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5.29 Amongst the following ions which one has the highest magnetic moment value? 3+ 2+ 2+ (i) [Cr(H O)2] 6 (ii) [Fe(H O)2] 6 (iii) [Zn(H O)2] 6 5 .30 Amongst the following, the most stable complex is 3+ 3+ 3– 3– (i) [Fe(H O)2] 6 (ii) [Fe(NH ) ]3 6 (iii) [Fe(C O 2 ]4 3 (iv) [FeCl ] 6 5.31 What will be the correct order for the wavelengths of absorption in the visible region for the following: 4– 2+ 2+ [Ni(NO ) ]2 6[Ni(NH ) ] , [3 6H O) ] ? 2 6 Answers to Some Intext Questions – 5.1 (i) [Co(NH ) (H O)3 4l 2 2 3 (iv) [Pt(NH )BrC3(NO )] 2 (ii) K 2Ni(CN) ] 4 (v) [PtCl (en2 ](NO2) 3 2 (iii) [Cr(en) ]Cl (vi) Fe [Fe(CN) ] 3 3 4 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: 2+ [Co(NH ) B3 5O 4 + Ba ﬁ BaSO (s) 4 2+ [Co(NH ) S3 5Br + 4a ﬁ No reaction + [Co(NH ) B3 5O 4 + Ag ﬁ No reaction [Co(NH ) SO ]Br + Ag ﬁ AgBr (s) + 3 5 4 2– 5.6 In Ni(CO) , Ni i4 in zero oxidation state whereas in NiCl , it is in +2 oxidation 4 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 2 3 unpaired electron. The hybridisation is d sp forming inner orbital complex. In the presence of H O, (a weak 2igand), 3d electrons do not pair up. The hybridisation is sp d forming an outer orbital complex containing five unpaired electrons, it is strongly paramagnetic. 5.8 In the presence of NH , the 3d elect3ons pair up leaving two d orbitals empty 2 3 3+ to be involved in d sp hybridisation forming inner orbital complex in case of [Co(NH ) ] . 3 6 2+ 8 In Ni(NH ) 3 6 , Ni is in +2 oxidation state and has d configuration, the 3 2 hybridisation involved is sp d forming outer orbital complex. 2 5.9 For square planar shape, the hybridisation is dsp . Hence the unpaired electrons 2 in 5d orbital pair up to make one d orbital empty for dsp hybridisation. Thus there is no unpaired electron. Chemistry 140 Reprint 2026-27