(iv) With reference to the diagram below, state, giving a reason for your answer, which equilibrium constant (K1 or K2) would be expected to be largest. Use this reasoning to explain why [Rh(PEt3)3CI] is not an active precatalyst Cl CI CI (0)Using a diagram, illustrate all of the symmetry elements present in BiF5 (ii) Discuss the origin of the inert pair effect and use this to explain why BiF3 is more stable than BiFs, and BiCls is not stable at all (c) Explain why the absorptions in UV-visible spectrum of [Cr(H20)s]3+ are very broad compared to those of [La(H20O)6)]3+ Solution ii) Electronic configuration of Bi = 4f 14 5d 10 6s 2 6p 3 In BiF 5 & BiCl 5 ; Bi(5+) = 4f 14 5d 10 6s 0 6p 0 In BiF 3 ; Bi(3+) = 4f 14 5d 10 6s 2 6p 0 Origin of Inert pair effect is that the electrons in the 6s orbitals are strongly attracted by the nucleus because of weaker shielding by inner (n-1) d and (n-2) f orditals. This s orbital is highly penetrating so it is close to the nucleus compared to other orbitals and electrons present in s orbitals do not involve in any bonding.Inert pair effect start from 6th period from 13 group onward within p-block element.Due to inert pair effect 6s electrons do not participate in the bonding or participate under drastic condition but that product is not stable so BiF 3 is stable and BiF 5 is unstable.In case of BiCl 5 ; 5 orbitals are not available for the acceptance of 5 eletcron pair from chlorine because of inert pair effect. c) Absorption UV-visible spectrum of [Cr(H 2 O) 6 ] 3+ are very broad as compared to those of [La(H 2 O) 6 ] 3+ because Cr(III) is d 3 (t 2g 3 e g 0 ) configuration and there is transition between d orbitals also known as d-d transitions. In [Cr(H 2 O) 6 ] 3+ is an octahedral complex and ligands of this complex interacts with d- orbitals electrons thats why spectrum is broad. Cr =[Ar]4d 1 3d 5 Cr(3+) =[Ar]4d 0 3d 3 But in case of [La(H 2 O) 6 ] 3+ there is no d -electron in the d- orbitals so,ligands do not interact with d- orbitals thats why spectrum is sharp. La = [Xe]5d 1 6s 2 La(3+) = [Xe]5d 0 6s 0 .