Question

The equilibrium
$2C{u}^I\rightleftharpoons C{u}^0+C{u}^{II}$

• A. $N{O}_3^{-}$
• B. $C{l}^{-}$
• C. $SC{N}^{-}$
• D. $C{N}^{-}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $C{l}^{-}$
C $SC{N}^{-}$
D $C{N}^{-}$

$C{u}^{2+}$ ions will react with $C{N}^{-}$ and $SC{N}^{-}$ forming $\left[Cu\right(CN{)}_4{]}^{3-}$ and $\left[Cu\right(SCN{)}_4{]}^{3-}$ leading the reaction in the backward direction.
$C{u}^{2+}+2C{N}^{-}\to Cu(CN{)}_2$
$2Cu(CN{)}_2\to 2CuCN+(CN{)}_2$
$CuCN+3C{N}^{-}\to \left[Cu\right(CN{)}_1{]}^{3-}$
$C{u}^{2+}+4SC{N}^{-}\to \left[Cu\right(SCN{)}_4{]}^{3-}$
$C{u}^{2+}$ also combines with $CuC{l}_2$ which reacts with Cu to produce CuCl pushing the reaction in the back ward direction.
$CuC{l}_2+Cu\to 2CuCl\downarrow$