Assertion : In a mixture of Cd( II ) and Cu( II ) , Cd^ 2+ gets...

$\mathrm{Cd}^{2+}+2 \mathrm{CN}^{-} \longrightarrow \mathrm{Cd}(\mathrm{CN})_2 \downarrow$ $\mathrm{Cd}(\mathrm{CN})_2 \downarrow+2 \mathrm{CN}^{-} \longrightarrow$\left[\mathrm{Cd}(\mathrm{CN})_4\right]^{2-} \left[\mathrm{Cd}(\mathrm{CN})_4\right]^{2-}$is colourless compound and not too stable. When hydrogen sulphide gas is added, cadmium sulphide is precipitated.$\left[\mathrm{Cd}(\mathrm{CN})_4\right]^{2-}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{CdS} \downarrow+2 \mathrm{H}^{+}+4 \mathrm{CN}^{-}$But in case of$\mathrm{Cu}^{2+}$,$\mathrm{Cu}^{2+}+2 \mathrm{CN}^{-} \longrightarrow \mathrm{Cu}(\mathrm{CN})_2 \downarrow 2 \mathrm{Cu}(\mathrm{CN})_2 \downarrow \frac{\text{ Quickly }}{\text{ decomposes }} 2 \mathrm{CuCN} \downarrow+(\mathrm{CN})_2 \uparrow \mathrm{CuCN} \downarrow+3 \mathrm{CN}^{-} \longrightarrow$\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$ This complex is so stable (i.e., $\left[\mathrm{Cu}^{+}\right]$is too low) that $\mathrm{H}_2 \mathrm{~S}$ cannot precipitate $\mathrm{Cu}($\mathrm{I}$)$ sulphide $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$