Match List - I with List - II.
|
List – I |
List – II |
|
(A) \(PbS\rightarrow PbO\) |
(I) Roasting |
|
(B) \(CaCO_3\rightarrow CaO\) |
(II) Calcination |
|
(C) \(ZnS\rightarrow Zn\) |
(III) Carbon reduction |
|
(D) \(Cu_2S\rightarrow Cu\) |
(IV) Self reduction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Before reduction an ore is converted to its oxide: carbonates and hydrated oxides by calcination (heating in limited air, volatile matter driven off) and sulphides by roasting (heating in a regular supply of air). The oxide is then reduced, usually by carbon; some sulphides reduce themselves (self reduction).
(A) PbS → PbO — This matches (I) Roasting. A sulphide ore heated in air:
2PbS + \( \ce{3O2} \) → 2PbO + \( \ce{2SO2} \)
(B) \( \ce{CaCO3} \) → CaO — This matches (II) Calcination. A carbonate decomposes on heating and loses \( \ce{CO2} \):
\( \ce{CaCO3} \) → CaO + \( \ce{CO2} \)
(C) ZnS → Zn — This matches (III) Carbon reduction. Zinc blende is roasted to ZnO, which is then reduced by coke:
2ZnS + \( \ce{3O2} \) → 2ZnO + \( \ce{2SO2} \); ZnO + C → Zn + CO
(D) \( \ce{Cu2S} \) → Cu — This matches (IV) Self reduction. Part of the \( \ce{Cu2S} \) is oxidised to \( \ce{Cu2O} \), which then reacts with the remaining \( \ce{Cu2S} \), with no outside reducing agent:
\( \ce{2Cu2O} \) + \( \ce{Cu2S} \) → 6Cu + \( \ce{SO2} \)
Shortcut: (A)-(I) appears only in Option (2).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.