$CuSO_4$ when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN, due to

Question

$CuSO_4$ when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN, due to the formation of the following complex:

Accepted Answer

When $CuSO_4$ reacts with $KCN$, cupric cyanide $Cu(CN)_2$ is initially formed, which being unstable rapidly decomposes to give a white precipitate of cuprous cyanide $CuCN$ and cyanogen gas $(CN)_2$.
$2CuSO_4 + 4KCN \rightarrow 2CuCN \downarrow + (CN)_2 \uparrow + 2K_2SO_4$

The insoluble cuprous cyanide ($CuCN$) then dissolves in excess of $KCN$ to form a highly stable and soluble complex, potassium tetracyanidocuprate(I), $K_3[Cu(CN)_4]$.
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$

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