JEE Main · 2024 · Shift-ImediumPB11-042

Consider the oxides of group 14 elements SiO2, GeO2, SnO2, PbO2, CO and GeO. The amphoteric oxides are

p-Block Elements (Class 11) · Class 11 · JEE Main Previous Year Question

Question

Consider the oxides of group 14 elements SiO2\mathrm{SiO_2}, GeO2\mathrm{GeO_2}, SnO2\mathrm{SnO_2}, PbO2\mathrm{PbO_2}, CO\mathrm{CO} and GeO\mathrm{GeO}. The amphoteric oxides are

Options
  1. a

    GeO, GeO2\mathrm{GeO,\ GeO_2}

  2. b

    SiO2, GeO2\mathrm{SiO_2,\ GeO_2}

  3. c

    SnO2, PbO2\mathrm{SnO_2,\ PbO_2}

  4. d

    SnO2, CO\mathrm{SnO_2,\ CO}

Correct Answerc

SnO2, PbO2\mathrm{SnO_2,\ PbO_2}

Detailed Solution

Strategy: Identify the amphoteric oxides from Group 14: SiO₂, GeO₂, SnO₂, PbO₂, CO, GeO.

Step 1: Classifying Oxides

  • CO: Neutral oxide (does not react with acid or base under normal conditions).
  • SiO₂: Acidic oxide (reacts with bases: \ceSiO2+2NaOH>Na2SiO3+H2O\ce{SiO2 + 2NaOH -> Na2SiO3 + H2O}, but not with acids easily).
  • GeO₂: Acidic oxide (similar to SiO₂; weakly amphoteric but predominantly acidic).
  • GeO: Basic/weakly amphoteric.
  • SnO₂: Amphoteric — reacts with both acid (\ceSnO2+4HCl>SnCl4+2H2O\ce{SnO2 + 4HCl -> SnCl4 + 2H2O}) and base (\ceSnO2+2NaOH>Na2SnO3+H2O\ce{SnO2 + 2NaOH -> Na2SnO3 + H2O}). ✅
  • PbO₂: Amphoteric — reacts with conc. HCl and with NaOH solution. ✅

Step 2: Amphoteric Pair \ceSnO2\ce{SnO2} and \cePbO2\ce{PbO2} are both amphoteric.

Answer: (C) \ceSnO2, PbO2\boxed{\text{Answer: (C) }\ce{SnO2,\ PbO2}}

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