Question

A solution of $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$ is green but a solution of $\left[Ni\right(CN{)}_4{]}^{2-}$ is colourless. Explaiin.

Answer 1

The configuration of $N{i}^{2+}$ ion is

In $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$, the $H_{2}O$ molecules are weak field ligands, they do not cause electron pairing. As a result, the complex has two unpaired electrons. Thus, d-d transition takes place due to absorption of radiation corresponding to red light and the emission of complementary green colour occurs. Hence, it is green. In $\left[Ni\right(CN{)}_4{]}^{2-},C{N}^{-}$ ions are strong field ligands. So, in the presence of $C{N}^{-}$ ions, the two unpaired electrons in the 3rd-orbitals pair up. Hence, there is no unpaired electron so, no d-d transition. Hence, it is colourless.