The oxidation state of the most electronegative atom in each of the product in the following reaction is
$H_{2} + O_{2} \to H_{2} O_{2} + H_{2} O$

- 2, - 2

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+ 1, - 2

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+ 2, - 1

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- 1, - 2

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Solution

The correct option is B $- 1, - 2$
In hydrogen peroxide, the oxidation number of O is $- 1$ as it is a peroxide (O-O) while in water it is $- 2$ as it is an oxide (O $^{2 -}$ ).

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