JEE Chemistry General Principles and Processes Of Isolation Of Elements MCQs Set 13

Answer: (a) copper pyrites

Explanation:
1. Copper pyrites is CuFeS₂, a sulphide.
2. Malachite is a carbonate, haematite Fe₂O₃ is an oxide and magnesite MgCO₃ is a carbonate.

Teacher's Note:
1. Sulphide ores include galena PbS, zinc blende ZnS, cinnabar HgS and copper pyrites CuFeS₂.

Answer: (c) Sulphides of copper(I) and iron(II)

Explanation:
1. Copper matte is a mixture of Cu₂S and FeS.
2. In Cu₂S, copper is in the +1 state and in FeS, iron is in the +2 state.

Teacher's Note:
1. Copper matte = Cu₂S + FeS. It is converted to blister copper in the Bessemer converter, where FeS is oxidised to slag and Cu₂S is self-reduced.

Answer: (b) Impurities are more soluble in the molten metal than in solid metal

Explanation:
1. When a molten zone travels along a rod, the pure metal solidifies behind it while impurities prefer to stay in the melt.
2. They are carried to one end, which is cut off. Option 1 is distillation, option 3 chromatography and option 4 the Van Arkel/Mond method.

Teacher's Note:
1. Zone refining is based on the greater solubility of impurities in the melt. It is repeated to give semiconductor-grade purity.

Answer: (b) Smelting

Explanation:
1. In self reduction a sulphide ore is partly roasted to give some oxide.
2. The oxide then reacts with the remaining sulphide at high temperature in the absence of air, i.e. during smelting: 2Cu₂O + Cu₂S → 6Cu + SO₂.

Teacher's Note:
1. Roasting makes some oxide, and the self reduction step follows once the air supply is cut off. No external carbon is needed.

Answer: (d) Below T₃, M is less reducing than M′

Explanation:
1. A kink in the line of an oxide appears where the metal changes state, so T₁ and T₂ (the slope increases at each) are the melting and boiling points of M. Statement 1 is correct.
2. Above T₃ the line of MₓO_y lies above that of M′ₓO_y, so the oxide of M′ is more stable. Hence M′ is the better reducing agent, and statement 2 is correct.
3. An oxide with ΔG > 0 does not form spontaneously, i.e. it is unstable, so statement 3 is correct.
4. Below T₃ the line of MₓO_y is lower, so its oxide is more stable and M is more reducing than M′. Statement 4 is therefore incorrect.

Teacher's Note:
1. In an Ellingham diagram the metal whose oxide line lies lower is the stronger reducing agent at that temperature. Where two lines cross, the order of reducing power reverses.

Answer: (b) To remove gangue

Explanation:
1. A flux combines with the gangue (impurity) to form a fusible slag.
2. For example CaO (basic flux) + SiO₂ (acidic gangue) → CaSiO₃ slag, which floats and is removed.

Teacher's Note:
1. Acidic gangue (SiO₂) needs a basic flux (CaO, limestone) and basic gangue (FeO, CaO) needs an acidic flux (SiO₂).

Answer: (b) (a)-R, (b)-Q, (c)-P, (d)-S

Explanation:
1. Copper pyrites is CuFeS₂ (R).
2. Malachite is CuCO₃·Cu(OH)₂ (Q).
3. Calamine is ZnCO₃ (P).
4. Sphalerite (zinc blende) is ZnS (S).

Teacher's Note:
1. Common ores: copper pyrites CuFeS₂, malachite CuCO₃·Cu(OH)₂, calamine ZnCO₃, zinc blende ZnS. Carbonate ores are calcined, sulphide ores are concentrated by froth flotation and roasted.

Answer: (b) Froth floatation – cerussite

Explanation:
1. Kroll's process (reduction of TiCl₄ by Mg) is used for Ti, so it is correct.
2. Froth flotation is for sulphide ores. Cerussite is PbCO₃, a carbonate ore, so this pair is incorrect.
3. Zn has a low boiling point and is refined by distillation, and NaCN is used as a depressant for ZnS in the PbS-ZnS mixture.

Teacher's Note:
1. Froth flotation works for sulphide ores such as galena, zinc blende and copper pyrites. Carbonate ores like cerussite and calamine are calcined, not floated.

Answer: (c) Fe₂O₃ + Al

Explanation:
1. Thermite is a mixture of iron(III) oxide and aluminium powder (about 3 : 1 by mass).
2. On ignition Fe₂O₃ + 2Al → Al₂O₃ + 2Fe with much heat, used in aluminothermic welding.

Teacher's Note:
1. In the aluminothermite (Goldschmidt) process Al reduces oxides of Cr, Mn and Fe because Al₂O₃ is very stable and the reaction is highly exothermic.

Answer: (b) Cyanide process – reduction of Pb

Explanation:
1. Al is obtained by electrolytic reduction of molten alumina in cryolite, so option 1 is right.
2. The cyanide process is used for Ag and Au, not for Pb, which is obtained by roasting and self or carbon reduction of galena. So option 2 is wrong.
3. Ag is leached with NaCN and Ge is zone refined, so options 3 and 4 are right.

Teacher's Note:
1. The cyanide process is for the noble metals Ag and Au. Lead comes from galena by partial roasting and self reduction.

Answer: (a) \( \ce{ZnS -> Zn} \)

Explanation:
1. Self reduction works when a partly roasted sulphide gives an oxide that is reduced by the remaining sulphide. This is true for Cu, Pb and Hg.
2. ZnO is stable enough that it is not reduced by the remaining ZnS; it needs carbon reduction at high temperature.
3. So self reduction is not possible for ZnS to Zn.

Teacher's Note:
1. Self reduction (auto reduction) works for sulphide ores of Cu, Pb and Hg: partial roasting gives some oxide, which reacts with the remaining sulphide to give the metal and SO₂. Zinc sulphide is roasted completely and then reduced with coke.

Answer: (b) Ag

Explanation:
1. Silver (and gold) ores are leached with NaCN, forming the soluble complex Na[Ag(CN)₂]. Zinc then displaces the silver.

Teacher's Note:
1. In the cyanide process the metal goes into solution as a cyanide complex, and the metal is later displaced with zinc dust.

Answer: (b) a-S, b-Q, c-R, d-P

Explanation:
1. Cyanide process extracts Au and Ag: a-S.
2. Pine oil is a collector or frother in froth flotation: b-Q.
3. Al is extracted by electrolytic reduction: c-R.
4. Zone refining yields ultrapure Ge: d-P.

Teacher's Note:
1. Link each process with its signature material: NaCN with Au and Ag, pine oil with froth flotation, cryolite with Al electrolysis, zone refining with semiconductors.

Answer: (b) Thermal decomposition

Explanation:
1. Impure Ni reacts with CO at about 330-350 K to form volatile Ni(CO)₄.
2. On heating to about 450-470 K, Ni(CO)₄ decomposes into pure Ni and CO. This is thermal decomposition.

Teacher's Note:
1. Mond's process is a vapour phase refining method like Van Arkel. In both, a volatile compound is formed and then decomposed on heating.

Answer: (d) galena

Explanation:
1. Froth flotation separates sulphide ores because they are wetted by oil, not water.
2. Galena (PbS) is a sulphide.
3. Cassiterite is SnO₂, magnetite Fe₃O₄ is concentrated magnetically and malachite is a carbonate.

Teacher's Note:
1. Sulphides are preferentially wetted by pine oil and rise with the froth while the gangue sinks. NaCN is a depressant to separate ZnS from PbS.

Answer: (a) Parkes process

Explanation:
1. Parkes process adds molten zinc to argentiferous lead; silver is more soluble in zinc and forms a zinc-silver alloy that separates.
2. Serpeck's process is for bauxite, Down's process makes sodium and the Castner-Kellner process makes NaOH from brine.

Teacher's Note:
1. Parkes process desilverises lead. The Zn-Ag alloy is then distilled to recover the silver, and cupellation is used for the final purification.

Answer: (c) Wolframite

Explanation:
1. Sheelite contains WO₄²⁻. Wolframite is (Fe,Mn)WO₄ and also has the tungstate ion.
2. Limonite is Fe₂O₃·3H₂O, dolomite is CaCO₃·MgCO₃, siderite is FeCO₃. None of them has tungstate.

Teacher's Note:
1. Tungsten ores are sheelite CaWO₄ and wolframite FeWO₄·MnWO₄, both tungstates. Magnetic separation is used to separate wolframite from cassiterite.

Answer: (b) Reducing agent

Explanation:
1. Al has a strong affinity for oxygen, since Al₂O₃ is very stable.
2. It reduces oxides such as Cr₂O₃ and Fe₂O₃ to the metals, so it acts as a reducing agent.

Teacher's Note:
1. Aluminium is oxidised (to Al₂O₃) and the metal oxide is reduced. The process is highly exothermic and the metal is obtained in the molten state.

Answer: (d) \( \ce{Cu2O + 1/2 Cu2S -> 3Cu + 1/2 SO2} \)

Explanation:
1. In autoreduction (self reduction) the oxide of the metal is reduced by the sulphide of the same metal, with no external reducing agent.
2. Only Cu₂O + ½Cu₂S → 3Cu + ½SO₂ fits this.
3. Option 1 and 2 use CO and C, option 3 uses Fe (displacement).

Teacher's Note:
1. Look for a reaction in which the oxide and sulphide of the same metal react together. This is typical of Cu, Pb and Hg.

Answer: (d) Ge – Zone refining

Explanation:
1. Ge is a semiconductor that is purified by zone refining, so (4) is correct.
2. Zr is purified by the Van Arkel method, Zn by distillation and Ni by the Mond process, so (1), (2) and (3) are wrong. Polling is used for Cu and Pb.

Teacher's Note:
1. Match refining methods to metals by the property used: volatility (distillation, Mond, Van Arkel), solubility of impurities in the melt (zone refining) or reduction of the oxide impurity (polling).

Answer: (c) FeSiO₃

Explanation:
1. Roasting converts FeS to FeO.
2. In the smelting and Bessemer steps FeO combines with silica flux: FeO + SiO₂ → FeSiO₃, which is the slag.

Teacher's Note:
1. Copper matte (Cu₂S + FeS) is separated from the slag, which is mainly iron silicate. This is why iron is oxidised and removed first.

Answer: (c) B is SO₂

Explanation:
1. The gas B from roasting a metal sulphide is SO₂.
2. SO₂ turns lime water milky (CaSO₃), decolourises acidified KMnO₄ (it is a reducing agent) and reduces Fe³⁺ to Fe²⁺.
3. CO₂ would not decolourise KMnO₄ or reduce Fe³⁺.
4. A, which burns to SO₂, is the amorphous solid sulphur, not FeO or ZnS.

Teacher's Note:
1. SO₂ is the usual by-product of roasting sulphide ores and is a reducing agent. Tests: milky lime water, decolourisation of acidified permanganate, and reduction of Fe³⁺.

Answer: (b) a complexing agent, a reducing agent

Explanation:
1. The metal dissolves in aerated cyanide solution as the soluble complex, e.g. Na[Ag(CN)₂]. So cyanide acts as a complexing agent.
2. Zn displaces the metal: Zn + 2[Ag(CN)₂]⁻ → [Zn(CN)₄]²⁻ + 2Ag. Zn loses electrons, so it is the reducing agent.

Teacher's Note:
1. Cyanide process: complex formation (leaching) is followed by displacement of the noble metal by the more electropositive zinc. Zinc is oxidised and the metal ion is reduced.

Answer: (c) Thermite process

Explanation:
1. Serpeck's process concentrates white bauxite.
2. Hall-Heroult is the electrolytic reduction of Al₂O₃ and Hoope's process refines Al electrolytically.
3. The thermite process uses Al as a reducing agent to extract other metals (Cr, Mn, Fe); it is not a process to extract Al.

Teacher's Note:
1. Aluminium: concentration by Baeyer, Hall or Serpeck process, electrolysis by Hall-Heroult, and purification by Hoope's process.

Answer: (b) Wrought iron

Explanation:
1. Wrought iron contains only about 0.12 to 0.25% carbon, the least of the common forms.
2. Cast iron and pig iron have 2.5 to 5% carbon and steel has 0.25 to 2%.

Teacher's Note:
1. Order of carbon content: pig iron > cast iron > steel > wrought iron. The melting point order is the reverse.

Answer: (c) a and c

Explanation:
1. Calcination is heating the ore strongly in the absence of air (or in limited air) to remove moisture and volatile matter, decompose carbonates and dehydrate hydrated oxides.
2. (a) decomposes a carbonate and (c) removes water of crystallisation, so both are calcination.
3. (b) and (d) use oxygen to convert sulphides to oxides, which is roasting.

Teacher's Note:
1. Calcination removes CO₂ and H₂O without oxygen, while roasting uses excess air on sulphide ores and releases SO₂. Check for oxygen on the left side of the equation to tell them apart.

Answer: (c) Ge

Explanation:
1. Zone refining is used for semiconductors needing very high purity, such as Ge, Si and Ga.
2. Cu and Ag are refined electrolytically and Zn by distillation.

Teacher's Note:
1. In zone refining a molten zone moves along a rod and impurities move into it because they are more soluble in the melt than in the solid.

Answer: (a) Oxide ore

Explanation:
1. Pyrolusite is MnO₂, the chief ore of manganese. It is an oxide ore.

Teacher's Note:
1. Oxide ores include haematite, magnetite, bauxite, cassiterite, pyrolusite and zincite.

Answer: (a) Ag

Explanation:
1. Ag and Au are dissolved in aerated NaCN/KCN as soluble complexes.
2. Na and Al are obtained by electrolysis and Cu by smelting and self reduction.

Teacher's Note:
1. Leaching with cyanide is limited to the noble metals Ag and Au, which are then displaced by zinc.

Answer: (a) CaO, K₂O

Explanation:
1. Oxides of highly electropositive metals such as Ca and K are too stable to be reduced by carbon.
2. Option 2 has Cu₂O, option 3 Fe₂O₃ and ZnO, and option 4 PbO and Fe₃O₄, all of which are reduced by coke.

Teacher's Note:
1. Carbon reduces oxides of metals below it in the Ellingham diagram: Fe, Zn, Pb, Cu. Oxides of Na, K, Ca, Mg and Al need electrolysis or a stronger reducing agent.