Question

${\text{CuS}}{{\text{O}}_{\text{4}}}$ when reacts with ${\text{KCN}}$ forms ${\text{CuCN}}$ which is insoluble in water. It is soluble in excess of ${\text{KCN}}$ due to the formation of the following complex:
A. ${{\text{K}}_{\text{2}}}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$
B. ${{\text{K}}_{\text{3}}}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$
C. ${\text{CuC}}{{\text{N}}_{\text{2}}}$
D. ${\text{Cu}}\left[ {{\text{KCu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$

Answer 1

Hint: Copper forms coordination compounds in which the coordination number of copper is four. The one copper atom is surrounded by four ligands. The oxidation state of copper in its coordination compounds is +1 or +2.

Complete step by step solution:
Copper sulphate when dissolved in potassium cyanide reacts with it to form potassium sulphate and cupric cyanide. The reaction of copper sulphate and potassium cyanide is as follows:
${\text{CuS}}{{\text{O}}_{\text{4}}} + 2{\text{KCN}} \to {{\text{K}}_2}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{Cu}}{\left( {{\text{CN}}} \right)_2}$
The cupric cyanide is unstable in nature. Thus, cupric cyanide further undergoes decomposition and forms cuprous cyanide and cyanogen gas. The decomposition reaction of cupric cyanide is as follows:
${\text{2Cu}}{\left( {{\text{CN}}} \right)_2} \to 2{\text{CuCN}} + {\left( {{\text{CN}}} \right)_2}$
In the above reaction, we can see that the oxidation state of copper decreases from +2 to +1. The decrease in oxidation state is known as reduction. Thus, we can say that cupric cyanide is reduced to cuprous cyanide.
Cuprous cyanide then reacts with potassium cyanide and forms a coordination complex. The reaction of cuprous cyanide with potassium cyanide is as follows:
${\text{CuCN}} + 3{\text{KCN}} \to {{\text{K}}_{\text{3}}}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$
In the above reaction, one mole of cuprous cyanide reacts with three moles of potassium cyanide to form the coordination complex. Thus, from the above reaction, it is clear that ${\text{CuS}}{{\text{O}}_{\text{4}}}$ reacts with ${\text{KCN}}$ forms ${\text{CuCN}}$ which again reacts with ${\text{KCN}}$ and forms a complex ${{\text{K}}_{\text{3}}}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$.

Thus, the correct option is (B) ${{\text{K}}_{\text{3}}}\left[ {{\text{Cu}}{{\left( {{\text{CN}}} \right)}_{\text{4}}}} \right]$.

Note:
Copper is a transition element. It is present in the d-block of the periodic table. The valence orbitals of copper are the d-orbitals. The d-orbitals are vacant. These vacant d-orbitals help copper to form coordination complexes and exhibit variable oxidation state.