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 :

Solution

Chemistry ma’am · 4 steps

Gold goes in as [Au(CN)₂]⁻, and zinc keeps the cyanide that gold gives up.

1. Leaching makes A. Gold is a noble metal, so oxygen can oxidise it only in the presence of CN⁻, which holds the Au⁺ formed as a stable linear complex. Gold is in the +1 state here, with two cyanides.

   4Au + 8CN⁻ + 2H₂O + O₂ → 4[Au(CN)₂]⁻ + 4OH⁻

2. Zinc displaces gold. Zinc is more reactive than gold, so it reduces Au⁺ back to the metal and becomes Zn²⁺: 2[Au(CN)₂]⁻ + Zn → 2Au + [Zn(CN)₄]²⁻.
3. Why four cyanides. Two gold complexes release four CN⁻ ions, and Zn²⁺ takes all four to form a tetrahedral complex. The charge works out as +2 + 4(-1) = -2.
4. So. A is [Au(CN)₂]⁻ and B is [Zn(CN)₄]²⁻, which is option (c).

The idea to keep. In the cyanide process, gold and silver dissolve as [M(CN)₂]⁻ with the metal in the +1 state. Zinc then displaces them and picks up the cyanide as [Zn(CN)₄]²⁻.

Answer from the NTA final answer key. Solution by PrepWiser.