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 ${\left[Ni\left(H}^{}$2O)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 $N{i}^{2+}$.

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{\left(H}_{}$2O)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$\left[Ni{\left(CN\right)}_{4}{\right]}^{2-}$. Hence, it is colourless.