$A$ solution of $[Ni(H_{2}O)_{6}]^{2+}$ is green but a solution of $[Ni(CN)_{4}]^{2-}$ is colourless. Explain.

(N/A) In $[Ni(H_{2}O)_{6}]^{2+}$,$H_{2}O$ is a weak field ligand. Therefore,there are unpaired electrons in $Ni^{2+}$. 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. Hence,$[Ni(H_{2}O)_{6}]^{2+}$ 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.