When chlorine reacts with cold and dilute solution of sodium hydroxide, the products obtained are:

C l^{-} + C l O^{-}

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C l^{-} + C l O_{2}^{-}

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C l^{-} + C l O_{3}^{-}

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C l^{-} + C l O_{4}^{-}

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Solution

The correct option is A $C l^{-} + C l O^{-}$
When chlorine reacts with cold and dilute sodium hydroxide, the following reaction takes place:
$2 N a O H$ $+$ $C l_{2}$ $⟶$ $N a C l$ $+$ $N a C l O$ $+$ $H_{2} O$ .
Hence the products are ${C l}^{-}$ and $C l O^{-}$
When chlorine reacts with hot and concentrated sodium hydoxide, the following reaction takes place:
$N a O H$ $+$ $C l_{2}$ $⟶$ $5 N a C l$ $+$ $N a C l O_{3}$ $+$ $3 H_{2} O$
Hence the products are ${C l}^{-}$ and ${C l O}_{3}^{-}$

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Similar questions

N a O H

C l - O

C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}

C l O_{4}^{-}

3 C l O^{-} ⟶ C l O_{2}^{-} + 2 {C l}^{-}

C l O^{-} + C l O^{-} K_{1} - \to C l O_{2}^{-} + {C l}^{-}

C l O_{2}^{-} + C l O^{-} K_{2} - \to C l O_{3}^{-} + {C l}^{-}

{C l}_{2}

3 C l O^{-} \to C l O_{3}^{-} + 2 C l

C l O^{-} + C l O^{-} K_{1} - \to C l O_{2}^{-} + C l^{-}

C l O_{2}^{+} C l O^{-} k_{2} - \to C l O_{3}^{-} + C l^{-}

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