What does “oxidizing behaviour of $H_2SO_4$” mean?
A substance shows oxidizing behaviour when it oxidizes another substance and itself gets reduced.

Check each reaction for redox change.

Option A: $NaCl + H_2SO_4 \to NaHSO_4 + HCl$ This is simply an acid-salt reaction. No oxidation numbers change. So this is not oxidizing behaviour.

Option B: $2PCl_5 + H_2SO_4 \to 2POCl_3 + 2HCl + SO_2Cl_2$ Here also there is no clear oxidation-reduction involving sulfuric acid as oxidizing agent. It is mainly a chemical transformation/chlorination-type reaction, not the usual oxidizing action of $H_2SO_4$.

Option C: $2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$ Now check oxidation states:

• In $HI$, iodine is $-1$.

• In $I_2$, iodine is $0$. So iodine is oxidized: $I^- \to I_2$

• In $H_2SO_4$, sulfur is $+6$.

• In $SO_2$, sulfur is $+4$. So sulfur is reduced: $S^{+6} \to S^{+4}$

Therefore, $H_2SO_4$ oxidizes $HI$ to $I_2$ and itself gets reduced to $SO_2$. Hence, this reaction clearly shows the oxidizing behaviour of $H_2SO_4$.

Option D: $Ca(OH)_2 + H_2SO_4 \to CaSO_4 + 2H_2O$ This is a neutralization reaction. No change in oxidation states. So this is not oxidizing behaviour.

Conclusion Only Option C represents the oxidizing behaviour of $H_2SO_4$.

Comparison with stored correct answer Stored correct answer: C
Derived answer: C
They match.