Question

Given below are two statements:
Statement-I: Reducing property of dioxide increases from $\text{SO}_2$ to $\text{TeO}_2$.}
Statement-II: $\text{Pb}_3\text{O}_4$ on heating gives lead dioxide and oxygen.}
The correct answer is:

• A. $\text{Both statements I and II are correct}$
• B. $\text{Statement I is correct, but statement II is not correct}$
• C. $\text{Statement I is not correct, but statement II is correct}$
• D. $\text{Both statements I and II are not correct}$

Hint:

To remember oxidation state trends under the inert pair effect: For heavier p-block elements, the lower oxidation state is more stable than the higher one. Thus:
• $\text{Pb}^{2+}$ is more stable than $\text{Pb}^{4+}$
• $\text{SO}_2$ is a strong reducing agent due to easy oxidation to $+6$

Answer 1

The Correct Option is D

Concept: This question requires a thorough analysis of two independent descriptive inorganic chemistry concepts:
• Inert Pair Effect and Group 16 Dioxide Trends: The stability of the $+6$ oxidation state decreases down Group 16, while the stability of the $+4$ oxidation state increases due to the progressive reluctance of the valence $s$-orbital electrons to participate in chemical bonding. A species acts as a reducing agent when it undergoes self-oxidation to a higher state.
• Thermal Decomposition of Mixed Valency Oxides: $\text{Pb}_3\text{O}_4$ (red lead or minium) is a mixed oxide. The thermal stability of lead oxides at elevated temperatures depends strictly on the stability of the $+2$ oxidation state over the $+4$ oxidation state for heavier group 14 elements.

Step 1: Evaluating Statement-I (Reducing property of Group 16 dioxides). Let us analyze the elements of Group 16 (Chalcogens) in their dioxide forms: $\text{SO}_2$, $\text{SeO}_2$, $\text{TeO}_2$. In all these dioxides, the central atom is in the $+4$ oxidation state. The maximum stable oxidation state for these group members is $+6$.
• For a dioxide to behave as a reducing agent, it must easily donate electrons and undergo oxidation from the $+4$ state to the $+6$ state: $$ \text{M}^{4+} \rightarrow \text{M}^{6+} + 2e^- $$
• As we move down the group from Sulfur ($\text{S}$) to Tellurium ($\text{Te}$), the inert pair effect becomes increasingly prominent. The $ns^2$ electron pair becomes more tightly bound by the nucleus and resists taking part in bonding.
• Consequently, the stability of the $+6$ oxidation state decreases dramatically down the group, whereas the stability of the $+4$ oxidation state increases. Since $\text{SO}_2$ can be easily oxidized to the $+6$ state ($\text{SO}_3$ or $\text{SO}_4^{2-}$), it is a very powerful reducing agent. On the other hand, because the $+4$ state is highly stable for Tellurium, $\text{TeO}_2$ resists oxidation to the $+6$ state and behaves instead as an oxidizing agent. Therefore, the reducing property decreases from $\text{SO}_2$ to $\text{TeO}_2$ ($\text{SO}_2 > \text{SeO}_2 > \text{TeO}_2$). Hence, Statement-I is incorrect.

Step 2: Evaluating Statement-II (Thermal decomposition of $\text{Pb}_3\text{O}_4$). Let us look closely at the structural nature of Red Lead ($\text{Pb}_3\text{O}_4$). It is a stoichiometric mixed crystalline oxide composed of Lead(II) oxide and Lead(IV) oxide: $$ \text{Pb}_3\text{O}_4 \equiv 2\text{PbO} \cdot \text{PbO}_2 $$ Due to the extreme manifestation of the inert pair effect in Group 14, the $+2$ oxidation state of lead is remarkably stable, while the $+4$ oxidation state is unstable. When $\text{Pb}_3\text{O}_4$ is heated strongly: $$ 2\text{Pb}_3\text{O}_4(\text{s}) \xrightarrow{\Delta} 6\text{PbO}(\text{s}) + \text{O}_2(\text{g}) $$ Statement-II claims formation of $\text{PbO}_2$ and oxygen, which is incorrect. Hence, Statement-II is incorrect.

Step 3: Conclusion. Since both Statement-I and Statement-II are incorrect, Option (4) is correct.