Which reaction represents the oxidizing behaviour of $H_{2} {SO}_{4}$ ?

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$2 {PCl}_{5} + H_{2} {SO}_{4} \to 2 {POCl}_{3} + 2 HCl + {SO}_{2} {Cl}_{2}$

$2 HI + H_{2} {SO}_{4} \to I_{2} + {SO}_{2} + 2 H_{2} O$

$NaCl + H_{2} {SO}_{4} \to {NaHSO}_{4} + HCl$

$2 NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + 2 H_{2} O$

answer is D.

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Detailed Solution

In option (D) sulphur decreases its oxidation state from $+ 6$ in $H_{2} {SO}_{4}$ to $+ 4$ in ${SO}_{2}$ and iodine increases its oxidation state from -1 in HI to o in $I_{2} \cdot {So}_{2}, H_{2} {SO}_{4}$ acts as oxidising agent and $I_{2}$ acts as reducing agent. In other reactions, sulphur does not decrease its oxidation state so $H_{2} {SO}_{4}$ does not act as oxidizing agent. $2 HI + H_{2} {SO}_{4} I_{2} + {SO}_{2} + 2 H_{2} O$ .