JEE Chemistry General Principles and Processes Of Isolation Of Elements MCQs Set 08

Answer: (b) A-S, B-P, C-Q, D-T, E-R

Explanation:
1. Malachite is CuCO₃·Cu(OH)₂, so A-S (Cu).
2. Siderite is FeCO₃, so B-P (Fe).
3. Cerussite is PbCO₃, so C-Q (Pb).
4. Pyrolusite is MnO₂, so D-T (Mn).
5. Wurtzite is ZnS, so E-R (Zn). The matching is A-S, B-P, C-Q, D-T, E-R.

Teacher's Note:
1. Learn ores by their anion: carbonates end in -ite (siderite, cerussite, malachite).
2. Wurtzite and zinc blende are both forms of ZnS.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: NaCN (with air) dissolves Ag₂S and Au as soluble cyanide complexes.
2. Reason is false: CN^- acts as a complexing ligand, not a reducing agent. Zinc dust is the reducing agent that precipitates the metal.
3. So A is true and R is false.

Teacher's Note:
1. Cyanide forms [Ag(CN)₂]^- and [Au(CN)₂]^-, which are soluble.
2. Zinc then displaces the noble metal from the complex.

Answer: (d) Dolomite

Explanation:
1. Magnesite is MgCO₃ (Mg only).
2. Calamine is ZnCO₃.
3. Carnallite is KCl·MgCl₂·6H₂O (K and Mg).
4. Dolomite is CaCO₃·MgCO₃, containing both Mg and Ca.

Teacher's Note:
1. Dolomite is a double carbonate of calcium and magnesium.
2. Carnallite is the source of both potassium and magnesium.

Answer: (d) Serpeck's process

Explanation:
1. Bauxite rich in silica is called white bauxite.
2. Bayer's and Hall's leaching processes work for red bauxite (Fe₂O₃ impurity), since silica also dissolves in alkali.
3. For white bauxite, Serpeck's process is used: Al₂O₃ + N₂ + 3C → 2AlN + 3CO, then AlN + 3H₂O → Al(OH)₃ + NH₃.

Teacher's Note:
1. Red bauxite means iron oxide impurity (Bayer or Hall); white bauxite means silica impurity (Serpeck).
2. Hoope's process is for refining aluminium, not for concentration.

Answer: (d) Gangue

Explanation:
1. The earthy and rocky impurities in an ore are called gangue (matrix).
2. Slag is the fusible product formed when flux combines with gangue.

Teacher's Note:
1. Gangue + flux gives slag.
2. Do not confuse the unwanted impurity (gangue) with the substance added to remove it (flux).

Answer: (a) Bauxite

Explanation:
1. Bauxite is leached with NaOH (Bayer's process) to dissolve alumina as sodium aluminate.
2. Copper pyrites is concentrated by froth flotation and cassiterite by magnetic or gravity separation.

Teacher's Note:
1. Leaching is used for Al, Ag and Au ores.
2. Sulphide ores such as copper pyrites go by froth flotation.

Answer: (c) Blister copper; of its appearance

Explanation:
1. In the Bessemer converter, 2Cu₂O + Cu₂S → 6Cu + SO₂.
2. The SO₂ escapes while the copper solidifies, leaving a blistered surface, so the product is blister copper.

Teacher's Note:
1. Blister copper is about 98 percent pure.
2. It is refined further by poling and electrolysis.

Answer: (b) Sulphide

Explanation:
1. Copper glance is Cu₂S (chalcocite).
2. It is a sulphide ore.

Teacher's Note:
1. Glance = shiny sulphide (copper glance Cu₂S, silver glance Ag₂S, lead glance PbS).
2. Malachite is the carbonate ore of copper.

Answer: (b) To remove gangue

Explanation:
1. Flux combines with the gangue, which is infusible, to form a fusible slag.
2. The slag melts and separates from the metal, so the flux removes gangue.

Teacher's Note:
1. Flux + gangue = slag.
2. Acidic gangue needs basic flux (CaO) and basic gangue needs acidic flux (SiO₂).

Answer: (b) Mond's process

Explanation:
1. Impure nickel forms volatile nickel tetracarbonyl with CO at low temperature.
2. On stronger heating the carbonyl decomposes to give pure nickel.

Teacher's Note:
1. This is the Mond process, a vapour phase refining.
2. Van Arkel uses iodine instead of CO (for Ti and Zr).

Answer: (d) Al₂O₃·2H₂O + Na₂CO₃ → 2NaAlO₂ + CO₂ + 2H₂O

Explanation:
1. Option 1 is Bayer's process (NaOH leaching).
2. Option 2 is the thermite reaction and option 3 is part of Serpeck's process.
3. Hall's process fuses bauxite with Na₂CO₃ to give soluble sodium aluminate, as in option 4.

Teacher's Note:
1. Hall's process uses sodium carbonate; Bayer's uses sodium hydroxide.
2. Do not confuse Hall's leaching process with the Hall-Heroult electrolysis.

Answer: (a) SiO₂

Explanation:
1. FeO is a basic impurity, so an acidic flux is needed.
2. SiO₂ combines with it to give FeSiO₃ slag.

Teacher's Note:
1. Basic gangue requires acidic flux; acidic gangue requires basic flux.
2. CaO and MgO are basic fluxes.

Answer: (d) The entropy change during the CO/CO₂ conversion is negative

Explanation:
1. In the Ellingham diagram, 2CO + O₂ → 2CO₂ has a negative ΔS, so its line slopes upward.
2. The line for 2C + O₂ → 2CO slopes downward. The lines cross near 983 K (about 710 °C).
3. Below that temperature CO is the better reducing agent, which is the cooler upper part of the furnace.

Teacher's Note:
1. Remember the three carbon lines: C to CO2 is flat, C to CO slopes down, CO to CO2 slopes up.
2. Coke is the effective reductant only above the crossover temperature.

Answer: (a) C at high temperature

Explanation:
1. Zn is more electropositive than Cu, so Cu cannot reduce ZnO.
2. Above about 1200 K the C to CO line lies below the Zn to ZnO line.
3. So carbon at high temperature reduces ZnO.

Teacher's Note:
1. Roasted zinc ore is heated with coke near 1673 K.
2. Zinc is volatile at that temperature and is condensed.

Answer: (b) −ΔS

Explanation:
1. ΔG = ΔH − TΔS.
2. Plotting ΔG against T gives a straight line of slope −ΔS.

Teacher's Note:
1. Oxide formation consumes gas, so ΔS is negative and the slope is positive.
2. The line bends only when a phase change occurs.

Answer: (d) All of these

Explanation:
1. Partial roasting of Cu₂S, HgS and PbS gives some oxide.
2. The oxide then reacts with the remaining sulphide, giving the metal and SO₂ without any external reducing agent.

Teacher's Note:
1. Self (auto) reduction applies to sulphide ores of Cu, Pb and Hg.
2. Cinnabar gives mercury even on simple roasting.

Answer: (c) (iii), (i) and (ii)

Explanation:
1. Sodium is obtained by electrolysis of fused NaCl with CaCl₂, which is (iii).
2. Nickel is purified through volatile Ni(CO)₄ (Mond process), which is (i).
3. Copper from Cu₂S is obtained by self-reduction, which is (ii).
4. So the order is (iii), (i), (ii).

Teacher's Note:
1. Ni(CO)₄ boils at 43 °C and decomposes near 450 K.
2. Down's process uses CaCl₂ to lower the melting point of NaCl.

Answer: (c) Cassiterite

Explanation:
1. Haematite is Fe₂O₃, limonite is Fe₂O₃·3H₂O and magnetite is Fe₃O₄.
2. Cassiterite is SnO₂, the ore of tin.

Teacher's Note:
1. Cassiterite = tin stone.
2. Iron ores: haematite, magnetite, limonite and siderite.

Answer: (b) SO₂ and ZnO

Explanation:
1. 2ZnS + 3O₂ → 2ZnO + 2SO₂.
2. The products are ZnO and SO₂.

Teacher's Note:
1. Roasting converts sulphide to oxide with release of SO₂.
2. Calcination of calamine gives ZnO and CO₂.

Answer: (c) CaO, K₂O

Explanation:
1. Oxides of Cu, Zn, Fe and Pb are reduced by carbon.
2. CaO and K₂O are very stable oxides of highly electropositive metals and are not reduced by coke.

Teacher's Note:
1. Very reactive metals (Na, K, Ca, Mg, Al) are obtained by electrolysis.
2. Check the Ellingham diagram: their lines lie far below the carbon lines.

Answer: (d) Cryolite

Explanation:
1. Galena is PbS and cinnabar is HgS (sulphides).
2. Bauxite is an oxide ore.
3. Cryolite Na₃AlF₆ is a fluoride, hence a halide ore.

Teacher's Note:
1. Halide ores: rock salt, horn silver, fluorspar, cryolite and carnallite.
2. Cryolite is also used as the solvent in aluminium extraction.

Answer: (a) Froth flotation

Explanation:
1. Iron pyrites FeS₂ is a sulphide ore.
2. Sulphide ores are concentrated by froth flotation.

Teacher's Note:
1. Sulphide ore means froth flotation.
2. Magnetic separation needs a magnetic ore or gangue.

Answer: (c) Sodium

Explanation:
1. Sodium is obtained by electrolysis of fused NaCl (Down's process).
2. Aluminium uses its oxide in cryolite, and Cu and Fe are obtained by chemical reduction.

Teacher's Note:
1. Highly electropositive metals are obtained by electrolysing fused chlorides.
2. CaCl₂ is added to lower the melting point of NaCl.

Answer: (d) Carbon monoxide

Explanation:
1. Coke burns to CO₂, which reacts with more coke to give CO.
2. CO reduces the iron oxides: Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Teacher's Note:
1. Coke acts mostly through the CO it generates.
2. Carbon dioxide is the oxidised product, not a reducer.

Answer: (d) Aluminothermite process

Explanation:
1. Aluminium is used as the reducing agent: Cr₂O₃ + 2Al → Al₂O₃ + 2Cr.
2. This is the aluminothermite (Goldschmidt thermite) process.

Teacher's Note:
1. The reaction is highly exothermic.
2. Carbon cannot reduce Cr₂O₃ without forming carbides.

Answer: (c) Electrolytic refining

Explanation:
1. Impure Al forms the anode and pure Al the cathode in a molten fluoride bath (Hoope's process).
2. Roasting and sublimation do not remove its impurities, and carbon cannot reduce Al₂O₃.

Teacher's Note:
1. Hoope's process uses three molten layers.
2. Purity obtained is about 99.99 percent.