Match List-I with List-IIList-IList-II(A) Zone refining(I) Titanium(B…

The correct answer is Option (3) → A-IV, B-III, C-II, D-I

Refining removes impurities from a crude metal. The method depends on the metal: zone refining for metals needed in very high purity (semiconductors), vapour-phase refining (Mond and van Arkel) for metals that form a volatile compound which decomposes easily, and electrolytic refining for metals such as copper and zinc.

• (A) Zone refining — This matches (IV) Germanium. A molten zone is moved along the rod; impurities are more soluble in the melt and are swept to one end. It gives semiconductors such as germanium, silicon, boron, gallium and indium of very high purity.

• (B) Mond’s process — This matches (III) Nickel. Nickel forms volatile nickel tetracarbonyl with CO, which is decomposed at a higher temperature to give pure nickel:

      \( \ce{Ni + 4CO} \xrightarrow{330-350\,\mathrm{K}} \ce{Ni(CO)4} \)

      \( \ce{Ni(CO)4} \xrightarrow{450-470\,\mathrm{K}} \ce{Ni + 4CO} \)

• (C) Electrolytic refining — This matches (II) Zinc. The impure metal is the anode, a strip of pure metal is the cathode, and a solution of a salt of the metal is the electrolyte; the pure metal deposits on the cathode. Copper and zinc are refined in this way.

      Anode: \( \ce{Zn -> Zn^2+ + 2e-} \)    Cathode: \( \ce{Zn^2+ + 2e- -> Zn} \)

• (D) van-Arkel method — This matches (I) Titanium. The crude metal is heated with iodine to form a volatile iodide, which is decomposed on a hot tungsten filament (about 1800 K); the pure metal deposits on the filament. Titanium and zirconium are purified in this way.

      \( \ce{Ti + 2I2 -> TiI4} \)

      \( \ce{TiI4} \xrightarrow{1800\,\mathrm{K}} \ce{Ti + 2I2} \)

Shortcut: Mond’s process is used only for nickel, so B-III; only Option (3) has it.

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.