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

This question is about the MacArthur-Forrest cyanide process for gold extraction.

Leaching reaction (formation of complex A): Gold reacts with dilute NaCN solution in the presence of oxygen:

$$4Au(s) + 8NaCN(aq) + O_2(g) + 2H_2O(l) \rightarrow 4Na[Au(CN)_2](aq) + 4NaOH(aq)$$

The gold complex formed is $$[Au(CN)_2]^-$$ (dicyanoaurate(I) ion). So complex A = $$[Au(CN)_2]^-$$.

Recovery with zinc (formation of complex B): Zinc displaces gold from the cyanide complex:

$$2[Au(CN)_2]^-(aq) + Zn(s) \rightarrow 2Au(s) + [Zn(CN)_4]^{2-}(aq)$$

The zinc complex formed is $$[Zn(CN)_4]^{2-}$$ (tetracyanozincate(II) ion). So complex B = $$[Zn(CN)_4]^{2-}$$.

Match with options: Option A: $$[Au(CN)_2]^-$$ and $$[Zn(CN)_4]^{2-}$$ — Matches both A and B. Correct.

Option B: $$[Au(CN)_4]^-$$ and $$[Zn(CN)_2(OH)_2]^{2+}$$ — Incorrect coordination numbers and charges.

Option C: $$[Au(CN)_2]^-$$ and $$[Zn(OH)_4]^{2-}$$ — Complex B is wrong.

Option D: $$[Au(CN)_4]^{2-}$$ and $$[Zn(CN)_6]^{4-}$$ — Both are incorrect.

The correct answer is Option A: $$[Au(CN)_2]^-$$ and $$[Zn(CN)_4]^{2-}$$.