9.20 A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.

Step 1:

In [Ni(H2O)6]2+, the oxidation state of Ni is +2. It is a 3d8 system. 

The orbital diagram is as follows:

Ni2+ = 3d8 ;  

Here, H2O is a weak field ligand. Hence, a pairing of 3d electron does not takes place. Therefore, there are unpaired electrons in Ni2+.

In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H2O)6]2+ is coloured. 

Step 2:

In [Ni(CN)4]2-, the oxidation state of Ni is +2. The electronic configuration is 3d8

Ni2+ = 3d8 ;  

But CN- is a strong field ligand hence, the pairing of electrons takes place present in 3d orbital. 

Thus, no possibility of d-d transition i[Ni(CN)4]2-. Hence, it is colourless.