In the process of extraction, gold combines with cyanide in the presence of oxygen and water to make a complex. The formula of the complex is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([Au(CN)_2]^{-}\)
Gold is extracted by complex formation. Gold does not dissolve in water on its own, but in the presence of oxygen and water, cyanide ions oxidise it to Au(I) and hold it in solution as the linear complex dicyanidoaurate(I):
\( \displaystyle \ce{4Au(s) + 8CN^-(aq) + 2H2O(l) + O2(g) -> 4[Au(CN)2]^-(aq) + 4OH^-(aq)} \)
The charge of the complex follows from the oxidation state of gold: \( +1 + 2(-1) = -1 \). Gold is then recovered by adding zinc, which displaces it from the complex:
\( \displaystyle \ce{2[Au(CN)2]^-(aq) + Zn(s) -> [Zn(CN)4]^2-(aq) + 2Au(s)} \)
So the complex is \( \ce{[Au(CN)2]^-} \).
Why the other options are incorrect
Option (1) \([Au(CN)_2]^{2-}\) — Incorrect. A charge of −2 with two \( \ce{CN-} \) ligands would need gold in the 0 state, but gold is oxidised by oxygen in this process.
Option (3) \([Au(CN)_6]^{2-}\) — Incorrect. This would need gold(+4) with six ligands; gold forms Au(I) and Au(III), and the cyanide complex formed here has two ligands.
Option (4) \([Au(CN)_4]^{2-}\) — Incorrect. This would need gold(+2), which gold does not show in this complex; Au(III) with four cyanides would be \( \ce{[Au(CN)4]^-} \).
NCERT (2026-27 edition):
Importance and Applications of Coordination Compounds – …
