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).

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

8.18 Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclo- hexanone does not. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.