Which of the following chemical reactions depicts the oxidizing behaviour of $$H_2SO_4$$?

The term “oxidizing behaviour” means that the substance itself gets reduced by gaining electrons while causing some other species to lose electrons (get oxidized).
For $$H_2SO_4$$, the sulphur atom is in the $$+6$$ oxidation state. To act as an oxidizing agent, sulphur must come down to a lower oxidation state (e.g. $$+4, +2, 0, -2$$) during the reaction.

Let us inspect each option by writing the oxidation numbers before and after reaction.

Case A:

$$2HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2H_2O$$

• In $$HI$$, iodine is $$-1$$. In $$I_2$$, iodine is $$0$$. Therefore iodine is oxidized from $$-1$$ to $$0$$ (loss of electrons).
• In $$H_2SO_4$$, sulphur is $$+6$$. In $$SO_2$$, sulphur is $$+4$$. Therefore sulphur is reduced from $$+6$$ to $$+4$$ (gain of electrons).
Because $$H_2SO_4$$ gets reduced while oxidizing $$HI$$, this reaction illustrates the oxidizing behaviour of $$H_2SO_4$$.

Case B:

$$Ca(OH)_2 + H_2SO_4 \longrightarrow CaSO_4 + 2H_2O$$

This is a simple acid-base neutralization. Oxidation states of all elements remain unchanged. No redox process occurs, so oxidizing behaviour is absent.

Case C:

$$NaCl + H_2SO_4 \longrightarrow NaHSO_4 + HCl$$

This reaction is an acid-salt displacement. Chlorine stays at $$-1$$, sulphur stays at $$+6$$. Again, no redox change; $$H_2SO_4$$ is not acting as an oxidizing agent here.

Case D:

$$2PCl_5 + H_2SO_4 \longrightarrow 2POCl_3 + 2HCl + SO_2Cl_2$$

This is mainly a dehydrating/chlorinating reaction. Check sulphur: it is $$+6$$ in both $$H_2SO_4$$ and $$SO_2Cl_2$$, so no reduction of sulphur. Hence no oxidizing action is displayed.

Only in Case A does sulphur in $$H_2SO_4$$ undergo reduction and thereby oxidizes another species. Therefore the reaction in Option A is the one that depicts the oxidizing behaviour of sulphuric acid.

Final Answer: Option A which is: $$2HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2H_2O$$