NCERT In-text Questions & Solutions

6.1. Which of the ores mentioned can be concentrated by magnetic separation method?
Answer:
Magnetic separation is used to separate magnetic ores from non-magnetic impurities. Ores like haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃), and iron pyrites (FeS₂) are magnetic in nature. These can be effectively separated from impurities like silica using magnetic separation.

6.2. What is the significance of leaching in the extraction of aluminium?
Answer:
Leaching is used to purify bauxite, the primary ore of aluminium, which is often mixed with iron oxides. The ore is treated with a concentrated NaOH solution that dissolves alumina as sodium aluminate, leaving behind iron oxides as impurities. This process helps in extracting pure aluminium oxide.

6.3. Why do some redox reactions used in metallurgy occur only at high temperatures?
Answer:
This is due to thermodynamic principles, specifically the Gibbs free energy change (ΔG). At room temperature, if ΔG is positive or not sufficiently negative, the reaction doesn’t occur. However, at high temperatures, entropy (ΔS) increases, making ΔG more negative, enabling the reaction to proceed.

6.4. Is it true that under certain conditions, Mg can reduce Al₂O₃ and Al can reduce MgO? What are those conditions?
Answer:
Yes. This depends on the Ellingham diagram which plots ΔG vs. T. At higher temperatures, the element lower in the diagram can reduce the oxide of the one above it. Mg, being below Al in the diagram, can reduce Al₂O₃. However, under normal conditions, Al cannot reduce MgO because its ΔG is less negative than that for the formation of MgO.

NCERT Exercise Questions & Solutions

6.1. Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper can be extracted from low-grade ores using hydrometallurgy because Cu²⁺ ions can be easily displaced by more reactive metals like Fe or Zn. However, zinc is a highly reactive metal and cannot be displaced from its salt solution by any metal placed above it in the reactivity series. Hence, zinc cannot be extracted by this method.

6.2. What is the role of depressant in the froth flotation process?
Answer:
Depressants help to suppress the flotation of certain ores. For instance, NaCN is used to prevent ZnS from forming froth in a mixture of PbS and ZnS by forming a complex Na₂[Zn(CN)₄]. This allows PbS to float and be separated from ZnS.

6.3. Why is extraction of copper from pyrites more difficult than from its oxide ore through reduction?
Answer:
The ΔG° of the formation of Cu₂S is less negative than that of other sulphides. Therefore, reducing copper from its sulphide ore (pyrite) requires a more negative ΔG, which is not feasible with commonly available reducing agents. Hence, copper is more easily extracted from its oxide ores.

6.4. Explain the following:

(i) Zone refining:
This technique is used to purify metals like Si, Ge, and Ga. A moving heater melts a small region of the impure metal. As it moves, the impurities concentrate in the molten zone and are pushed to one end, which is then removed.

(ii) Column chromatography:
It separates components of a mixture based on their adsorption to a stationary phase (like Al₂O₃ or SiO₂). The components move at different speeds when a solvent (mobile phase) is passed through the column, allowing for their separation.

6.5. Out of C and CO, which is a better reducing agent at 673 K?
Answer:
At 673 K, CO is a better reducing agent. According to the Ellingham diagram, CO’s line lies below that of carbon at this temperature, indicating CO can reduce metal oxides more effectively.

6.6. Name the common elements present in anode mud in the electro-refining of copper. Why are they so present?
Answer:
Anode mud contains silver (Ag), gold (Au), and platinum (Pt). These metals are less reactive than copper and do not dissolve during electrolysis. They settle down as anode mud during the electro-refining of copper.

6.7. Write the reactions occurring in different zones of the blast furnace during iron extraction.
Answer:

• At the bottom (1200–1700 K):
  Fe₂O₃ + 3C → 2Fe + 3CO

• Middle zone (900–1500 K):
  Fe₂O₃ + 3CO → 2Fe + 3CO₂

• Upper zone (500–900 K):
  3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

6.8. Write chemical reactions in the extraction of zinc from zinc blende.
Answer:

• Roasting:
  2ZnS + 3O₂ → 2ZnO + 2SO₂

• Reduction:
  ZnO + C → Zn + CO

• Refining:
  ZnSO₄ (electrolyte); Zn (anode) → Zn²⁺ + 2e⁻; Zn²⁺ + 2e⁻ → Zn (cathode)

6.9. What is the role of silica in the metallurgy of copper?
Answer:
Silica (SiO₂) acts as a flux to remove FeO impurities by forming iron silicate (FeSiO₃), which floats as slag and is removed.

6.10. What is meant by the term “chromatography”?
Answer:
Chromatography is a separation and purification technique based on the different adsorption or solubility levels of substances in a mixture. Originally used for coloured compounds, it is now widely used for colourless substances too.

6.11. What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected based on the ability of components in the mixture to get adsorbed on it at different rates. For instance, in column chromatography, materials like alumina (Al₂O₃) or silica gel (SiO₂) are used. The substance that gets adsorbed more strongly moves slower, aiding separation.

6.12. Describe a method for refining nickel.
Answer:
Nickel is refined by Mond’s Process, which involves the following steps:

1. Impure Ni is treated with CO at 330–350 K, forming volatile nickel tetracarbonyl:
   Ni + 4CO → Ni(CO)₄

2. This gas is then heated to 450–470 K, decomposing to give pure Ni:
   Ni(CO)₄ → Ni + 4CO

6.13. How can you separate alumina from silica in a bauxite ore associated with silica? Give equations.
Answer:
Baeyer’s process is used:

• The powdered bauxite is digested with concentrated NaOH under pressure: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

• Silica remains insoluble and is filtered out.

• On cooling and seeding, Al(OH)₃ precipitates:
  NaAlO₂ + 2H₂O → Al(OH)₃ ↓ + NaOH

6.14. Giving examples, differentiate between roasting and calcination.
Answer:

• Roasting: Ore is heated in presence of air. Used for sulphide ores.
  Example:
  2ZnS + 3O₂ → 2ZnO + 2SO₂

• Calcination: Ore is heated without air. Used for carbonate ores.
  Example:
  CaCO₃ → CaO + CO₂

6.15. How is the extraction of aluminium different from iron?
Answer:

• Aluminium is extracted by electrolysis of alumina in molten cryolite.

• Iron is extracted via reduction in a blast furnace using coke as a reducing agent.

• Aluminium is more electropositive, hence cannot be reduced by carbon like iron.

6.16. Explain the significance of leaching in the extraction of aluminium.
Answer:
Leaching removes impurities like Fe₂O₃ and SiO₂ from bauxite. By reacting bauxite with NaOH, alumina forms soluble sodium aluminate, and impurities remain undissolved. It enables the purification of aluminium ore before electrolysis.

6.17. What is the role of cryolite in the extraction of aluminium?
Answer:
Cryolite (Na₃AlF₆) serves two main roles:

1. Lowers melting point of alumina, reducing energy consumption.

2. Increases electrical conductivity of the molten mixture, making electrolysis efficient.

6.18. How is cast iron different from pig iron?
Answer:

• Pig iron: Directly obtained from blast furnace; contains ~4% carbon and other impurities.

• Cast iron: Made by melting pig iron with scrap iron and coke; slightly lower carbon content (~3%). More malleable and tough than pig iron.

6.19. Differentiate between ‘minerals’ and ‘ores’.
Answer:

• Minerals: Naturally occurring compounds of metals in the earth’s crust.

• Ores: Minerals from which metals can be economically extracted.
  All ores are minerals, but not all minerals are ores.

6.20. Why copper matte is put in a silica-lined converter during extraction?
Answer:
Copper matte contains Cu₂S and FeS. When blown with air in a silica-lined converter:

• FeS is oxidized to FeO and reacts with silica to form slag (FeSiO₃).

• Cu₂S is converted to metallic Cu by reacting with Cu₂O: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
  2Cu₂O + Cu₂S → 6Cu + SO₂

6.21. What is the role of zinc in the extraction of silver?
Answer:
In cyanide process, Ag is extracted as a soluble complex [Ag(CN)₂]⁻. Zinc is added to displace silver:

2[Ag(CN)₂]⁻ + Zn → 2Ag + [Zn(CN)₄]²⁻

Zinc being more reactive than silver, reduces the complex to metallic Ag.

6.22. What type of metals are generally extracted by electrolytic methods? Why?
Answer:
Highly electropositive metals like Na, K, Al, Ca, Mg are extracted using electrolysis. These metals cannot be reduced using chemical reducing agents like carbon, as their metal oxides are very stable (ΔG° is highly negative). Only electric current can decompose their molten salts.

6.23. The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer:
Yes, the choice of a reducing agent is strongly governed by thermodynamic factors, particularly the change in Gibbs free energy (ΔG°). This can be understood using the Ellingham diagram, which plots ΔG° for the formation of various metal oxides against temperature.

• A metal can reduce the oxide of another metal only if the ΔG° of formation of its own oxide is more negative than that of the other metal’s oxide.

• This means the reducing metal must lie below the metal to be reduced in the Ellingham diagram.

Examples:

1. Aluminium can reduce FeO to iron since ΔG° for Al₂O₃ formation is more negative than for FeO.

2. Zinc can reduce CuO, but copper cannot reduce ZnO because Zn lies below Cu in the Ellingham diagram.

Thus, only a reducing agent that leads to an overall negative ΔG° for the redox reaction will be thermodynamically favorable.

6.24. Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:

• Down’s Process is used for the extraction of sodium metal, where chlorine gas is obtained as a by-product. The process involves the electrolysis of a fused mixture of NaCl and CaCl₂ at 873 K. Sodium is deposited at the cathode and chlorine gas is released at the anode.

• If an aqueous solution of NaCl is electrolyzed:

  • Hydrogen gas (H₂) is evolved at the cathode.

  • Chlorine gas (Cl₂) is evolved at the anode.

  • This occurs because water is also present, and its reduction potential is higher than that of sodium ions.

6.25. What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, oxygen gas is liberated at the anode. If a metal anode were used, oxygen would react with the freshly produced aluminium, forming aluminium oxide, which would reduce the yield.

Hence, graphite rods are used as the anode because:

• Oxygen reacts with graphite to form CO and CO₂, preventing it from attacking aluminium.

• Graphite is cheaper than aluminium, so its slow consumption is economically acceptable.

6.26. Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour-phase refining

Answer:
(i) Zone Refining:

• Based on the principle that impurities are more soluble in molten metal than in its solid state.

• A circular heater is moved along a metal rod; pure metal crystallizes, and impurities move with the molten zone.

• Repeated movement concentrates impurities at one end, which is then removed.

• Used for ultra-pure metals like Ge, Si, B, Ga, In.

(ii) Electrolytic Refining:

• Used for metals like Cu, Ag, Au, Pb, Al.

• Anode: Impure metal

• Cathode: Pure metal strip

• Electrolyte: A salt solution of the metal

• Metal dissolves at anode and deposits at cathode. Impurities settle as anode mud.

(iii) Vapour-Phase Refining:

• The metal is converted to a volatile compound, which is then decomposed to give pure metal.

Examples:

• Mond’s Process:

  • Nickel + CO → Ni(CO)₄ (at 330–350 K)

  • Ni(CO)₄ → Ni + CO (at 450–470 K)

• Van Arkel Process:

  • Used for Zirconium and Titanium.

  • Crude Zr reacts with I₂ to form volatile ZrI₄, which is decomposed on a hot tungsten filament to give ultra-pure Zr.

6.27. Predict conditions under which Al might be expected to reduce MgO.
Answer:

• From the Ellingham diagram, we can observe the ΔG° vs. temperature plots for formation of Al₂O₃ and MgO.

• These plots intersect at a certain temperature. At this intersection point, ΔG° for the reduction of MgO by Al becomes zero.

• Above this temperature, the ΔG° for the reaction:
  3MgO + 2Al → Al₂O₃ + 3Mg
  becomes negative, and thus thermodynamically favorable.

• So, aluminium can reduce MgO to magnesium above the intersection temperature seen in the Ellingham diagram.