The oxidation state of the most electronegative element in the reactant and product of the reaction, $B a O_{2}$ with dil $H_{2} S O_{4}$ are

O and – 1

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– 2 only

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– 2 and O

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– 2 and – 1

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Solution

The correct option is D – 2 and – 1
$B a O_{2} + H_{2} S O_{4} \to B a S O_{4} + H_{2} O$
$O x i d a t i o n s t a t e o f O - a t o m i n B a S O_{4}$
$O x i d a t i o n s t a t e o f O - a t o m i n B a_{2} O_{2} = - 1$

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