Question

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. \([Au(CN)_4]^–\) and \([Zn(CN)_2 (OH)_2]^{2-}\)
• B. \([Au(CN)_2]^–\) and \([Zn (OH)_4]^{2-}\)
• C. \([Au(CN)_2]^–\) and \([Zn (CN)_4]^{2-}\)
• D. \([Au(CN)_4]^{2-}\) and \([Zn (CN)_6]^{4-}\)

Answer 1

The Correct Option is C

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

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