Match List-I with List-II.
| List-I Process | List-II Metal |
| A. Van Arkel method | I. Ga |
| B. Mond process | II. Al |
| C. Hall-Heroult process | III. Ni |
| D. Zone refining | IV. Zr |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-III, C-II, D-I
A. Van Arkel method — This matches IV. Zr. Crude zirconium is heated with iodine to form volatile \( \ce{ZrI4} \), which decomposes on a hot tungsten filament:
\( \displaystyle \ce{Zr + 2I2 -> ZrI4} \xrightarrow{1800\ \mathrm{K}} \ce{Zr + 2I2} \)
B. Mond process — This matches III. Ni. Nickel forms volatile nickel tetracarbonyl, which decomposes on stronger heating:
\( \displaystyle \ce{Ni + 4CO} \xrightarrow{330-350\ \mathrm{K}} \ce{Ni(CO)4} \xrightarrow{450-470\ \mathrm{K}} \ce{Ni + 4CO} \)
C. Hall-Heroult process — This matches II. Al. Purified alumina mixed with cryolite (\( \ce{Na3AlF6} \)) is electrolysed:
Cathode: \( \ce{Al^3+(melt) + 3e^- -> Al(l)} \)
Anode: \( \ce{C(s) + O^2-(melt) -> CO(g) + 2e^-} \) and \( \ce{C(s) + 2O^2-(melt) -> CO2(g) + 4e^-} \)
Net: \( \ce{2Al2O3 + 3C -> 4Al + 3CO2} \)
D. Zone refining — This matches I. Ga. It gives very pure semiconductors and metals such as Ge, Si, B, Ga and In, because impurities are more soluble in the melt than in the solid.
Shortcut: The pair B-III appears only in Option (3).
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.