CBSE Class 10 Science 2024: question 28(a)

Hints

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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. (i) Mercury – Process: Roasting followed by heating alone. Cinnabar is roasted: $2HgS+3{\mathrm{O}}_2\stackrel{\mathrm{heat}}{\to }2HgO+2S{\mathrm{O}}_2$, and then $2HgO\stackrel{\mathrm{heat}}{\to }2Hg+{\mathrm{O}}_2$.

   ½ mark

2. Reason: Mercury is low in the activity series (low reactivity), so heating alone is enough to reduce its oxide.

   ½ mark

3. (ii) Copper – Process: Roasting with self-reduction. $2C{\mathrm{u}}_2\mathrm{S}+3{\mathrm{O}}_2\to 2C{\mathrm{u}}_2\mathrm{O}+2S{\mathrm{O}}_2$, then $2C{\mathrm{u}}_2\mathrm{O}+\mathrm{C}{\mathrm{u}}_2\mathrm{S}\to 6Cu+\mathrm{S}{\mathrm{O}}_2$.

   ½ mark

4. Reason: Copper also has low reactivity, so heating in air is enough to obtain the metal.

   ½ mark

5. (iii) Sodium – Process: Electrolytic reduction of molten sodium chloride. Sodium is deposited at the cathode: $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na}$.

   ½ mark

6. Reason: Sodium is highly reactive, near the top of the activity series. Carbon cannot reduce its compounds, so electrolysis is needed.

   ½ mark

Answer: (i) Mercury: roasting/heating alone, because it has low reactivity. (ii) Copper: roasting/heating alone (self-reduction), because it has low reactivity. (iii) Sodium: electrolytic reduction of molten NaCl, because it is highly reactive.

Revise the chapter

• Chapter: Metals and Non-metals (notes, formulas, where marks are lost and every past question)