JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
Among statements (a) - (d), the correct ones are
(a) Lime stone is decomposed to CaO during the extraction of iron from its oxides.
(b) In the extraction of silver, silver is extracted as an anionic complex.
(c) Nickel is purified by Mond’s process.
(d) Zr and Ti are purified by Van Arkel method.
- (a) (a), (c) and (d) only
- (b) (c) and (d) only
- (c) (b), (c) and (d) only
- (d) (a), (b), (c) and (d)
Answer and solution
Answer (d) (a), (b), (c) and (d)
Solution
Chemistry ma’am · 2 steps
Match each statement to its real process; all four hold, so not one of them can be dropped.
- Test one by one. Each statement names a real process. Match it to what actually happens in that process, then keep only the true ones.
Statement What actually happens Verdict (a) Limestone gives CaO in iron extraction In the blast furnace, CaCO₃ → CaO + CO₂. The CaO is the flux and removes silica as slag: CaO + SiO₂ → CaSiO₃. True (b) Silver extracted as an anionic complex Silver ore is leached with NaCN and air to form [Ag(CN)₂]⁻. Zinc then displaces the silver. True (c) Nickel purified by Mond's process Ni + 4CO → Ni(CO)₄ at about 330 to 350 K. The volatile carbonyl decomposes at about 450 to 470 K to give pure Ni. True (d) Zr and Ti purified by van Arkel method Heating with I₂ gives volatile ZrI₄ or TiI₄. This decomposes on a hot tungsten filament to deposit pure metal. True - So. All four statements are correct, so the answer is option (d).
The idea to keep. Link each metal to its signature step: CaO flux for iron, cyanide complex for Ag and Au, CO for Ni (Mond), iodine for Zr and Ti (van Arkel).
Answer from the NTA final answer key. Solution by PrepWiser.