- and Polybasic Acids and Di- pK values of the conjugate acid and base are and Polyacidic Bases related to each other by the equation: Some of the acids like oxalic acid, sulphuric pKa + pKb = pKw = 14 (at 298K) acid and phosphoric acids have more than one ionizable proton per molecule of the Problem 6.23 acid. Such acids are known as polybasic or polyprotic acids. Determine the degree of ionization and pH of The ionization reactions for example for a 0.05M of ammonia solution. The ionization a dibasic acid H2X are represented by the constant of ammonia can be taken from equations: Table 6.7. Also, calculate the ionization H2X(aq) H+(aq) + HX–(aq) constant of the conjugate acid of ammonia. HX–(aq) H+(aq) + X2–(aq) Solution And the corresponding equilibrium The ionization of NH3 in water is represented constants are given below: by equation: Ka1= {[H+][HX–]} / [H2X] and Reprint 2025-26 200 chemistry Ka2 = {[H+][X2-]} / [HX-] In general, when strength of H-A bond decreases, that is, the energy required to break Here, Ka1and Ka2are called the first and second the bond decreases, HA becomes a stronger ionization constants respectively of the acid H2 acid. Also, when the H-A bond becomes more X. Similarly, for tribasic acids like H3PO4 we polar i.e., the electronegativity difference have three ionization constants. The values between the atoms H and A increases and of the ionization constants for some common there is marked charge separation, cleavage polyprotic acids are given in Table 6.8. of the bond becomes easier thereby increasing Table 6.8 The Ionization Constants of Some the acidity. Common Polyprotic Acids (298K) But it should be noted that while comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases. For example, Size increases HF << HCl << HBr << HI It can be seen that higher order ionization Acid strength increasesconstants (Ka2, Ka3) are smaller than the Similarly, H2S is stronger acid than H2O.lower order ionization constant (Ka1) of a polyprotic acid. The reason for this is that But, when we discuss elements in the same it is more difficult to remove a positively row of the periodic table, H-A bond polarity charged proton from a negative ion due to becomes the deciding factor for determining electrostatic forces. This can be seen in the the acid strength. As the electronegativity case of removing a proton from the uncharged of A increases, the strength of the acid also H2CO3 as compared from a negatively charged increases. For example, HCO3–. Similarly, it is more difficult to remove 2– Electronegativity of A increasesa proton from a doubly charged HPO4 anion as compared to H2PO4–. CH4 < NH3 < H2O < HF Polyprotic acid solutions contain a Acid strength increases mixture of acids like H2A, HA– and A2– in case 6.11.8 Common Ion Effect in theof a diprotic acid. H2A being a strong acid, the Ionization of Acids and Basesprimary reaction involves the dissociation of H2 A, and H3O+ in the solution comes mainly Consider an example of acetic acid dissociation from the first dissociation step. equilibrium represented as: CH3COOH(aq) H+(aq) + CH3COO– (aq)6.11.7 Factors Affecting Acid Strength or HAc(aq) H+ (aq) + Ac– (aq)Having discussed quantitatively the strengths of acids and bases, we come to a stage where Ka = [H+][Ac– ] / [HAc] we can calculate the pH of a given acid Addition of acetate ions to an aceticsolution. But, the curiosity rises about why acid solution results in decreasing theshould some acids be stronger than others? concentration of hydrogen ions, [H+]. Also,What factors are responsible for making if H+ ions are added from an external sourcethem stronger? The answer lies in its being a then the equilibrium moves in the directioncomplex phenomenon. But, broadly speaking of undissociated acetic acid i.e., in a directionwe can say that the extent of dissociation of of reducing the concentration of hydrogenan acid depends on the strength and polarity ions, [H+]. This phenomenon is an exampleof the H-A bond. Reprint 2025-26 EQUILIBRIUM 201 of common ion effect. It can be defined as Thus, x = 1.33 × 10–3 = [OH–] a shift in equilibrium on adding a substance that provides more of an ionic species already Therefore, [H+] = Kw / [OH–] = 10–14 / present in the dissociation equilibrium. (1.33 × 10–3) = 7.51 × 10–12 Thus, we can say that common ion effect is pH = –log (7.5 × 10–12) = 11.12a phenomenon based on the Le Chatelier’s principle discussed in section 6.8. On addition of 25 mL of 0.1M HCl solution (i.e., 2.5 mmol of HCl) to 50 In order to evaluate the pH of the solution mL of 0.1M ammonia solution (i.e., 5resulting on addition of 0.05M acetate ion to mmol of NH3), 2.5 mmol of ammonia0.05M acetic acid solution, we shall consider molecules are neutralized. The resultingthe acetic acid dissociation equilibrium once 75 mL solution contains the remaining again, unneutralized 2.5 mmol of NH3 molecules HAc(aq) H+(aq) + Ac–(aq) and 2.5 mmol of NH4+. Initial concentration (M) NH3 + HCl → NH4+ + Cl– 0.05 0 0.05 2.5 2.5 0 0 At equilibrium Let x be the extent of ionization of acetic acid. 0 0 2.5 2.5 Change in concentration (M) The resulting 75 mL of solution contains 2.5 mmol of NH4+ ions (i.e., 0.033 M) and –x +x +x 2.5 mmol (i.e., 0.033 M ) of uneutralised Equilibrium concentration (M) NH3 molecules. This NH3 exists in the 0.05-x x 0.05+x following equilibrium: Therefore, NH4OH   NH4+ + OH– 0.033M – y y yKa= [H+][Ac– ]/[H Ac] = {(0.05+x)(x)}/(0.05-x) where, y = [OH–] = [NH4+]As Ka is small for a very weak acid, x<<0.05. The final 75 mL solution afterHence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05 neutralisation already contains Thus, + 2.5 m mol NH4 ions (i.e. 0.033M), thus 1.8 × 10–5 = (x) (0.05 + x) / (0.05 – x) total concentration of NH4+ ions is given as:= x(0.05) / (0.05) = x = [H+] = 1.8 × 10–5M [NH4+] = 0.033 + ypH = – log(1.8 × 10–5) = 4.74 As y is small, [NH4OH]  0.033 M and Problem 6.24 [NH4+]  0.033M. We know, Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL Kb = [NH4+][OH–] / [NH4OH] of this solution is treated with 25.0 mL of = y (0.033)/(0.033) = 1.77 × 10–5 M 0.10M HCl. The dissociation constant of Thus, y = 1.77 × 10–5 = [OH–] ammonia, Kb = 1.77 × 10–5 [H+] = 10–14 / 1.77 × 10–5 = 0.56 × 10–9 Solution + Hence, pH = 9.24 NH3 + H2O → NH4 + OH– Kb = [NH4 +][OH–] / [NH3] = 1.77 × 10–5 6.11.9 Hydrolysis of Salts and the pH of Before neutralization, their Solutions [NH4 +] = [OH–] = x Salts formed by the reactions between acids [NH3] = 0.10 – x  0.10 and bases in definite proportions, undergo x2 / 0.10 = 1.77 × 10–5 ionization in water. The cations/anions Reprint 2025-26 202 chemistry formed on ionization of salts either exist as increased of H+ ion concentration in solution hydrated ions in aqueous solutions or interact making the solution acidic. Thus, the pH of with water to reform corresponding acids/ NH4Cl solution in water is less than 7. bases depending upon the nature of salts. Consider the hydrolysis of CH3COONH4The later process of interaction between salt formed from weak acid and weak base. water and cations/anions or both of salts The ions formed undergo hydrolysis as follow: is called hydrolysis. The pH of the solution + gets affected by this interaction. The cations CH3COO– + NH4 + H2O CH3COOH + (e.g., Na+, K+, Ca2+, Ba2+, etc.) of strong bases NH4OH and anions (e.g., Cl–, Br–, NO3–, ClO4– etc.) of CH3COOH and NH4OH, also remain into strong acids simply get hydrated but do not partially dissociated form: hydrolyse, and therefore the solutions of CH3COOH CH3COO– + H+ salts formed from strong acids and bases are + NH4OH NH4 + OH–neutral i.e., their pH is 7. However, the other category of salts do undergo hydrolysis. H2O H+ + OH– We now consider the hydrolysis of the Without going into detailed calculation, salts of the following types : it can be said that degree of hydrolysis is (i) salts of weak acid and strong base e.g., independent of concentration of solution, and CH3COONa. pH of such solutions is determined by their pK values:(ii) salts of strong acid and weak base e.g., NH4Cl, and pH = 7 + ½ (pKa – pKb) (6.38) (iii) salts of weak acid and weak base, e.g., The pH of solution can be greater than 7, CH3COONH4. if the difference is positive and it will be less In the first case, CH3COONa being a salt of than 7, if the difference is negative. weak acid, CH3COOH and strong base, NaOH Problem 6.25gets completely ionised in aqueous solution. The pKa of acetic acid and pKb of ammoniumCH3COONa(aq) → CH3COO– (aq)+ Na+(aq) hydroxide are 4.76 and 4.75 respectively. Acetate ion thus formed undergoes Calculate the pH of ammonium acetate hydrolysis in water to give acetic acid and solution. OH– ions Solution CH3COO–(aq)+H2O(l) CH3COOH(aq)+OH–(aq) pH = 7 + ½ [pKa – pKb] Acetic acid being a weak acid = 7 + ½ [4.76 – 4.75] (Ka = 1.8 × 10–5) remains mainly unionised in = 7 + ½ [0.01] = 7 + 0.005 = 7.005solution. This results in increase of OH– ion concentration in solution making it alkaline. The pH of such a solution is more than 7. 6.12 BUFFER SOLUTIONS Many body fluids e.g., blood or urine have Similarly, NH4Cl formed from weak definite pH and any deviation in their pHbase, NH4OH and strong acid, HCl, in water indicates malfunctioning of the body. Thedissociates completely. + control of pH is also very important in NH4Cl(aq) → NH 4(aq) +Cl– (aq) many chemical and biochemical processes. Ammonium ions undergo hydrolysis with Many medical and cosmetic formulations water to form NH4OH and H+ ions require that these be kept and administered NH +4 (aq) + H2O (1) NH4OH(aq) + H+(aq) at a particular pH. The solutions which Ammonium hydroxide is a weak base resist change in pH on dilution or with (Kb = 1.77 × 10–5) and therefore remains the addition of small amounts of acid or almost unionised in solution. This results in alkali are called Buffer Solutions. Buffer Reprint 2025-26 EQUILIBRIUM 203 solutions of known pH can be prepared from acid present in the mixture. Since acid is a the knowledge of pKa of the acid or pKb of base weak acid, it ionises to a very little extent and and by controlling the ratio of the salt and acid concentration of [HA] is negligibly different or salt and base. A mixture of acetic acid and from concentration of acid taken to form sodium acetate acts as buffer solution around buffer. Also, most of the conjugate base, [A—], pH 4.75 and a mixture of ammonium chloride comes from the ionisation of salt of the acid. and ammonium hydroxide acts as a buffer Therefore, the concentration of conjugate around pH 9.25. You will learn more about base will be negligibly different from the buffer solutions in higher classes. concentration of salt. Thus, equation (6.40) takes the form:6.12.1 Designing Buffer Solution [Salt]Knowledge of pKa, pKb and equilibrium pH=pKa + log constant help us to prepare the buffer solution [Acid] of known pH. Let us see how we can do this. In the equation (6.39), if the concentration Preparation of Acidic Buffer of [A—] is equal to the concentration of [HA], then pH = pKa because value of log 1 is zero.To prepare a buffer of acidic pH we use weak Thus if we take molar concentration of acidacid and its salt formed with strong base. and salt (conjugate base) same, the pH of theWe develop the equation relating the pH, the buffer solution will be equal to the pKa of theequilibrium constant, Ka of weak acid and acid. So for preparing the buffer solution ofratio of concentration of weak acid and its the required pH we select that acid whose pKaconjugate base. For the general case where is close to the required pH. For acetic acidthe weak acid HA ionises in water, pKa value is 4.76, therefore pH of the buffer HA + H2O  H3O+ + A– solution formed by acetic acid and sodium For which we can write the expression acetate taken in equal molar concentration will be around 4.76. A similar analysis of a buffer made with a weak base and its conjugate acid leads to Rearranging the expression we have, the result, [Conjugate acid,BH+] pOH= p K b +log [Base,B] Taking logarithm on both the sides and (6.41) rearranging the terms we get — pH of the buffer solution can be calculated by using the equation pH + pOH =14. We know that pH + pOH = pKw and Or pKa + pKb = pKw. On putting these values in equation (6.41) it takes the form as follows:  (6.39) [Conjugate acid,BH ] p K w - pH= p K w  p Ka  log [Base,B] or + [Conjugate acid,BH ] pH= p Ka + log (6.40) [Base,B] (6.42) The expression (6.40) is known as If molar concentration of base and its Henderson–Hasselbalch equation. The conjugate acid (cation) is same then pH of the buffer solution will be same as pKa for the quantity is the ratio of concentration base. pKa value for ammonia is s9.25; therefore a buffer of pH close to 9.25 can be obtained of conjugate base (anion) of the acid and the by taking ammonia solution and ammonium Reprint 2025-26 204 chemistry chloride solution of same molar concentration. We shall now consider the equilibrium For a buffer solution formed by ammonium between the sparingly soluble ionic salt and chloride and ammonium hydroxide, equation its saturated aqueous solution. (6.42) becomes: 6.13.1 Solubility Product Constant + [Conjugate acid,BH ] pH= 9 .25 + log Let us now have a solid like barium sulphate [Base,B] in contact with its saturated aqueous solution. pH of the buffer solution is not affected by The equilibrium between the undisolved solid dilution because ratio under the logarithmic and the ions in a saturated solution can be term remains unchanged. represented by the equation: 6.13 SOLUBILITY EQUILIBRIA OF BaSO4(s)  Ba2+(aq) + SO42–(aq), SPARINGLY SOLUBLE SALTS We have already known that the solubility of The equilibrium constant is given by the ionic solids in water varies a great deal. Some of equation: these (like calcium chloride) are so soluble that K = {[Ba2+][SO42–]} / [BaSO4] they are hygroscopic in nature and even absorb For a pure solid substance thewater vapour from atmosphere. Others (such concentration remains constant and we canas lithium fluoride) have so little solubility writethat they are commonly termed as insoluble. The solubility depends on a number of factors Ksp = K[BaSO4] = [Ba2+][SO42–] (6.43) important amongst which are the lattice We call Ksp the solubility product constant enthalpy of the salt and the solvation enthalpy or simply solubility product. The experimental of the ions in a solution. For a salt to dissolve value of Ksp in above equation at 298K is in a solvent the strong forces of attraction 1.1 × 10–10. This means that for solid barium between its ions (lattice enthalpy) must be sulphate in equilibrium with its saturated overcome by the ion-solvent interactions. The solution, the product of the concentrations solvation enthalpy of ions is referred to in of barium and sulphate ions is equal terms of solvation which is always negative i.e. to its solubility product constant. The energy is released in the process of solvation. concentrations of the two ions will be equal to The amount of solvation enthalpy depends on the molar solubility of the barium sulphate. the nature of the solvent. In case of a non- If molar solubility is S, then polar (covalent) solvent, solvation enthalpy is 1.1 × 10–10 = (S)(S) = S2small and hence, not sufficient to overcome or S = 1.05 × 10–5.lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent. As Thus, molar solubility of barium sulphate a general rule, for a salt to be able to dissolve will be equal to 1.05 × 10–5 mol L–1. in a particular solvent its solvation enthalpy A salt may give on dissociation two or must be greater than its lattice enthalpy so more than two anions and cations carrying that the latter may be overcome by former. different charges. For example, consider a salt Each salt has its characteristic solubility which like zirconium phosphate of molecular formula depends on temperature. We classify salts on (Zr4+)3(PO43–)4. It dissociates into 3 zirconium the basis of their solubility in the following cations of charge +4 and 4 phosphate anions of charge –3. If the molar solubility ofthree categories. zirconium phosphate is S, then it can be seen Category I Soluble Solubility > 0.1M from the stoichiometry of the compound that Category II Slightly 0.01M<Solubility< 0.1M [Zr4+] = 3S and [PO43–] = 4S Soluble Category III Sparingly Solubility < 0.01M and Ksp = (3S)3 (4S)4 = 6912 (S)7 Soluble or S = {Ksp / (33 × 44)}1/7 = (Ksp / 6912)1/7 Reprint 2025-26 EQUILIBRIUM 205 A solid salt of the general formula M px  X qy  Table 6.9 The Solubility Product Constants, with molar solubility S in equilibrium with Ksp of Some Common Ionic Salts at its saturated solution may be represented by 298K. the equation: MxXy(s) xMp+(aq) + yXq– (aq) (where x × p+ = y × q–) And its solubility product constant is given by: Ksp = [Mp+]x[Xq– ]y = (xS)x(yS)y (6.44) = xx . yy . S(x + y) S(x + y) = Ksp / xx . yy S = (Ksp / xx . yy)1 / x + y (6.45) The term Ksp in equation is given by Qsp (section 6.6.2) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions Ksp = Qsp but otherwise it gives the direction of the processes of precipitation or dissolution. The solubility product constants of a number of common salts at 298K are given in Table 6.9. Problem 6.26 Calculate the solubility of A2X3 in pure water, assuming that neither kind of ion reacts with water. The solubility product of A2X3, Ksp = 1.1 × 10–23. Solution A2X3 → 2A3+ + 3X2– Ksp = [A3+]2 [X2–]3 = 1.1 × 10–23 If S = solubility of A2X3, then [A3+] = 2S; [X2–] = 3S therefore, Ksp = (2S)2(3S)3 = 108S5 = 1.1 × 10–23 thus, S5 = 1 × 10–25 S = 1.0 × 10–5 mol/L. Problem 6.27 The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10–15 and 6 × 0–17 respectively. Which salt is more soluble? Explain. Solution AgCN Ag+ + CN– Reprint 2025-26 206 chemistry Ksp = [Ag+][CN–] = 6 × 10–17 Dissolution of S mol/L of Ni(OH)2 provides S mol/L of Ni2+ and 2S mol/L of OH–, but Ni(OH)2 Ni2+ + 2OH– the total concentration of OH– = (0.10 + 2S) Ksp = [Ni2+][OH–]2 = 2 × 10–15 mol/L because the solution already contains Let [Ag+] = S1, then [CN-] = S1 0.10 mol/L of OH– from NaOH. Let [Ni2+] = S2, then [OH–] = 2S2 2 Ksp = 2.0 × 10–15 = [Ni2+] [OH–]2 S1 = 6 × 10–17 , S1 = 7.8 × 10–9 = (S) (0.10 + 2S)2 (S2)(2S2)2 = 2 × 10–15, S2 = 0.58 × 10–4 As Ksp is small, 2S << 0.10, Ni(OH)2 is more soluble than AgCN. thus, (0.10 + 2S) ≈ 0.10 6.13.2 Common Ion Effect on Solubility Hence, of Ionic Salts 2.0 × 10–15 = S (0.10)2 It is expected from Le Chatelier’s principle S = 2.0 × 10–13 M = [Ni2+] that if we increase the concentration of any one of the ions, it should combine with the ion of its opposite charge and some of the The solubility of salts of weak acids like salt will be precipitated till once again Ksp = phosphates increases at lower pH. This is Qsp. Similarly, if the concentration of one of because at lower pH the concentration of the the ions is decreased, more salt will dissolve anion decreases due to its protonation. This to increase the concentration of both the ions in turn increase the solubility of the salt so till once again Ksp = Qsp. This is applicable that Ksp = Qsp. We have to satisfy two equilibria even to soluble salts like sodium chloride simultaneously i.e., except that due to higher concentrations of the ions, we use their activities instead Ksp = [M+] [X–], of their molarities in the expression for Qsp. Thus if we take a saturated solution of sodium chloride and pass HCl gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride [X–] / [HX] = Ka/[H+]ion available from the dissociation of HCl. Sodium chloride thus obtained is of very Taking inverse of both side and adding 1 high purity and we can get rid of impurities we get Hlike sodium and magnesium sulphates. The  HX   common ion effect is also used for almost 1   1   acomplete precipitation of a particular ion  X  K as its sparingly soluble salt, with very low    HX  H   H  K avalue of solubility product for gravimetric   aestimation. Thus we can precipitate silver ion  X  K as silver chloride, ferric ion as its hydroxide Now, again taking inverse, we get(or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations. [X–] / {[X–] + [HX]} = f = Ka/(Ka + [H+]) and it can be seen that ‘f’ decreases as pH decreases. Problem 6.28 If S is the solubility of the salt at a given Calculate the molar solubility of Ni(OH)2 in pH then 0.10 M NaOH. The ionic product of Ni(OH)2 is 2.0 × 10–15. Ksp = [S] [f S] = S2 {Ka/(Ka + [H+])} and S = {Ksp ([H+] + Ka)/Ka}1/2 (6.46) Solution Thus solubility S increases with increase Let the solubility of Ni(OH)2 be equal to S. in [H+] or decrease in pH. Reprint 2025-26 EQUILIBRIUM 207 SUMMARY When the number of molecules leaving the liquid to vapour equals the number of molecules returning to the liquid from vapour, equilibrium is said to be attained and is dynamic in nature. Equilibrium can be established for both physical and chemical processes and at this stage rate of forward and reverse reactions are equal. Equilibrium constant, Kc is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient. For reaction, a A + b B c C +d D Kc = [C]c[D]d/[A]a[B]b Equilibrium constant has constant value at a fixed temperature and at this stage all the macroscopic properties such as concentration, pressure, etc. become constant. For a gaseous reaction equilibrium constant is expressed as Kp and is written by replacing concentration terms by partial pressures in Kc expression. The direction of reaction can be predicted by reaction quotient Qc which is equal to Kc at equilibrium. Le Chatelier’s principle states that the change in any factor such as temperature, pressure, concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce or counteract the effect of the change. It can be used to study the effect of various factors such as temperature, concentration, pressure, catalyst and inert gases on the direction of equilibrium and to control the yield of products by controlling these factors. Catalyst does not effect the equilibrium composition of a reaction mixture but increases the rate of chemical reaction by making available a new lower energy pathway for conversion of reactants to products and vice-versa. All substances that conduct electricity in aqueous solutions are called electrolytes. Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous solutions is due to anions and cations produced by the dissociation or ionization of electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In weak electrolytes there is equilibrium between the ions and the unionized electrolyte molecules. According to Arrhenius, acids give hydrogen ions while bases produce hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one proton. Lewis further generalised the definition of an acid as an electron pair acceptor and a base as an electron pair donor. The expressions for ionization (equilibrium) constants of weak acids (Ka) and weak bases (Kb) are developed using Arrhenius definition. The degree of ionization and its dependence on concentration and common ion are discussed. The pH scale (pH = –log[H+]) for the hydrogen ion concentration (activity) has been introduced and extended to other quantities (pOH = – log[OH–]); pKa = –log[Ka]; pKb = –log[Kb]; and pKw = –log[Kw] etc.). The ionization of water has been considered and we note that the equation: pH + pOH = pKw is always satisfied. The salts of strong acid and weak base, weak acid and strong base, and weak acid and weak base undergo hydrolysis in aqueous solution. The definition of buffer solutions, and their importance are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed and the equilibrium constant is introduced as solubility product constant (Ksp). Its relationship with solubility of the salt is established. The conditions of precipitation of the salt from their solutions or their dissolution in water are worked out. The role of common ion and the solubility of sparingly soluble salts is also discussed. Reprint 2025-26 208 chemistry SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT (a) The student may use pH paper in determining the pH of fresh juices of various vegetables and fruits, soft drinks, body fluids and also that of water samples available. (b) The pH paper may also be used to determine the pH of different salt solutions and from that he/she may determine if these are formed from strong/weak acids and bases. (c) They may prepare some buffer solutions by mixing the solutions of sodium acetate and acetic acid and determine their pH using pH paper. (d) They may be provided with different indicators to observe their colours in solutions of varying pH. (e) They may perform some acid-base titrations using indicators. (f) They may observe common ion effect on the solubility of sparingly soluble salts. (g) If pH meter is available in their school, they may measure the pH with it and compare the results obtained with that of the pH paper. EXERCISES 6.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure? 6.2 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) 2SO3(g) 6.3 At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms I2 (g) 2I (g) Calculate Kp for the equilibrium. 6.4 Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) 2NO (g) + Cl2 (g) (ii) 2Cu(NO3)2 (s) 2CuO (s) + 4NO2 (g) + O2 (g) (iii) CH3COOC2H5(aq) + H2O(l) CH3COOH (aq) + C2H5OH (aq) (iv) Fe3+ (aq) + 3OH– (aq) Fe(OH)3 (s) (v) I2 (s) + 5F2 2IF5 6.5 Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K (ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K Reprint 2025-26 EQUILIBRIUM 209 6.6 For the following equilibrium, Kc= 6.3 × 1014 at 1000 K NO (g) + O3 (g) NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction? 6.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? 6.8 Reaction between N2 and O2– takes place as follows: 2N2 (g) + O2 (g) 2N2O (g) If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture. 6.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br2 (g) 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 . 6.10 At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) 2SO3 (g) What is Kc at this temperature ? 6.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI (g) H2 (g) + I2 (g) 6.12 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction? 6.13 The equilibrium constant expression for a gas reaction is, NH 3 4 O 2 5 Kc  4  NO  H 2 O 6 Write the balanced chemical equation corresponding to this expression. 6.14 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction. 6.15 At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) 2HI (g) is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K? Reprint 2025-26 210 chemistry 6.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14 6.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6 (g) C2H4 (g) + H2 (g) 6.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached? 6.19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium? PCl5 (g) PCl3 (g) + Cl2(g) 6.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2. FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and = 0.80 atm? 6.21 Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium? 6.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium? 6.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO2 (g) 2CO (g) Calculate Kc for this reaction at the above temperature. Reprint 2025-26 EQUILIBRIUM 211 6.24 Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + ½ O2 (g) NO2 (g) where ∆fG (NO2) = 52.0 kJ/mol ∆fG (NO) = 87.0 kJ/mol ∆fG (O2) = 0 kJ/mol 6.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl5 (g) PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) CaCO3 (s) (c) 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) 6.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl2 (g) CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) 2CO (g) (iv) 2H2 (g) + CO (g) CH3OH (g) (v) CaCO3 (s) CaO (s) + CO2 (g) (vi) 4 NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g) 6.27 The equilibrium constant for the following reaction is 1.6 ×105 at 1024K H2(g) + Br2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K. 6.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4 (g) + H2O (g) CO (g) + 3H2 (g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst? 6.29 Describe the effect of: a) addition of H2 b) addition of CH3OH c) removal of CO d) removal of CH3OH on the equilibrium of the reaction: 2H2(g) + CO (g) CH3OH (g) 6.30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as, Reprint 2025-26 212 chemistry PCl5 (g) PCl3 (g) + Cl2 (g) ∆rH = 124.0 kJ mol–1 a) write an expression for Kc for the reaction. b) what is the value of Kc for the reverse reaction at the same temperature? c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ? 6.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H2O (g) CO2 (g) + H2 (g) If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C 6.32 Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl2 (g) 2Cl (g) Kc = 5 ×10–39 b) Cl2 (g) + 2NO (g) 2NOCl (g) Kc = 3.7 × 108 c) Cl2 (g) + 2NO2 (g) 2NO2Cl (g) Kc = 1.8 6.33 The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3? 6.34 The reaction, CO(g) + 3H2(g) CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90. 6.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN–, HClO4, F –, OH–, CO , and S2– 6.36 Which of the followings are Lewis acids? H2O, BF3, H+, and NH4+ 6.37 What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO–3? 6.38 Write the conjugate acids for the following Brönsted bases: NH2–, NH3 and HCOO–. 6.39 The species: H2O, HCO3–, HSO4– and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base. 6.40 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH– (b) F– (c) H+ (d) BCl3 . 6.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH? 6.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. 6.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base. Reprint 2025-26 EQUILIBRIUM 213 6.44 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate? 6.45 The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions. 6.46 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH. 6.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa. 6.48 Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH 6.49 Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution. 6.50 The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid. 6.51 The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb. 6.52 What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline. 6.53 Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ? 6.54 The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH? 6.55 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4. 6.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each. 6.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH? 6.58 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution. Reprint 2025-26 214 chemistry 6.59 The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also? 6.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution. 6.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. 6.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine. 6.63 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF 6.64 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution? 6.65 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature? 6.66 Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2 c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH 6.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions. 6.68 The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions. 6.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10–8 ). 6.70 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water? 6.71 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18). 6.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6). 6.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place? Reprint 2025-26

6.9 Which compound in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane. 6.11 How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene (ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl. 6.12 Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water? (iii) Grignard reagents should be prepared under anhydrous conditions? 6.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform. 6.14 Write the structure of the major organic product in each of the following reactions: (i) CH3CH2CH2Cl + NaI (ii) (CH3)3CBr + KOH (iii) CH3CH(Br)CH2CH3 + NaOH (iv) CH3CH2Br + KCN (v) C6H5ONa + C2H5Cl (vi) CH3CH2CH2OH + SOCl2 (vii) CH3CH2CH = CH2 + HBr (viii) CH3CH = C(CH3)2 + HBr 6.15 Write the mechanism of the following reaction: nBuBr + KCN nBuCN 6.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane. 6.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH. 6.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss. 6.19 How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane Chemistry 190 Reprint 2025-26 (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide

6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN? Answers to Some Intext Questions 6.1 6.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2. 6.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CHCH2CH3 (iv) CH3CCl2CH3 191 Haloalkanes and Haloarenes Reprint 2025-26

6.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.