Match List I with List II
| List I | List II |
| A. Hall Heroult Process | I. \(Ni + 4CO\overset{350K}{\longrightarrow} Ni(CO)_4\underset{450K}{\overset{\Delta }{\longrightarrow}} Ni + 4CO\) |
| B. Mond Process | II. \(Zr + 2I_2 \longrightarrow ZrI_4\) |
| C. Leaching | III. \(4Au + 8CN^- + 2H_2O + O_2 \longrightarrow 4[Au(CN)_2]^- + 4OH^-\) |
| D. Van Arkel Method | IV. \(2Al_2O_3 + 3C \longrightarrow 4Al + 3CO_2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-I, C-III, D-II
Each method is used at a different stage of getting a metal: leaching concentrates the ore, the Hall–Héroult process reduces alumina by electrolysis, and the Mond and van Arkel methods refine a metal through a volatile compound that is decomposed again.
(A) Hall Heroult Process — This matches (IV) 2Al2O3 + 3C → 4Al + 3CO2. Alumina mixed with cryolite is electrolysed; Al forms at the cathode and the oxygen burns the carbon anode.
(B) Mond Process — This matches (I) Ni + 4CO → Ni(CO)4 → Ni + 4CO. Nickel forms volatile nickel tetracarbonyl at 330–350 K, which decomposes at 450–470 K to give pure nickel.
(C) Leaching — This matches (III) 4Au + 8CN− + 2H2O + O2 → 4[Au(CN)2]− + 4OH−. Gold is dissolved out of the ore by dilute cyanide solution in the presence of air.
(D) Van Arkel Method — This matches (II) Zr + 2I2 → ZrI4. Crude zirconium forms volatile \( \ce{ZrI4} \), which is decomposed on a hot tungsten filament to pure Zr.
\( \ce{ZrI4 -> Zr + 2I2} \)
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.