Question

The oxidation state of the most electronegative element in the products of the reaction between $Ba{O}_2$ and $H_{2}S{O}_4$ is:

• A. 0 and -1
• B. -1 and -2
• C. -2 and 0
• D. -2 and +1

Answer 1

The correct option is B -1 and -2

Reaction between $Ba{O}_2$ and $H_{2}S{O}_4$ is as follows:
$Ba{O}_2+H_{2}S{O}_4\to H_2{O}_2+BaS{O}_4$
$-1$ $-2$
The most electronegative element is oxygen in both the products and its oxidation state is $-1$ and $-2$ respectively, in $H_2{O}_2$ and $BaS{O}_4$.