(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