JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
The reaction representing the Mond process for metal refining is _________.
- (a) ZnO+C→(Δ) Zn+CO
- (b) 2K[Au(CN)₂]+Zn→(Δ) K₂[Zn(CN)₄]+2Au
- (c) Zr+2I₂ →(Δ) ZrI₄
- (d) Ni+4CO→(Δ) Ni(CO)₄
Answer and solution
Answer (d) Ni+4CO→(Δ) Ni(CO)₄
Solution
Chemistry ma’am · 3 steps
Mond process means nickel plus carbon monoxide: a volatile carbonyl forms, then breaks back into pure nickel.
- Idea of Mond process. Impure nickel is heated in a stream of CO at about 330 to 350 K. Nickel forms volatile nickel tetracarbonyl, Ni(CO)₄, and the impurities stay behind as solids.
Ni + 4CO →(330-350 K) Ni(CO)₄ →(450-470 K) Ni + 4CO
- Why it purifies. The carbonyl vapour is carried away and heated more strongly. It decomposes, so pure nickel deposits and the CO is released to be reused.
Option Reaction Process Purpose (a) ZnO + C → Zn + CO Reduction with coke Extraction of zinc (b) 2K[Au(CN)₂] + Zn → K₂[Zn(CN)₄] + 2Au Cyanide process Extraction of gold (c) Zr + 2I₂ → ZrI₄ van Arkel method Refining of zirconium (d) Ni + 4CO → Ni(CO)₄ Mond process Refining of nickel - So. The Mond process is the formation of nickel tetracarbonyl, option (d).
The idea to keep. Vapour phase refining needs a volatile compound that later decomposes. Mond uses CO with nickel; van Arkel uses I₂ with Zr or Ti. Match the reagent to the name.
Answer from the NTA final answer key. Solution by PrepWiser.