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 primarily determined by the ligand field and the d-electron transitions within the metal ion.

For \(\text{Ni(H}_2\text{O})_6^{2+}\), water acts as a weak field ligand. This results in minimal splitting of the d-orbitals in \(\text{Ni}^{2+}\), enabling the absorption of visible light and imparting a green hue to the solution.

Conversely, in \(\text{Ni(CN)}_4^{2-}\), cyanide functions as a strong field ligand, inducing significant splitting of the d-orbitals in \(\text{Ni}^{2+}\). This extensive splitting leaves no electronic transitions accessible within the visible spectrum. Consequently, the \(\text{Ni(CN)}_4^{2-}\) solution appears colorless due to the absence of visible light absorption.