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)

The color of a coordination complex depends on the d–d transitions in the visible region of the spectrum, which are influenced by the ligand field strength.

- In \(\text{Ni(H}_2\text{O})_6^{2+}\), Ni\(^{2+}\) has an electronic configuration of \([Ar]3d^8\). Water (\(\text{H}_2\text{O}\)) is a weak field ligand, causing a small crystal field splitting (\(\Delta\)), allowing d–d transitions in the visible region, which results in a green color.
- In \(\text{Ni(CN)}_4^{2-}\), cyanide (\(\text{CN}^-\)) is a strong field ligand,

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.