Question

Match the refining methods (Column I) with metals (Column II).

Column I (Refining Methods)	Column II (Metals)
(I) Liquation	(a) Zr
(II) Zone Refining	(b) Ni
(III) Mond Process	(c) Sn
(IV) Van Arkel Method	(d) Ga

• A. (I)-(c) ; (II)-(a) ; (III)-(b) ; (IV)-(d)
• B. (I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)
• C. (I)-(b) ; (II)-(d) ; (III)-(a) ; (IV)-(c)
• D. (I)-(b) ; (II)-(c) ; (III)-(d) ; (IV)-(a)

Hint:

Liquation suits low-melting metals like Sn; zone refining gives ultrapure semiconductors and metals like Ga; the Mond process purifies Ni; the Van Arkel method refines Zr and Ti.

Answer 1

The Correct Option is B

(I) Liquation is used for low-melting metals that separate from high-melting impurities, e.g. tin (Sn): (I)-(c).
(II) Zone refining is used for very high purity of semiconductors and metals such as Ge, Si, B, Ga, In: (II)-(d).
(III) Mond process: impure nickel is converted to volatile \(\mathrm{Ni(CO)_4}\), which is decomposed to pure Ni: (III)-(b).
(IV) Van Arkel method: crude Zr (or Ti) is heated with \(\mathrm{I_2}\) to form volatile \(\mathrm{ZrI_4}\), which is decomposed on a hot tungsten filament: (IV)-(a).
The combination (I)-(c); (II)-(d); (III)-(b); (IV)-(a) is option (2).
Answer: (2) (I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)