Question

Why is a solution of \(\text{Ni(H}_2\text{O})_6^{2+}\) green while a solution of \(\text{Ni(CN)}_4^{2-}\) is colourless? (At. No. of Ni = 28) 

Hint:

The color of a complex is influenced by the strength of the ligand and the extent of splitting of the metal€™s d-orbitals.

Answer 1

The color of a coordination compound is largely influenced by the ligand field and the d-electron transitions of the metal ion.

In the case of \(\text{Ni(H}_2\text{O})_6^{2+}\), water is a weak field ligand that causes a small splitting of the d-orbitals in \(\text{Ni}^{2+}\). This allows the absorption of light in the visible spectrum, giving the solution a green color.

In contrast, in \(\text{Ni(CN)}_4^{2-}\), cyanide is a strong field ligand that causes a large splitting of the d-orbitals in \(\text{Ni}^{2+}\), leading to no available electronic transitions in the visible region. As a result, the solution of \(\text{Ni(CN)}_4^{2-}\) is colorless because no visible light is absorbed.