Question
$\ce{CuSO_{4}}$ when reacts with $\ce{KCN}$ forms $\ce{CuCN}$ which is insoluble in water. It is soluble in excess of $\ce{KCN},$ due to formation of the following complex:
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Answer
Here, $\ce{CuCN}$ is insoluble in water. This means $Cu$ in $\ce{CuCN}$ is in $+1$ state.
Because $Cu$ is unstable in water only when it is in $+1$ state.
Co$-$ordination sphere exactly contains $\ce{4CN^−}$ ions.
So, its structure would be $\ce{[Cu(CN)_4]^x}$
Calculating $x, Cu$ is $+1$ state
$1 + 4(−1) = x$
$1 − 4 = x$
$x = −3$
So, $\ce{[Cu(CN)4]^{−3}}$
It combines with $3K^+$ ions to form $\ce{K_3[Cu(CN)_{4}]}.$
Because $Cu$ is unstable in water only when it is in $+1$ state.
Co$-$ordination sphere exactly contains $\ce{4CN^−}$ ions.
So, its structure would be $\ce{[Cu(CN)_4]^x}$
Calculating $x, Cu$ is $+1$ state
$1 + 4(−1) = x$
$1 − 4 = x$
$x = −3$
So, $\ce{[Cu(CN)4]^{−3}}$
It combines with $3K^+$ ions to form $\ce{K_3[Cu(CN)_{4}]}.$
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