Which of the following chemical reactions depicts the oxidising behaviour of $H_{2} S O_{4}$ ?

2 H I + H_{2} S O_{4} \to I_{2} + S O_{2} + 2 H_{2} O

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C a (O H)_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H_{2} O

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N a H S O_{4} + H C l \to N a C l + H_{2} S O_{4}

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2 P C l_{5} + H_{2} S O_{4} \to 2 P O C l_{3} + 2 H C l + S O_{2} C l_{2}

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Solution

The correct option is A $2 H I + H_{2} S O_{4} \to I_{2} + S O_{2} + 2 H_{2} O$
(a) In the reaction,
$2 + 1 H - 1 I + {- + 1 H}_{2} + 6 S {- 2 O}_{4} \to {0 I}_{2} + + 4 S {- 2 O}_{2} + 2 {+ 1 H}_{2} - 2 O$
1st half reaction: $2 H - 1 I \to_{2}$
In this reaction oxidation number of $I$ increases by one, thus this is an oxidation reaction and $H I$ behaves as a reducing agent.
IInd half reaction : $H_{2} + 6 S O_{4} \to + 4 S O_{2}$
In this reaction oxidation number of S decreases by two, thus this is a reduction reaction and $H_{2} S O_{4}$ behaves as oxidising agent.
(b) $H_{2} S O_{4} \to C a S O_{4}$
Oxidation state of Sulphur changes from $+ 6 \to + 6$
Hence, it will not show oxidising behaviour.
(c) $N a H S O_{4} \to H_{2} S O_{4}$
Oxidation state of Sulphur changes from $+ 6 \to + 6$ , hence $H_{2} S O_{4}$ will not show oxidising or reducing behavouir.
(d) $H_{2} S O_{4} \to S O_{2} C l_{2}$
Oxidation state of sulphur changes from $+ 6 \to + 6$ , hence it will not show oxidising or reducing behavouir.

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