When ${CuSO}_{4}$ reacts with KI the oxidation number of Cu changes by

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answer is C.

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Detailed Solution

$\begin{array}{l} {C \overset{}{u} SO}_{4} + 2 KI ⇌ K_{2} {SO}_{4} + {CuI}_{2} \\ 2 {CuI}_{2} ⟶ {Cu}_{2} I_{2} + I_{2} \\ {In CuSO}_{4} {Cu o.s is +2, in Cu}_{2} I_{2} is +1 \end{array}$

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