Question

Copper sulphate dissolved in excess of ${\text{KCN}}$ to give:
A. ${\text{CuCN}}$
B. ${\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{3 - }}$
C. ${\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{2 - }}$
D. ${\text{Cu}}{\left( {{\text{CN}}} \right)_2}$

Answer 1

Usually copper forms coordination compounds in which copper has coordination number of four. Four ligands surround one copper atom. In the coordination compounds of copper, its oxidation state is either +1 or +2.

Complete step by step answer:
Copper sulphate is dissolved in potassium cyanide solution. Copper sulphate reacts with potassium cyanide to form potassium sulphate and cupric cyanide.
${\text{CuS}}{{\text{O}}_4}{\text{ + 2 KCN }} \to {\text{ }}{{\text{K}}_2}{\text{S}}{{\text{O}}_4}{\text{ + Cu}}{\left( {{\text{CN}}} \right)_2}$
Cupric cyanide is unstable. It decomposes to form cuprous cyanide and cyanogen gas.
${\text{2 Cu}}{\left( {{\text{CN}}} \right)_2}{\text{ }} \to {\text{ 2 CuCN + }}{\left( {{\text{CN}}} \right)_2}$
In the above reaction, the oxidation state of copper decreases from +2 to +1. The oxidation state of copper decreases by one during the reaction. Decrease in the oxidation state is called reduction. Thus, cupric cyanide is reduced to cuprous cyanide.
One mole of cuprous cyanide reacts with three moles of potassium cyanide to form the complex compound.
${\text{CuCN + 3 KCN }} \to {\text{ }}{{\text{K}}_3}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]$
In the complex ${\text{CuCN + 3 KCN }} \to {\text{ }}{{\text{K}}_3}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]$ the cation is potassium ion and the anion is ${\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{3 - }}$.

Hence, the option B ) ${\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{3 - }}$ is the correct answer.

Additional Information: The option ${\text{A ) CuCN}}$ is incorrect option because, the initially formed cuprous cyanide reacts with excess of potassium cyanide to form the complex anion ${\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{3 - }}$.

The option ${\text{C ) }}{\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{2 - }}$ is incorrect option because, the oxidation state of copper in this complex is +2. But in the actual complex, copper has an oxidation state of +1.
The option ${\text{D ) Cu}}{\left( {{\text{CN}}} \right)_2}$ is incorrect option because, the initially formed cupric cyanide is unstable and decomposes to form cuprous cyanide and cyanogen gas.

Note: Copper is the transition metal. It has incompletely filled d orbitals in its valence shell. Due to this, it shows variable oxidation states and complex formation tendency.