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

4.3 Bond Parameters c = 99 pm 1984.3.1 Bond Length r pmBond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques about which you will learn in higher classes. Each atom of the bonded pair contributes to the rvdw= bond length (Fig. 4.1). In the case of a covalent 180 bond, the contribution from each atom is pm called the covalent radius of that atom. pm 360 The covalent radius is measured approximately as the radius of an atom’s Fig. 4.2 Covalent and van der Waals radii incore which is in contact with the core of an a chlorine molecule. The inner circles adjacent atom in a bonded situation. The correspond to the size of the chlorine covalent radius is half of the distance between atom (rvdw and rc are van der Waals and two similar atoms joined by a covalent bond covalent radii respectively). Reprint 2025-26 108 chemistry Some typical average bond lengths for Table 4.2 Average Bond Lengths for Some single, double and triple bonds are shown in Single, Double and Triple Bonds Table 4.2. Bond lengths for some common Covalent Bondmolecules are given in Table 4.3. Bond Type Length (pm) The covalent radii of some common O–H 96elements are listed in Table 4.4. C–H 107 4.3.2 Bond Angle N–O 136 C–O 143It is defined as the angle between the orbitals C–N 143 containing bonding electron pairs around the C–C 154 central atom in a molecule/complex ion. Bond C=O 121 angle is expressed in degree which can be N=O 122 experimentally determined by spectroscopic C=C 133 methods. It gives some idea regarding the C=N 138 distribution of orbitals around the central C≡N 116 atom in a molecule/complex ion and hence it C≡C 120 helps us in determining its shape. For Table 4.3 Bond Lengths in Some Commonexample H–O–H bond angle in water can be Moleculesrepresented as under : Molecule Bond Length (pm) H2 (H – H) 74 F2 (F – F) 144 4.3.3 Bond Enthalpy Cl2 (Cl – Cl) 199 It is defined as the amount of energy required Br2 (Br – Br) 228 to break one mole of bonds of a particular I2 (I – I) 267 type between two atoms in a gaseous state. N2 (N ≡ N) 109 The unit of bond enthalpy is kJ mol–1. For O2 (O = O) 121 example, the H – H bond enthalpy in hydrogen HF (H – F) 92 molecule is 435.8 kJ mol–1. HCl (H – Cl) 127 HBr (H – Br) 141H2(g) → H(g) + H(g); ∆aH = 435.8 kJ mol–1 HI (H – I) 160 Similarly the bond enthalpy for molecules Table 4.4 Covalent Radii, *rcov/(pm)containing multiple bonds, for example O2 and N2 will be as under : O2 (O = O) (g) → O(g) + O(g); ∆aH = 498 kJ mol–1 N2 (N ≡ N) (g) → N(g) + N(g); ∆aH = 946.0 kJ mol–1 It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule. For a heteronuclear diatomic molecules like HCl, we have HCl (g) → H(g) + Cl (g); ∆aH = 431.0 kJ mol–1 In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O * The values cited are for single bonds, except where molecule, the enthalpy needed to break the otherwise indicated in parenthesis. (See also Unit 3 for two O – H bonds is not the same. periodic trends). Reprint 2025-26 Chemical Bonding And Molecular Structure 109 H2O(g) → H(g) + OH(g); ∆aH1 = 502 kJ mol–1 OH(g) → H(g) + O(g); ∆aH2 = 427 kJ mol–1 The difference in the ∆aH value shows that the second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in different molecules like C2H5OH (ethanol) and water. Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule, 502 + 427 Fig. 4.3 Resonance in the O3 molecule Average bond enthalpy = 2 (structures I and II represent the two canonical = 464.5 kJ mol–1 forms while the structure III is the resonance hybrid) 4.3.4 Bond Order In the Lewis description of covalent bond, In both structures we have a O–O single the Bond Order is given by the number bond and a O=O double bond. The normal of bonds between the two atoms in a O–O and O=O bond lengths are 148 pm molecule. The bond order, for example in and 121 pm respectively. Experimentally H2 (with a single shared electron pair), in O2 determined oxygen-oxygen bond lengths in (with two shared electron pairs) and in N2 the O3 molecule are same (128 pm). Thus the (with three shared electron pairs) is 1,2,3 oxygen-oxygen bonds in the O3 molecule are respectively. Similarly in CO (three shared intermediate between a double and a single electron pairs between C and O) the bond bond. Obviously, this cannot be represented order is 3. For N2, bond order is 3 and its by either of the two Lewis structures shown is 946 kJ mol–1; being one of the highest above. for a diatomic molecule. The concept of resonance was introduced Isoelectronic molecules and ions have to deal with the type of difficulty experienced identical bond orders; for example, F2 and in the depiction of accurate structures of O22– have bond order 1. N2, CO and NO+ have molecules like O3. According to the concept bond order 3. of resonance, whenever a single Lewis structure cannot describe a molecule A general correlation useful for accurately, a number of structures withunderstanding the stablities of molecules is that: with increase in bond order, similar energy, positions of nuclei, bonding bond enthalpy increases and bond length and non-bonding pairs of electrons are decreases. taken as the canonical structures of the hybrid which describes the molecule 4.3.5 Resonance Structures accurately. Thus for O3, the two structures It is often observed that a single Lewis structure shown above constitute the canonical is inadequate for the representation of a structures or resonance structures and molecule in conformity with its experimentally their hybrid i.e., the III structure represents determined parameters. For example, the the structure of O3 more accurately. This is ozone, O3 molecule can be equally represented also called resonance hybrid. Resonance is by the structures I and II shown below: represented by a double headed arrow. Reprint 2025-26 110 chemistry Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule. Fig. 4.5 Resonance in CO2 molecule, I, II Problem 4.3 and III represent the three canonical Explain the structure of CO32– ion in forms. terms of resonance. Solution In general, it may be stated that The single Lewis structure based on • Resonance stabilizes the molecule as the the presence of two single bonds and energy of the resonance hybrid is less one double bond between carbon than the energy of any single cannonical and oxygen atoms is inadequate to structure; and, represent the molecule accurately as it • R e s o n a n c e a v e r a g e s t h e b o n d represents unequal bonds. According characteristics as a whole. to the experimental findings, all carbon to oxygen bonds in CO32– are equivalent. Thus the energy of the O3 resonance Therefore the carbonate ion is best hybrid is lower than either of the two described as a resonance hybrid of the cannonical froms I and II (Fig. 4.3). canonical forms I, II, and III shown below. Many misconceptions are associated with resonance and the same need to be dispelled. You should remember that : • The cannonical forms have no real existence. • The molecule does not exist for a certain fraction of time in one cannonical form and for other fractions of time in other Fig. 4.4 Resonance in CO32–, I, II and III cannonical forms. represent the three canonical • There is no such equilibrium between forms. the cannonical forms as we have Problem 4.4 between tautomeric forms (keto and enol) in tautomerism. Explain the structure of CO2 molecule. • The molecule as such has a single Solution structure which is the resonance The experimentally determined carbon hybrid of the cannonical forms and to oxygen bond length in CO 2 is which cannot as such be depicted by 115 pm. The lengths of a normal a single Lewis structure. carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) 4.3.6 Polarity of Bonds are 121 pm and 110 pm respectively. The existence of a hundred percent ionic or The carbon-oxygen bond lengths in covalent bond represents an ideal situation. CO2 (115 pm) lie between the values for C=O and C≡O. Obviously, a single In reality no bond or a compound is either Lewis structure cannot depict this completely covalent or ionic. Even in case of position and it becomes necessary to covalent bond between two hydrogen atoms, write more than one Lewis structures there is some ionic character. and to consider that the structure of When covalent bond is formed between CO2 is best described as a hybrid of two similar atoms, for example in H2, O2, the canonical or resonance forms I, II and III. Cl2, N2 or F2, the shared pair of electrons is equally attracted by the two atoms. As a result Reprint 2025-26 Chemical Bonding And Molecular Structure 111 electron pair is situated exactly between the In case of polyatomic molecules the dipole two identical nuclei. The bond so formed is moment not only depend upon the individual called nonpolar covalent bond. Contrary to dipole moments of bonds known as bond this in case of a heteronuclear molecule like dipoles but also on the spatial arrangement HF, the shared electron pair between the two of various bonds in the molecule. In such atoms gets displaced more towards fluorine case, the dipole moment of a molecule is the since the electronegativity of fluorine (Unit 3) vector sum of the dipole moments of various is far greater than that of hydrogen. The bonds. For example in H2O molecule, which resultant covalent bond is a polar covalent has a bent structure, the two O–H bonds are bond. oriented at an angle of 104.50. Net dipole As a result of polarisation, the molecule moment of 6.17 × 10–30 C m (1D = 3.33564 possesses the dipole moment (depicted × 10–30 C m) is the resultant of the dipole below) which can be defined as the product of moments of two O–H bonds. the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter ‘µ’. Mathematically, it is expressed as follows : Dipole moment (µ) = charge (Q) × distance of separation (r) Dipole moment is usually expressed in Net Dipole moment, µ = 1.85 D Debye units (D). The conversion factor is = 1.85 × 3.33564 × 10–30 C m = 6.17 ×10–30 C m 1 D = 3.33564 × 10–30 C m The dipole moment in case of BeF2 is zero.where C is coulomb and m is meter. This is because the two equal bond dipoles Further dipole moment is a vector quantity point in opposite directions and cancel the and by convention it is depicted by a small effect of each other. arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow ( ) put on Lewis structure of the molecule. The In tetra-atomic molecule, for example in cross is on positive end and arrow head is on BF3, the dipole moment is zero although thenegative end. For example the dipole moment B – F bonds are oriented at an angle of 120o of HF may be represented as : to one another, the three bond moments give a net sum of zero as the resultant of any two H F is equal and opposite to the third. This arrow symbolises the direction of the shift of electron density in the molecule. Note that the direction of crossed arrow is opposite to the conventional direction of dipole moment vector. Peter Debye, the Dutch chemist received Nobel prize in 1936 for Let us study an interesting case of NH3 his work on X-ray diffraction and and NF3 molecule. Both the molecules have dipole moments. The magnitude pyramidal shape with a lone pair of electrons of the dipole moment is given in Debye units in order to honour him. on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant Reprint 2025-26 112 chemistry dipole moment of NH3 (4.90 × 10–30 C m) is • The smaller the size of the cation and the greater than that of NF3 (0.8 × 10–30 C m). This larger the size of the anion, the greater is because, in case of NH3 the orbital dipole the covalent character of an ionic bond. due to lone pair is in the same direction as • The greater the charge on the cation, the the resultant dipole moment of the N – H greater the covalent character of the ionic bonds, whereas in NF3 the orbital dipole is in bond. the direction opposite to the resultant dipole • For cations of the same size and charge,moment of the three N–F bonds. The orbital the one, with electronic configurationdipole because of lone pair decreases the effect (n-1)dnnso, typical of transition metals, isof the resultant N – F bond moments, which more polarising than the one with a noble results in the low dipole moment of NF3 as gas configuration, ns2 np6, typical of alkalirepresented below : and alkaline earth metal cations. The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per Dipole moments of some molecules are cent covalent character of the ionic bond. shown in Table 4.5. Just as all the covalent bonds have 4.4 The Valence Shell Electron some partial ionic character, the ionic Pair Repulsion (VSEPR) Theory bonds also have partial covalent character. As already explained, Lewis concept is unable The partial covalent character of ionic to explain the shapes of molecules. This bonds was discussed by Fajans in terms of theory provides a simple procedure to predict the following rules: the shapes of covalent molecules. Sidgwick Table 4.5 Dipole Moments of Selected Molecules Dipole Type of Molecule Example Geometry Moment, µ(D) Molecule (AB) HF 1.78 linear HCl 1.07 linear HBr 0.79 linear Hl 0.38 linear H2 0 linear Molecule (AB2) H2O 1.85 bent H2S 0.95 bent CO2 0 linear Molecule (AB3) NH3 1.47 trigonal-pyramidal NF3 0.23 trigonal-pyramidal BF3 0 trigonal-planar Molecule (AB4) CH4 0 tetrahedral CHCl3 1.04 tetrahedral CCl4 0 tetrahedral Reprint 2025-26 Chemical Bonding And Molecular Structure 113 and Powell in 1940, proposed a simple theory result in deviations from idealised shapes and based on the repulsive interactions of the alterations in bond angles in molecules. electron pairs in the valence shell of the For the prediction of geometrical shapes atoms. It was further developed and redefined of molecules with the help of VSEPR theory, by Nyholm and Gillespie (1957). it is convenient to divide molecules into The main postulates of VSEPR theory are two categories as (i) molecules in which as follows: the central atom has no lone pair and • The shape of a molecule depends upon (ii) molecules in which the central atom the number of valence shell electron pairs has one or more lone pairs. (bonded or nonbonded) around the central Table 4.6 (page114) shows the atom. arrangement of electron pairs about a • Pairs of electrons in the valence shell repel central atom A (without any lone pairs) and one another since their electron clouds are geometries of some molecules/ions of the type negatively charged. AB. Table 4.7 (page 115) shows shapes of some simple molecules and ions in which the central• These pairs of electrons tend to occupy atom has one or more lone pairs. Table 4.8 such positions in space that minimise (page 116) explains the reasons for the repulsion and thus maximise distance distortions in the geometry of the molecule. between them. As depicted in Table 4.6, in the• The valence shell is taken as a sphere compounds of AB2, AB3, AB4, AB5 and AB6, with the electron pairs localising on the the arrangement of electron pairs and the spherical surface at maximum distance B atoms around the central atom A are : from one another. linear, trigonal planar, tetrahedral, • A multiple bond is treated as if it is a single trigonal-bipyramidal and octahedral, electron pair and the two or three electron respectively. Such arrangement can be seen pairs of a multiple bond are treated as a in the molecules like BF3 (AB3), CH4 (AB4) and single super pair. PCl5 (AB5) as depicted below by their ball and • Where two or more resonance structures stick models. can represent a molecule, the VSEPR model is applicable to any such structure. The repulsive interaction of electron pairs decrease in the order: Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp) Fig. 4.6 The shapes of molecules in which central atom has no lone pair Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important The VSEPR Theory is able to predict difference between the lone pairs and bonding geometry of a large number of molecules, pairs of electrons. While the lone pairs are especially the compounds of p-block elements localised on the central atom, each bonded accurately. It is also quite successful in pair is shared between two atoms. As a result, determining the geometry quite-accurately the lone pair electrons in a molecule occupy even when the energy difference between more space as compared to the bonding pairs possible structures is very small. The of electrons. This results in greater repulsion theoretical basis of the VSEPR theory regarding between lone pairs of electrons as compared the effects of electron pair repulsions on to the lone pair - bond pair and bond pair - molecular shapes is not clear and continues bond pair repulsions. These repulsion effects to be a subject of doubt and discussion. Reprint 2025-26 114 chemistry Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons Reprint 2025-26 Chemical Bonding And Molecular Structure 115 Table 4.7 Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or More Lone Pairs of Electrons(E). Reprint 2025-26 116 chemistry Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB2E 4 1 Bent Theoretically the shape should have been triangular planar but actually it is found to be bent or v-shaped. The reason being the lone pair- bond pair repulsion is much more as compared to the bond pair-bond pair repulsion. So the angle is reduced to 119.5° from 120°. AB3E 3 1 Trigonal Had there been a bp in place of lp the shape would have pyramidal been tetrahedral but one lone pair is present and due to the repulsion between lp-bp (which is more than bp-bp repulsion) the angle between bond pairs is reduced to 107° from 109.5°. Bent The shape should have been AB2E2 2 2 tetrahedral if there were all bp but two lp are present so the shape is distorted tetrahedral or angular. The reason is lp-lp repulsion is more than lp-bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°. AB4E 4 1 See- In (a) the lp is present at axial saw position so there are three lp—bp repulsions at 90°. In(b) the lp is in an equatorial position, and there are two lp—bp repulsions. Hence, arrangement (b) is more stable. The shape shown in (b) is described as a distorted tetrahedron, a folded square (More stable) or a see-saw. Reprint 2025-26 Chemical Bonding And Molecular Structure 117 Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB3E2 3 2 T-shape In (a) the lp are at equatorial position so there are less lp- bp repulsions as compared to others in which the lp are at axial positions. So structure (a) is most stable. (T-shaped).

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.