If $C l_{2}$ is passed into hot NaOH solution, oxidation number of chlorine changes from:

0 to +5

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0 to - 2

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0 to +1

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0 to +7

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Solution

The correct option is A 0 to +5
$0 C l_{2} + O H^{-} \to + 5 C l O_{3}^{-} + C l^{-}$

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When $H_{2} S O_{3}$ is converted into $H_{2} S O_{4}$ the change in the oxidation state of sulphur is from:

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N a O H

C l_{2} + H_{2} S ⟶ 2 H C l + S

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