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The oxidation states of the most electronegative element in the products of the reaction between ${B a O}_{2}$ and dilute $H_{2} {S O}_{4}$ are

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- 2

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Correct option is B

The reaction between ${B a O}_{2}$ and dilute $H_{2} {S O}_{4}$ is as follows:

${B a O}_{2} + H_{2} {S O}_{4} \to {B a S O}_{4} + H_{2} O_{2}$

From the above reaction it is clear that the most electronegative element in the product is oxygen.

The oxidation state of oxygen in ${B a S O}_{4}$ is $- 2$ and in $H_{2} O_{2}$ is $- 1$ .

Hence the correct option is (B).

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The oxidation states of the most electronegative element in the products of the reaction between BaO2 and dilute H2SO4 are

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The oxidation states of the most electronegative element in the products of the reaction between ${B a O}_{2}$ and dilute $H_{2} {S O}_{4}$ are