In the chemical reaction $C l_{2} + H_{2} S ⟶ 2 H C l + S$ , the oxidation number of sulphur changes from

0 to 2

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2 to 0

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2 to 0

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-2 to -1

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Solution

The correct option is C 2 to 0
The oxidation state of sulphur in $H_{2} S i s - 2$ , while it is zero in 'S' i.e. in this reaction oxidation of sulphur and reduction of chlorine takes place. The oxidation number of chlorine changes from 0 to -1.

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C l_{2} + H_{2} S ⟶ 2 H C l + S

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