Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

$CuSO_4$ 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:

$K_2[Cu(CN)_4]$

$K_3[Cu(CN)_4]$

$CuCN_2$

$Cu[KCu(CN)_4]$

Step-by-Step Solution

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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