Question

A solution of $[Ni(H_2O)_6]^{2+}$ is green but a solution of $[Ni(CN)_4]^{2-}$ is colourless. Explain.

Answer 1

In$[Ni(H_2O)_6]^{ 2+}$ ,$H_2\ddot{O}$ is a weak field ligand. Therefore, there are unpaired electrons in $Ni^{2+}$. In this complex, the delectrons from the lower energy level can be excited to the higher energy level i.e., the possibility of $d - d$ transition is present. Hence, $Ni(H_2O)_6]^{ 2+}$ 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.