When chlorine water is exposed to sunlight $O_{2}$ is liberated. Hence :

hydrogen has little affinity to

O_{2}

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hydrogen has more affinity to

O_{2}

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hydrogen has more affinity to chlorine

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it is a reducing agent

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Solution

The correct option is C hydrogen has more affinity to chlorine
When chlorine water is exposed to sunlight $O_{2}$ is liberated. Hence hydrogen has more affinity to chlorine. $C l_{2} + H_{2} O \to 2 H C l + \frac{1}{2} O_{2}$

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