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  4. CuSO4 reacts with excess KCN to form

Q. $CuSO_{4}$ reacts with excess $KCN$ to form

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Solution:

$CuSO_{4}+2KCN \rightarrow Cu\left(\right.CN\left(\left.\right)_{2}+K_{2}SO_{4}$
$2Cu\left(\right.CN\left(\left.\right)_{2} \rightarrow Cu_{2}\left(\right.CN\left(\left.\right)_{2}+\left(\right.CN\left(\left.\right)_{2}$
$Cu_{2}\left(\right.CN\left(\left.\right)_{2}+KCN\left(\right.excess\left.\right) \rightarrow K_{3}\left[\right.Cu\left(\right.CN\left(\left.\right)_{4}\left]\right.$

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