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Chemistry Question on Concentration of ores

Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are :

A

[Au(CN)4]–[Au(CN)_4]^– and [Zn(CN)2(OH)2]2−[Zn(CN)_2 (OH)_2]^{2-}

B

[Au(CN)2]–[Au(CN)_2]^– and [Zn(OH)4]2−[Zn (OH)_4]^{2-}

C

[Au(CN)2]–[Au(CN)_2]^– and [Zn(CN)4]2−[Zn (CN)_4]^{2-}

D

[Au(CN)4]2−[Au(CN)_4]^{2-} and [Zn(CN)6]4−[Zn (CN)_6]^{4-}

Answer

[Au(CN)2]–[Au(CN)_2]^– and [Zn(CN)4]2−[Zn (CN)_4]^{2-}

Explanation

Solution

In the metallurgy of gold
4Au+8CN–\+2H2O+O2→4[Au(CN)2]–\+4OH–4Au + 8CN^– \+ 2H_2O + O_2 → 4[Au(CN)_2]^– \+ 4OH^–
2[Au(CN)2]–\+Zn→[Zn(CN)4]2–\+2Au2[Au(CN)_2]^– \+ Zn → [Zn(CN)_4]^{2–} \+ 2Au

So, the correct option is (C): [Au(CN)2]–[Au(CN)_2]^– and [Zn(CN)4]2−[Zn (CN)_4]^{2-}