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p-Block Elements appeared 43 times across 3 years — 5% of Chemistry. This question is from Oxidation States of Group 14 Elements.

Year 2026 2025 2024 Total
Questions 16 14 13 43

It is noticed that Pb²⁺ is more stable than Pb⁴⁺ but Sn²⁺ is less stable than Sn⁴⁺. Observe the following reactions PbO₂ + Pb arrow 2PbO; ΔᵣG°(1) SnO₂ + Sn arrow 2SnO; ΔᵣG°(2) Identify the correct set from the following.

Solution & Explanation

Related Formula
Δ G° = -nFE°cell

For a spontaneous reaction, Δ G° < 0.

Core Logic

Due to the inert pair effect, the stability of the lower oxidation state (+2) increases down the group in Group 14. Therefore, Pb²⁺ is more stable than Pb⁴⁺. Conversely, for lighter elements like Sn, the +4 oxidation state is more stable than +2.

Let's evaluate the first reaction:

PbO₂ + Pb arrow 2PbO

Here, Pb⁴⁺ (PbO₂) is being reduced to Pb²⁺ (PbO), and Pb⁰ is being oxidized to Pb²⁺. Since Pb²⁺ is the more stable state, this reaction is thermodynamically favorable (spontaneous). Thus, Δᵣ G°(1) < 0.

Step 1: Evaluate Second Reaction

Now evaluate the second reaction:

SnO₂ + Sn arrow 2SnO

Here, Sn⁴⁺ (SnO₂) is being converted to Sn²⁺ (SnO). However, we know that Sn⁴⁺ is more stable than Sn²⁺. This means the formation of Sn²⁺ from Sn⁴⁺ is not favored. The reaction is non-spontaneous. Thus, Δᵣ G°(2) > 0.

Pattern Recognition

Inert pair effect drives heavier elements to lower oxidation states (spontaneous, Δ G < 0) and lighter elements to higher oxidation states (making reduction to lower state non-spontaneous, Δ G > 0).

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 12 Chemistry: Thermodynamics

Oxidation States of Group 14 Elements diagram for Q52 - JEE Main 2026 Evening
Oxidation States of Group 14 Elements diagram for Q52 - JEE Main 2026 Evening

Reference Study Guides

More p-Block Elements Previous-Year Questions — Page 9

Q64 jee_main_2024_31_jan_evening Properties of Sulphur and Halogens
Given below are two statements: Statement I: S₈ solid undergoes disproportionation reaction under alkaline conditions to form S²⁻ and S₂O₃²⁻ Statement II: ClO₄^- can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Statement I is correct but statement II is incorrect.
  • B. Statement I is incorrect but statement II is correct
  • C. Both statement I and statement II are incorrect.
  • D. Both statement I and statement II are correct.

Solution

Core Logic

Statement I: S₈ disproportionates in alkaline medium to give sulphide (S²⁻) and thiosulphate (S₂O₃²⁻). The reaction is:

S₈ + 12OH^- arrow 4S²⁻ + 2S₂O₃²⁻ + 6H₂O

This is a correct statement.

Statement II: ClO₄^- contains chlorine in its highest possible oxidation state of +7. A species can undergo disproportionation only if the element can exist in both higher and lower oxidation states than the present one. Since chlorine cannot exceed +7, ClO₄^- cannot undergo disproportionation. This is an incorrect statement.

Step 1: Final Conclusion

Statement I is correct, and Statement II is incorrect. Therefore, option (1) is the correct answer.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p-Block Elements

Q61 jee_main_2024_31_jan_morning Group 18 Elements
Give below are two statements: Statement-I: Noble gases have very high boiling points. Statement-II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points. In the light of the above statements, choose the correct answer from the options given below:
  • A. Statement I is false but Statement II is true.
  • B. Both Statement I and Statement II are true.
  • C. Statement I is true but Statement II is false.
  • D. Both Statement I and Statement II are false.

Solution

Core Logic

Statement I and II are False.

Noble gases have low boiling points. Noble gases are held together by weak dispersion forces.

Pattern Recognition

Noble gases are characterized by extremely weak intermolecular forces (London dispersion forces) because they are monoatomic and non-polar, which directly results in very low boiling and melting points.

Chapter Mix

Class 12 Chemistry: The p-Block Elements

Q79 jee_main_2024_31_jan_morning Group 14 Elements
Consider the oxides of group 14 elements SiO₂, GeO₂, SnO₂, PbO₂, CO and GeO. The amphoteric oxides are
  • A. GeO, GeO₂
  • B. SiO₂, GeO₂
  • C. SnO₂, PbO₂
  • D. SnO₂, CO

Solution

Core Logic

In Group 14 elements, the nature of oxides varies down the group: CO is neutral. CO₂ and SiO₂ are acidic. GeO₂ is distinctly acidic. SnO₂ and PbO₂ are amphoteric.

Pattern Recognition

Tin (Sn) and Lead (Pb) mostly form amphoteric oxides (e.g., SnO, SnO₂, PbO, PbO₂), distinguishing them from the acidic oxides of Carbon and Silicon.

Chapter Mix

Class 11 Chemistry: The p-Block Elements

More p-Block Elements Questions — jee_main_2026_23_january_evening

Practice all p-Block Elements previous-year questions →

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