Question
$$CuS{O_4}$$ when reacts with $$KCN$$ forms $$CuCN$$ which is insoluble in water. It is soluble in excess of $$KCN$$ due to the formation of the complex
A.
$${K_2}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
B.
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
C.
$$Cu{\left( {CN} \right)_2}$$
D.
$$Cu\left[ {KCu{{\left( {CN} \right)}_4}} \right]$$
Answer :
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
Solution :
$$CuS{O_4}$$ reacts with $$KCN$$ to give a white precipitate of cuprous cyanide and cyanogen gas. The cuprous cyanide dissolves in excess of $$KCN$$ forming $${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right].$$
$$\eqalign{
& CuS{O_4} + 2KCN \to {K_2}S{O_4} + \mathop {Cu{{\left( {CN} \right)}_2}}\limits_{{\text{Unstable}}} \cr
& 2Cu{\left( {CN} \right)_2} \to \mathop {2CuCN}\limits_{{\text{Insoluble}}} + \mathop {CN - CN}\limits_{{\text{cyanogen}}} \cr
& CuCN + 3KCN \to \mathop {{K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]}\limits_{{\text{Soluble}}} \cr} $$