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

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

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