A solution contains 0.10 mol Cu^2+ and 0.10 mol Cd^2+. It is treated with just enough KCN to convert copper to [Cu(CN)4]^3- (with (CN)2 as the oxidation product, 2Cu^2+ + 10CN- -> 2[Cu(CN)4]^3- + (CN)2) and cadmium to [Cd(CN)4]^2-, and then H2S is passed. The minimum amount of KCN and the precipitate formed are

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc045

A solution contains $0.10 mol$ $Cu X^{2 +}$ and $0.10 mol$ $Cd X^{2 +}$ . It is treated with just enough $KCN$ to convert copper to $[Cu (CN) X_{4}] X^{3 -}$ (with $(CN) X_{2}$ as the oxidation product, $2 Cu X^{2 +} + 10 CN X^{-} 2 [Cu (CN) X_{4}] X^{3 -} + (CN) X_{2}$ ) and cadmium to $[Cd (CN) X_{4}] X^{2 -}$ , and then $H X_{2} S$ is passed. The minimum amount of $KCN$ and the precipitate formed are

Copper needs

10/2 = 5

mol

CN X^{-}

per mol

Cu X^{2 +}

, so

0.10 \times 5 = 0.50 mol

; cadmium needs

4

mol per mol, i.e.

0.40 mol

. Total

= 0.50 + 0.40 = 0.90 mol

. Since

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

is far more stable than

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

, only yellow

CdS

precipitates.