(A) The magnitude of splitting energy increases down the group. Thus Pt2+−5d8,Pd2+−4d8 and Ni2+−3d8 (B) NH3 is a borderline ligand which forces pairing of electrons in Co3+, whereas H2O is a weakfield ligand and it can not force the pairing in Ni2+. Co3+ in [Co(NH3)6]3+−3d6,4s0
Ni2+ in [Ni(H2O)6]2+−3d8,4s0
So, there is no unpaired electron in [Co(NH3)6]3+ whereas there are unpaired electron in [Ni(H2O)6]2+. Thus [Co(NH3)6]3+ is colourless and [Ni(H2O)6]2+ is coloured due to d−d transition.
As en is symmetrical ligand thus there is no geometrical isomerism. (D) In K3[Fe(CN)6], the oxidation state of Fe is +3 . [Fe(CN)6]3−−3d5,4s0
Number of unpaired electron =1 ∴µ=√n(n+2)B⋅M=√1(1+2)B⋅M=√3B⋅M