The equilibrium
$2 C u^{I} ⇌ C u^{0} + C u^{I I}$

N O_{3}^{-}

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C l^{-}

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S C N^{-}

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C N^{-}

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Solution

The correct option is D $C N^{-}$
$C u^{2 +}$ ions will react with $C N^{-}$ and $S C N^{-}$ forming $[C u (C N)_{4}]^{3 -}$ and $[C u (S C N)_{4}]^{3 -}$ leading the reaction in the backward direction.
$C u^{2 +} + 2 C N^{-} \to C u (C N)_{2}$
$2 C u (C N)_{2} \to 2 C u C N + (C N)_{2}$
$C u C N + 3 C N^{-} \to [C u (C N)_{1}]^{3 -}$
$C u^{2 +} + 4 S C N^{-} \to [C u (S C N)_{4}]^{3 -}$
$C u^{2 +}$ also combines with $C u C l_{2}$ which reacts with Cu to produce CuCl pushing the reaction in the back ward direction.
$C u C l_{2} + C u \to 2 C u C l ↓$

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2 C u^{I} ⇌ C u^{0} + C u^{I I}

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