A solution of

In $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \mathrm{H}_{2} \ddot{\mathrm{O}}$ is a weak field ligand. Therefore, there are unpaired electrons in $\mathrm{Ni}^{2+} .$ 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. $\left.\operatorname{Hence}, \mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is coloured.

In [Ni(CN)4]2−, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2−. Hence, it is colourless.