JEE Chemistry General Principles and Processes Of Isolation Of Elements MCQs Set 21

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Multiple Choice Questions

Question 1: With respect to an ore, an Ellingham diagram helps to predict the feasibility of its:
(a) electrolysis
(b) zone refining
(c) vapour phase refining
(d) thermal reduction
Show Answer & Explanation

Answer: (d) thermal reduction

Explanation:
1. An Ellingham diagram plots ΔG° against temperature for the formation of oxides.
2. It shows which reducing agent can reduce a metal oxide at a given temperature, i.e. thermal reduction.

Teacher's Note:
1. A metal whose oxide line lies lower can reduce the oxide of a metal whose line lies above it.

Question 2: The ore that contains the metal in the form of fluoride is:
(a) magnetite
(b) sphalerite
(c) malachite
(d) cryolite
Show Answer & Explanation

Answer: (d) cryolite

Explanation:
1. Cryolite is Na₃AlF₆.
2. Magnetite is Fe₃O₄, sphalerite is ZnS and malachite is CuCO₃·Cu(OH)₂.

Teacher's Note:
1. Fluorspar CaF₂ and cryolite are the common fluoride minerals.

Question 3: The one that is not a carbonate ore is:
(a) siderite
(b) calamine
(c) malachite
(d) bauxite
Show Answer & Explanation

Answer: (d) bauxite

Explanation:
1. Siderite is FeCO₃, calamine is ZnCO₃, malachite is CuCO₃·Cu(OH)₂.
2. Bauxite is an oxide ore of aluminium, Al₂O₃·xH₂O.

Teacher's Note:
1. Learn the formula of each common ore.

Question 4: Match the refining methods in Column I with the metals in Column II: I. Liquation, II. Zone refining, III. Mond process, IV. van Arkel method; (A) Zr, (B) Ni, (C) Sn, (D) Ga. The correct match is:
(a) I-C, II-D, III-B, IV-A
(b) I-B, II-C, III-D, IV-A
(c) I-C, II-A, III-B, IV-D
(d) I-B, II-D, III-A, IV-C
Show Answer & Explanation

Answer: (a) I-C, II-D, III-B, IV-A

Explanation:
1. Liquation suits low-melting metals such as Sn.
2. Zone refining gives ultra-pure semiconductors such as Ga, Ge and Si.
3. Mond is for Ni and van Arkel for Zr.

Teacher's Note:
1. Match each refining method with its typical metals.

Question 5: The correct statement is:
(a) zone refining is used for the refining of titanium
(b) zincite is a carbonate ore
(c) sodium cyanide cannot be used in the metallurgy of silver
(d) aniline is a froth stabiliser
Show Answer & Explanation

Answer: (d) aniline is a froth stabiliser

Explanation:
1. Titanium is refined by the van Arkel method, zincite is ZnO (an oxide ore), and NaCN is used in the cyanide process for silver.
2. Aniline and cresol stabilise the froth in froth floatation.

Teacher's Note:
1. Froth stabilisers are cresols and aniline; collectors are pine oil and xanthates.

Question 6: The correct statement is:
(a) leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate
(b) the Hall-Heroult process is used for the production of aluminium and iron
(c) pig iron is obtained from cast iron
(d) the blistered appearance of copper during the metallurgical process is due to the evolution of CO₂
Show Answer & Explanation

Answer: (a) leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate

Explanation:
1. In Bayer's process concentrated NaOH dissolves Al₂O₃ as sodium aluminate and SiO₂ as sodium silicate.
2. Hall-Heroult is for aluminium only, cast iron is made from pig iron, and blistering of copper is due to SO₂.

Teacher's Note:
1. Fe₂O₃ and TiO₂ impurities do not dissolve in NaOH.

Question 7: Hall-Heroult's process is given by:
(a) ZnO + C → Zn + CO (coke, 1673 K)
(b) Cr₂O₃ + 2Al → Al₂O₃ + 2Cr
(c) 2Al₂O₃ + 3C → 4Al + 3CO₂
(d) Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)
Show Answer & Explanation

Answer: (c) 2Al₂O₃ + 3C → 4Al + 3CO₂

Explanation:
1. Pure alumina is electrolysed in molten cryolite with fluorspar between graphite electrodes.
2. The overall reaction is 2Al₂O₃ + 3C → 4Al + 3CO₂.

Teacher's Note:
1. Cryolite lowers the melting point and increases conductivity.

Question 8: The electrolytes usually used in the electroplating of gold and silver, respectively, are:
(a) [Au(OH)₄]⁻ and [Ag(OH)₂]⁻
(b) [Au(NH₃)₂]⁺ and [Ag(CN)₂]⁻
(c) [Au(CN)₂]⁻ and [Ag(CN)₂]⁻
(d) [Au(CN)₂]⁻ and [AgCl₂]⁻
Show Answer & Explanation

Answer: (c) [Au(CN)₂]⁻ and [Ag(CN)₂]⁻

Explanation:
1. Gold and silver are plated from their cyanide complexes.
2. These are Na[Au(CN)₂] and Na[Ag(CN)₂].

Teacher's Note:
1. Cyanide complexes give a smooth, adherent deposit.

Question 9: The reaction that does not define calcination is:
(a) Fe₂O₃·xH₂O → Fe₂O₃ + xH₂O
(b) ZnCO₃ → ZnO + CO₂
(c) CaCO₃·MgCO₃ → CaO + MgO + 2CO₂
(d) 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
Show Answer & Explanation

Answer: (d) 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Explanation:
1. Calcination is heating in limited or no air to remove water or CO₂ from hydrated or carbonate ores.
2. Heating a sulphide in excess air is roasting.

Teacher's Note:
1. Calcination is for carbonates and hydrated oxides; roasting is for sulphides.

Question 10: The pair that does not require calcination is:
(a) ZnO and MgO
(b) ZnO and Fe₂O₃·xH₂O
(c) ZnCO₃ and CaO
(d) Fe₂O₃ and CaCO₃·MgCO₃
Show Answer & Explanation

Answer: (a) ZnO and MgO

Explanation:
1. Calcination is applied to carbonates and hydrated oxides.
2. ZnO and MgO are already oxides, so neither needs calcination.

Teacher's Note:
1. Check whether the ore contains water or carbonate.

Question 11: Which series of reactions correctly represents chemical relations related to iron and its compounds?
(a) Fe → FeSO₄ (dil. H₂SO₄) → Fe₂(SO₄)₃ (H₂SO₄, O₂) → Fe (heat)
(b) Fe → FeO (O₂, heat) → FeSO₄ (dil. H₂SO₄) → Fe (heat)
(c) Fe → FeCl₃ (Cl₂, heat) → FeCl₂ (heat, air) → Fe (Zn)
(d) Fe → Fe₃O₄ (O₂, heat) → FeO (CO, 600°C) → Fe (CO, 700°C)
Show Answer & Explanation

Answer: (d) Fe → Fe₃O₄ (O₂, heat) → FeO (CO, 600°C) → Fe (CO, 700°C)

Explanation:
1. Iron burns in oxygen to give Fe₃O₄, which is reduced to FeO by CO at about 600°C and to Fe at about 700°C, as in the blast furnace.
2. In the others heating Fe₂(SO₄)₃ or FeSO₄ gives oxides rather than iron, and FeCl₃ does not change to FeCl₂ on heating in air.

Teacher's Note:
1. The blast furnace reduces oxides stepwise as the temperature rises.

Question 12: Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?
(a) Metal sulphides are thermodynamically more stable than CS₂
(b) CO₂ is thermodynamically more stable than CS₂
(c) Metal sulphides are less stable than the corresponding oxides
(d) CO₂ is more volatile than CS₂
Show Answer & Explanation

Answer: (c) Metal sulphides are less stable than the corresponding oxides

Explanation:
1. Carbon reduction of MS to CS₂ has positive ΔG, whereas reduction of MO to CO₂ has negative ΔG.
2. Metal sulphides are in fact more stable than the corresponding oxides, so the statement that they are less stable is not relevant, and false.

Teacher's Note:
1. Sulphides are roasted to oxides because reduction of oxides with carbon is feasible.

Question 13: Which one of the following ores is best concentrated by the froth floatation method?
(a) Magnetite
(b) Cassiterite
(c) Galena
(d) Malachite
Show Answer & Explanation

Answer: (c) Galena

Explanation:
1. Froth floatation is used for sulphide ores.
2. Galena is PbS.

Teacher's Note:
1. Magnetite is concentrated by magnetic separation and cassiterite by gravity separation.

Question 14: The cyanide process is used for the extraction of:
(a) barium
(b) silver
(c) boron
(d) zinc
Show Answer & Explanation

Answer: (b) silver

Explanation:
1. Argentite Ag₂S is leached with NaCN to form Na[Ag(CN)₂].
2. The silver is then precipitated with zinc dust.

Teacher's Note:
1. The cyanide process is also used for gold.

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