Question

A nickel salt solution containing ${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^ + }$ is green in color on addition of KCN the green color disappears and the complex ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^ - }$ is formed. Explain this observation.

Answer 1

In order to solve this question, we are going to first see the mechanism of the formation of the two compounds and the nature of the two ligands of the compounds. Due to the nature of the ligands, the pairing and the unpairing of electrons is decided and thus, the presence of colour to solution also.

Complete step by step solution:
In the nickel salt solution containing ${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^ + }$, the ligand${H_2}O$ is unpaired which causes the electrons present also to be unpaired. Now these electrons make$d - d$transition and emit the energy so as to give colour to the compounds. However in${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^ - }$, the ligand present is $CN$which a strong ligand in nature is. Thus, there are no unpaired electrons present in the compound and only the paired ones as the pairing is done by the presence of this ligand. Thus, there is no unpaired electron to make the$d - d$transition. Thus, this doesn’t cause any colour to the compound.
Therefore, the green coloured compound containing${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^ + }$, on addition of $KCN\ ]forms the complex \[{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^ - }$ and makes the green colour to disappear from the solution.

Note: Ligands that produce a large splitting are called strong field ligands, and those that produce a small splitting are called weak field ligands. The strong ligands cause all the unpaired electrons to get paired and no $d - d$ transitions, however, weak ligands have unpaired electrons and so $d - d$transitions are there.