(a) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_O < P.

(b) [Ni(CN)₄]^2– is colorless whereas [Ni(H₂O)₆]²⁺ is green. Why? (At. no. of Ni = 28)

(a) The splitting of the degenerate d-orbitals due to presence of ligands in a definite geometry is called crystal field splitting and the energy separation between the two splitted orbitals i.e. t_2g and e_g is called crystal field splitting energy (CFSE) denoted by Δ_o.

Electronic configuration for d⁴ ion if Δ_O < P is t_2g³ e_g¹_. Since, Δ_O < P thus the 4^th electron can go in upper e_g orbital.

(b) [Ni(CN)₄]^2– is colorless whereas [Ni(H₂O)₆]²⁺ is green. This can be explained as follows-

Ni is in +2 oxidation state in both [Ni(CN)₄]^2– and[Ni(H₂O)₆]²⁺.

Electronic configuration of Ni⁺²= 4s⁰ 3d⁸

In [Ni(H₂O)₆]²⁺, weak ligands i.e. H₂O are present which are not able to pair up the unpaired electrons. Therefore, the presence of unpaired electrons in d orbital participate in d-d transition and impart color.

In [Ni(CN)₄]^2–, strong ligands i.e. CN^- are present. These are able to pair the two unpaired electrons in the d oribital and thus no unpaired electrons are left to participate in d-d transition and hence the compound is colorless.