Match Unbalanced Reactions with Missing Reagents and Conditions

To determine the correct matching between List I and List II, we analyze each chemical reaction in detail:

1. Reaction P: Lead dioxide ($\text{PbO}_2$) reacts with concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) when warmed to produce lead(II) sulfate ($\text{PbSO}_4$), oxygen gas ($\text{O}_2$), and water ($\text{H}_2\text{O}$): $2\text{PbO}_2 + 2\text{H}_2\text{SO}_4 \xrightarrow{\text{Warm}} 2\text{PbSO}_4 + 2\text{H}_2\text{O} + \text{O}_2 \uparrow$ Hence, P matches with 3 ($\text{Warm}$).

2. Reaction Q: Sodium thiosulfate ($\text{Na}_2\text{S}_2\text{O}_3$) acts as an "antichlor" in aqueous solution to remove excess chlorine gas ($\text{Cl}_2$). Chlorine oxidizes thiosulfate to sodium hydrogen sulfate ($\text{NaHSO}_4$): $\text{Na}_2\text{S}_2\text{O}_3 + 4\text{Cl}_2 + 5\text{H}_2\text{O} \rightarrow 2\text{NaHSO}_4 + 8\text{HCl}$ Hence, Q matches with 4 ($\text{Cl}_2$).

3. Reaction R: Hydrazine ($\text{N}_2\text{H}_4$) is oxidized by iodine ($\text{I}_2$) to produce nitrogen gas ($\text{N}_2$) and hydrogen iodide ($\text{HI}$): $\text{N}_2\text{H}_4 + 2\text{I}_2 \rightarrow \text{N}_2 + 4\text{HI}$ Hence, R matches with 2 ($\text{I}_2$).

4. Reaction S: Xenon difluoride ($\text{XeF}_2$) is a strong fluorinating agent that oxidizes nitric oxide ($\text{NO}$) to nitrosyl fluoride ($\text{NOF}$), while being reduced to elemental xenon ($\text{Xe}$): $\text{XeF}_2 + 2\text{NO} \rightarrow \text{Xe} + 2\text{NOF}$ Hence, S matches with 1 ($\text{NO}$).

Summary of Matches:

• $\text{P} \rightarrow 3$
• $\text{Q} \rightarrow 4$
• $\text{R} \rightarrow 2$
• $\text{S} \rightarrow 1$

This corresponds to Option D.