When Barium chlorite is treated with sulphuric acid ,the product formed is:

H C l O

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H C l O_{2}

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H C l O_{3}

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H C l O_{4}

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Solution

The correct option is B $H C l O_{2}$
When Barium chlorite is treated with sulphuric acid ,the product formed is chlorous acid $H C l O_{2}, B a S O_{4}$

$B a (C l O_{2})_{2} + H 2 S O 4 \to 2 H C l O_{2} + B a S O_{4}$ .
Hence option $B$ is correct.

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Similar questions

Arrange the following compounds according to the increasing order of oxidation state of chlorine.

$H C l O_{4}, H C l O_{3}, H C l O_{2}, H C l O$

H C l O_{4}, H C l O_{3}, H C l O_{2}, H C l O

H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}

H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}

H C l O, H C l O_{2}, H C l O_{3}, H C l O_{4}

H C l O_{4} < H C l O_{3} < H C l O_{2} < H C l O

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