CBSE Class 10 Science 2026: question 28
Set in the CBSE Class 10 board exam of 2026, paper 31/1/1, section B, question 28, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q28 · 4 marks · Case-basedCompetency-basedMetals and Non-metals
Read the following passage and answer the questions given below:
Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like – Al, Mg are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.
- (a)1 mark
Why carbonate or sulphide ores are converted to oxides before extraction of metal from it?
- (b)1 mark
Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.
- (c)(i)2 marks
How is copper obtained from its ore (Cu2S)? Give equations of the reactions.
- (c)(ii)(I)1 mark
Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent?
- (c)(ii)(II)1 mark
Why solder, an alloy of lead and tin, is used for welding electrical wires together?
Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 31/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Which compound is the easiest for carbon to reduce?
Hint 2
For copper, part of the sulphide is first turned into oxide, which then reacts with the rest of the sulphide.
Hint 3
Compare how strongly reactive metals and carbon hold on to oxygen.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(a) It is easier to obtain a metal from its oxide than from its carbonate or sulphide, because a metal oxide can be reduced readily by carbon. So carbonate ores are calcined and sulphide ores are roasted to convert them into oxides first.
- 1 mark
(b) Thermit reaction:
- 1 mark
(c)(i) Copper glance () is roasted in air, which converts part of it to copper(I) oxide:
- 1 mark
(c)(i) The formed then reacts with the remaining (self-reduction) to give copper metal:
- 1 mark
(c)(ii)(I) [OR alternative] Highly reactive metals such as Na, Mg and Al have a greater affinity for oxygen than carbon has. So carbon cannot remove oxygen from their oxides, and these metals are obtained by electrolytic reduction instead.
- 1 mark
(c)(ii)(II) [OR alternative] Solder has a low melting point, lower than that of the pure metals lead and tin. So it melts easily and can be used to join electrical wires.
Answer: (a) Oxides are easier to reduce to metal; (b) Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat; (c)(i) 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then 2Cu₂O + Cu₂S → 6Cu + SO₂; OR (c)(ii) (I) these metals have more affinity for oxygen than carbon; (II) solder has a low melting point
Where the marks are
- Oxides are easier to reduce to metal (1 mark)
- Any correct thermit-type equation (1 mark)
- Roasting equation for Cu₂S (1 mark)
- Self-reduction equation giving Cu (1 mark)
- OR: greater affinity for oxygen than carbon (1 mark); low melting point of solder (1 mark)
Revise the chapter
- Chapter: Metals and Non-metals (notes, formulas, where marks are lost and every past question)
- Topic: Metallurgy: Extraction of Metals from Ores
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The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.