When $C u S O_{4}$ is treated with excess KCN solution, what compound(s) is/are formed?

K [C u (C N)_{2}]

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K_{3} [C u (C N)_{4}]

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K_{2} [C u (C N)_{4}]

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N \equiv C - C \equiv N

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Solution

The correct option is D $N \equiv C - C \equiv N$
$(C u S O_{4} + 2 K C N ⟶ C u (C N)_{2} + K_{2} S O_{4}) \times 2$
$2 C u (C N)_{2} ⟶ C u_{2} (C N)_{2} + (C N)_{2}$
$C u_{2} (C N)_{2} + 6 K C N ⟶ 2 K_{3} [C u (C N)_{4}]$
So,adding these reactions we get,
$2 C u S O_{4} + 10 K C N ⟶ 2 K_{3} [C u (C N)_{4}] + 2 K_{2} S O_{4} + (C N)_{2}$

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