Match List I with List II
| List I | List II |
| A. van Arkel Method | I. Copper |
| B. Zone Refining | II. Titanium |
| C. Mond Process | III. Germanium |
| D. Electrolytic refining | IV. Nickel |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-III, C-IV, D-I
Each refining method suits metals with a particular property: vapour phase refining (van Arkel, Mond) needs a volatile compound of the metal that decomposes easily; zone refining needs impurities more soluble in the melt than in the solid metal; electrolytic refining uses impure metal as anode and pure metal as cathode.
(A) van Arkel Method — This matches (II) Titanium. Ti is heated with iodine to form volatile \( \ce{TiI4} \), which is decomposed on a tungsten filament electrically heated to about 1800 K.
\( \ce{Ti + 2I2} \xrightarrow{\text{heat}} \ce{TiI4} \xrightarrow{\text{W filament, about 1800 K}} \ce{Ti + 2I2} \)
(B) Zone Refining — This matches (III) Germanium. Semiconductors such as Ge, Si, B and Ga are obtained in very high purity this way.
(C) Mond Process — This matches (IV) Nickel. Nickel is heated in a stream of carbon monoxide to form volatile nickel tetracarbonyl, which decomposes at a higher temperature to give pure nickel.
\( \ce{Ni + 4CO} \xrightarrow{\text{330-350 K}} \ce{Ni(CO)4} \xrightarrow{\text{450-470 K}} \ce{Ni + 4CO} \)
(D) Electrolytic refining — This matches (I) Copper. Impure Cu is the anode, a strip of pure Cu the cathode, in acidified \( \ce{CuSO4} \).
Anode: \( \ce{Cu -> Cu^{2+} + 2e-} \) Cathode: \( \ce{Cu^{2+} + 2e- -> Cu} \)
Shortcut: (A)-(II) 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.