Copper sulphate dissolves in excess of KCN to give :-
A
$CuCN$
B
$[Cu(CN)_4]^{3-}$
C
$[Cu(CN)_4]^{2-}$
D
$Cu(CN)_2$
The answer is B.
Solution
Copper sulphate first reacts with KCN to give cupric cyanide: $2KCN + CuSO_4 \rightarrow K_2SO_4 + Cu(CN)_2$ Cupric cyanide is unstable and decomposes into cuprous cyanide and cyanogen: $2Cu(CN)_2 \rightarrow 2CuCN + (CN)_2$ In this step copper is reduced from +2 to +1. Cuprous cyanide dissolves in excess of KCN to form a soluble complex: $CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$ The complex ion formed is $[Cu(CN)_4]^{3-}$, in which copper is in the +1 oxidation state ($x - 4 = -3$). So copper sulphate dissolves in excess of KCN to give $[Cu(CN)_4]^{3-}$.
ConceptReduction of Cu(II) by cyanide
The d- and f-Block Elements in past papers
76 questions from this chapter have appeared across 27 exam years.
