A mixture of Cu^2+ and Cd^2+ is treated with excess KCN. Copper(II) is reduced by cyanide: 2Cu^2+ + 10CN- -> 2[Cu(CN)4]^3- + (CN)2. When H2S is then passed, the moles of CN- consumed per mole of Cu^2+ and the precipitate obtained are

5 mol; black CuS and yellow CdS

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc177

A mixture of $Cu X^{2 +}$ and $Cd X^{2 +}$ is treated with excess KCN. Copper(II) is reduced by cyanide: $2 Cu X^{2 +} + 10 CN X^{-} 2 [Cu (CN) X_{4}] X^{3 -} + (CN) X_{2}$ . When $H X_{2} S$ is then passed, the moles of $CN X^{-}$ consumed per mole of $Cu X^{2 +}$ and the precipitate obtained are

The balanced equation uses 10 cyanide per 2 Cu, i.e. 5 per Cu. The very stable

[Cu (CN) X_{4}] X^{3 -}

gives too little

Cu X^{+}

to precipitate with

H X_{2} S

, whereas

[Cd (CN) X_{4}] X^{2 -}

is less stable and releases enough

Cd X^{2 +}

to give yellow CdS. This is how Cd is detected in presence of Cu.