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, Explain.

Answer 1

In both complexes, $Ni$ is in +2 oxidation state with a valence shell electronic configuration of $3d^8$.

In the presence of weak field water ligands, two unpaired electrons do not pair up.

Hence, the complex $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$ has two unpaired electrons which result in green colour.

Due to d-d transition, red light is absorbed and complimentary light emitted is green.

In presence of strong field cyanide ligand, the unpaired electrons in 3d orbital pair up.

Due to an absence of unpaired electrons, no d-d transitions are possible and the complex $\left[Ni\right(CN{)}_4{]}^{2-}$ is colourless.