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

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

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