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

The leaching of gold involves the use of a dilute aqueous solution of sodium cyanide \((\text{NaCN})\) in presence of oxygen. This process is known as cyanidation and it's an important step in the extraction of gold.

1. During the leaching process, the gold metal reacts with cyanide ions and oxygen to form a water-soluble complex. The equation for this reaction is: \[ 4 \, \text{Au} + 8 \, \text{CN}^- + \text{O}_2 + 2 \, \text{H}_2\text{O} \rightarrow 4 \, [\text{Au(CN)}_2]^- + 4 \, \text{OH}^- \] The complex formed here is \([Au(CN)_2]^-\), which represents complex [A].
2. In the next step, the gold complex \([Au(CN)_2]^-\) is treated with zinc. This is known as the Merrill-Crowe process, wherein zinc reduces the gold complex to elemental gold, and itself forms a new complex. The reaction can be represented as: \[ 2 \, [\text{Au(CN)}_2]^- + \text{Zn} \rightarrow 2 \, \text{Au} + [\text{Zn(CN)}_4]^{2-} \] Here, the complex \([Zn(CN)_4]^{2-}\) is formed, which is complex [B].
3. From the options given, the correct answer is \([Au(CN)_2]^-\) for complex [A] and \([Zn(CN)_4]^{2-}\) for complex [B].

This process is highly significant industrially because it allows for the efficient extraction of gold from its ores.

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