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.
Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding stimulus
Model Answer
(a) Carbonate and sulphide ores are converted to oxides because oxides are easier to reduce using reducing agents like carbon to obtain the free metal.
(b) Thermite reaction — Aluminium reduces iron(III) oxide:
$$Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe$$
(c) Cu₂S (copper glance) is roasted in the presence of air (oxygen) to obtain copper directly:
Step 1 — Partial roasting:
$$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$
Step 2 — Reduction of Cu₂O by remaining Cu₂S:
$$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$$
Thus, copper metal is obtained.
Source: Metals and Non-metals, Extraction of Metals
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Explanation
- (a) Key point: oxides are easier to reduce than carbonates/sulphides — one line is enough.
- (b) Any thermite-type reaction works; Fe₂O₃ + Al is the standard NCERT example. Write the balanced equation correctly.
- (c) This is a 2-mark question — examiners expect both steps with balanced equations. Cu₂S roasting is a special case where no external reducing agent is needed; the ore itself acts as the reducing agent in the second step. Award marks are split: 1 for each correct equation.