62) a s and p both form π -bonding with complex and π-bonding capacity of sulphur is greater than phosphorus due to smaller size of d-orbital of sulphur. Hence, in presence of sulphur trans to phosphorus donor atom phosphorus-metal bond will be weak hence they do not lie trans to each other in the complex. As nitrogen does not involment in π -bonding with complex hence when nitrogen atom is trans to phosphorus, phosphorus become able to form efficient π -bond with metal hence become stable thats why P and N are trans to each other 63) c 64) a Terminal NO=1672 cm¹ Bridging NO=1505 cm 65) d N=1s 2s 2p L= +1 0 -1 S= 1/2+1/2+1/2=3/2 L=+1+0-1=0 66) a - - Trans effect NO2 > Cl > NH3 67) c Total isomers = geometrical isomer + ½(optical isomer) Geometrical isomer = 2 isomers (trans form + cis form) Optical isomer = 2 isomers (d –cis-form + l-cis form) Total isomers= 3 68) a Explanation There is spin change during electron transition. Therefore, transition is spin disallowed 69) a [CoCl(NH3)5]+2 + OH- → [Co(NH3)4(NH2)Cl]+1 +H2O [Co(NH3)4(NH2)Cl]+1 → [Co(NH3) 4(NH2)]+2 + Cl-1 [Co(NH3) 4(NH2)]+2 + OH- → [Co(NH3)5OH]+2 Base hydrolysis of [CoCl(NH3)5]+2 depends on the concentration of both [CoCl(NH3)5]+2 and base. In this reaction OH- abstract proton from coordinated NH3. The base hydrolysis of [Co(CN)]-3 is independent of base because it has no ionizable proton. 70) c [Ne] 2p1 3p1 2s+1=3; 2s=2; s=1 For, D, L=2 J=|L+s|..... |L-s|=|2+1|... |2-1|= 3,2,1 Levels for 3D are 3D3 , 3D2 , 3D1 71) a [MnF6] -3 , Mn+3→3d4 → 𝒕𝟑𝟐𝒈 𝒆𝟏𝒈 (HS) 𝜇 = 4 4 + 2 = 24 = 4.9 S =2 →2s+1 = 2×2+1=5 L=2→D Ground state term symbol ³D 72) a It is first inorganic optically active compound discovered by Werner. Number of Co-O bonds = 12 Number of Co-N bonds 12 73) a [CrCl6] -3 Cr 3+ →3d3 𝜇 s = 𝑛(𝑛 + 2) = 3(3 + 2) = 3.87 𝜇 L+S = 𝑛 𝑛 + 2 + 𝑙(𝑙 + 1) BM 𝜇 L+S = 𝑛 𝑛 + 2 + 𝑙(𝑙 + 1) = 3 3 + 2 + 3(3 + 1) = 15 + 12 = 27 = 5.20 BM 74) a The bonding in phosphine ligands, like that of carbonyls having two components. The primary component is sigma donation of the phosphine lone pain to on empty orbital on the metal. The second component is back donation from filled metal orbital to an empty orbital on the phosphine ligand. This empty phosphorous orbital has been described as being either a d-orbital or an antibonding sigma orbital (ϭ*). So, in transition metal phosphine (M-PR3) complexes, the back bonding involves donation of electrons from M(t2g )→ PR3(σ*) 75) b For allowed transition : a)Spin multiplicity should be same or ΔS = 0 Option c and d (incorrect) as ΔS ≠ 0 b) ΔL = ±1 , for allowed transition 3S→3P ΔS = 3-3=0 ΔL= 1-0= 1 76) d ns1 np1 L=1 ≡ 𝑃 S=1 hence, 2S+1=3 For J value, if orbital is less than half filled than J=|L-S|=|1-1|=0 Hence, 3P0 77) a Fe(II) protoporphyrin → π →π* transition [Mn(H₂O) 6]Cl₂ → Mn+2→d5 (high spin) spin forbidden (d-d) transition. ∆S=1. Hence, not allowed. [Co(H₂O)6]Cl2 →Co+2→ d7 (high spin) (d-d) ∆S= 0. hence allowed 78) c K4[Cr(CN) 6]= Cr+2→ d4 (low spin) For a compound to show John Teller Distortion eg or t2g set should be electronically degenerate. (1) Cr+2 → d4 (low spin) Electronically degenerate hence show John Teller Distortion. (2) K4[Fe(CN) 6] = Fe+2 (low spin) = d6 (low spin) Electronically non-degenerate hence, no Jahn-Teller distrotion (3) K3[Co(CN)6] = Co+3 → d6 (low spin) Same as above hence, no Jahn-Teller Distrotion. (4) K4[Mn(CN) 6] = Mn+2 (low spin) electronically degenerate Hence, complex will show Jahn-Teller Distortion. 79) a [Fe(Phen) 2 (NCS) 2] Fe+2 = d6 complex At high temperature high spin and at low temperature low spin behaviour at 250 K CFSE = 1.6 Δ0 +1.2 Δ0 =0.44, 𝜇 = 𝑛(𝑛 + 2) = 4(4 + 2) = 4.90 at 150 K CFSE -0.4 x 6 Δ0 = -2.4∆ 𝜇 =0 80) a Taking the valence shell configuration Be 2s1 3s1 s=1/2+1/2 =1 Multiplicity =2S+1= 2x1+1=3 L= 0+0= 0 , S term J= (L+S) = (0+1) = 1 Hence, terms is 3S1 81) a [CrO4]2-< [MnO4]2- < [FeO4]2- All have +6 oxidation state but due to smaller size on going from Cr to Fe, Fe has very high charge density. Hence, it has very high tendency to accept electron. Hence, strongest oxidising agent. 82) c Number of unpaired electrons (A) [CoF6] 3- → d6 (high spin) = t42g e2g 4 (B) [IrCl6] 3- →d6 (low spin) = t62g e0g 0 (C) [Fe(H2O) 6] 2+ →d6 (high spin) = t42g e2g 4 Since, A and C has same number of unpaired electron. Hence, they have same magnetic moment. As A and Care 3d-block metal with weak ligand. Hence, they are high spin while Ir being 5d-metal is low spin. Because 5d has greater splitting power than 3d. 83) a [Mn (H2O) 6] 2+ = d6 (HS) As all the levels are electronically non-degenerate. Hence, No Jahn-Teller distortion. Therefore, all the Mn-O bond length will be equal. 84) a [Fe(phen) 3] 2+ → [Fe(phen) 3] 3+ + e- ( Red) (Blue) In presence of oxidising agent like K2Cr2O7 , [Fe(phen) 3] 2+ changes it colours from red to blue hence, they are used in redox titration. 85) a The stability (formation constant for complexation of cryptate complex depends upon) (1) size of cavity (2) size of metal cation Hence, K+ ion will form most stable complex with cryptand-222. 86) c E2 > E1 hence, energy required for transfer of electron from oxygen to Re falls in UV region. Hence, [ReO4] - colourless while in [MnO4] - due to less energy difference it falls in visible region. Hence, [MnO4] - coloured. Also as both Mn and Fe have +7 oxidation state i.e. they have no d-electron hence, no d-d transition and colour arises due to LMCT. 87) a [Co(CN) 5Cl] 3- + OH- →[Co(CN)5(OH)] 3- +Cl- The reaction proceed through dissociative pathway and rate is dependent only concentration of substrates. As substate do not have acidic hydrogen. Hence, it does not undergo SN 1CB mechanism. [Co(CN) 5Cl] 3- → [Co(CN) 5] 2- + OH- → [Co(CN) 5OH] 3- (slow) 88) a Explanation 89) c Tanabe-Sugano are useful in interpretation of spectra of both high spin and low spin complexes of d2- d8 metal cation. In Tanabe-Sugano, the energy of excited sate (expressed as E/B) are plotted against ligand field strength (expressed as (∆0 /B). Zero energy is taken for the lowest term and also two form of same symmetry never cross each other and they bent for apart from each other due to repulsion. 90) b [CrF6] -3 = d3 Orgel diagram for d3. ground state= 4F First excited state = 4P 4 A2g →4T2g (Δ0) v1 = 14900 cm-1 ; v2 = 22700 cm-1 ; v3 = 34400 cm-1 Since, 4T1g (P) and 4T1g (F) have no fix energy therefore, they will not provide accurate value of Δ0. Thus, energy difference between 4A2g → 4T2g 14900 cm-1 will be correspond to Δ0 91) b For [1s22s2p6 3p1] O 2s+1=2× + 1 = 2 W J=|l+s|..... |l-s| =|1+½|.....|1-½| = 3/2 1/2 Term symbol 2P3/2 2 P1/2 92) b 3 F→ 3D Since, for allowed transition (atomic), ∆S=0, ∆L=0,±1 In option a and c spin multiplicity is not same therefore incorrect In option d , ∆L=2 therefore incorrect. 93) d [NiIIL6] n+ or n- has shows absorption bands at 8500, 15400, and 26000 cm-1 [NiII Lʹ6] n+ or n- at 10750, 17500 and 28200 cm-1 as for first complex absorptions bands are at low energy i.e. it has weak splitting therefore, it has weak ligand and for II complex high energy absorption bands corresponds to strong ligand.Thus, L is weak and L' is strong ligand. 94) b Number of microstate in 3F is calculated as (2S + 1) (2L+1) For F, L=3. Hence, (3) (2x3+1) =21 95) b Chelate effect is predominately due to entropy change ∆H nearly same but entropy change is more. 96) b D6 = L= ∑ML = 5-3 = 2 = D S=2 , 2S+1=5 Hence, lowest energy form = 5D 97) d H2 (excited state) L=0=Σ S=1/2+1/2= 1 , 2s+1=3 gxu=u = 3∑. 𝒖 For half filled. 98) c [Cr(bipyridyl)3]+3→ d3 = 4F (ground state term) As phosphorescence is spin-forbidden transition and also occurs when electron comes from excited to ground state. Hence, transition 4A2g ← 2Eg is responsible for the phosphorescence. 99) c For orbital contribution the set should be unsymmetrically filled [Cu(H2O)6]+2→ t62g e3g →No contribution [Ni(H2O)6]+2→ t62g e2g →No contribution [Co(H2O)6]+2→ t52g e2g → Orbital contribution [Cr(H2O)6]+2→ t32g e1g → No contribution
Explanation as extracted from the printed page; notation may be imperfect.