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)
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)
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Solution and Explanation
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.
Top Questions on Coordination chemistry
Werner’s coordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. It must be remembered that this theory was put forward before the electron had been discovered by J.J. Thomson in 1897, and before the electronic theory of valency. Werner did not have any of the modern instrumental techniques and all his studies were made using simple experimental techniques. Werner was able to explain the nature of bonding in complexes and he concluded that in complexes, the metal shows two different sorts of valency: primary and secondary. Primary valences are normally ionisable whereas secondary valences are non-ionisable.
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