When silver or gold is leached as follows:
\(4M + 8CN^- + 2H_2O + O_2 \longrightarrow 4\,X + 4OH^-\), the product X is:
(M = Ag or Au)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → [M(CN)2]–
In the metallurgy of silver and gold, the metal is leached with a dilute solution of NaCN or KCN in the presence of air. Oxygen oxidises the metal to the +1 state, and cyanide holds the \( \ce{M+} \) ion in solution as the soluble dicyanido complex. Counting the cyanide ions: 8 CN− for 4 M gives two CN− per metal atom, so X = [M(CN)2]−, with charge (+1) + 2(−1) = −1.
\( \displaystyle \ce{4M(s) + 8CN^-(aq) + 2H2O(aq) + O2(g) -> 4[M(CN)2]^-(aq) + 4OH^-(aq)} \)
The metal is then recovered by displacement with zinc:
\( \displaystyle \ce{2[M(CN)2]^-(aq) + Zn(s) -> [Zn(CN)4]^2-(aq) + 2M(s)} \)
Option (2) M2O — Incorrect. An oxide is insoluble and would not carry the metal into solution. Also, the 8 CN− on the left must end up in the product, and an oxide has none.
Option (3) [M(CN)4]3– — Incorrect. Four cyanides per metal would need 16 CN− for 4 M; the equation supplies only 8.
Option (4) M2O3 — Incorrect. Silver and gold are oxidised only to +1 here, and an oxide does not dissolve in the leaching solution.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Other Examples – “In the metallurgy of silver and that of gold, the respective metal is leached with a dilute solution of NaCN or KCN in the presence of air (for O2) from which the metal is obtained later by replacement:
4M(s) + 8CN–(aq) + 2H2O(aq) + O2(g) → 4[M(CN)2]–(aq) + 4OH–(aq) (M = Ag or Au)”