Question

The oxidation states of the most electronegative elements in the products of the reaction between $Ba{O}_2$ and $H_{2}S{O}_4$ are, respectively :

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

Answer 1

Answer: (B)

$-1$ and $-2$

The reaction between $Ba{O}_2$ and $H_{2}S{O}_4$ is as follows:
$Ba{O}_2+H_{2}S{O}_4\left(dil\right)⟶BaS{O}_4+H_2{O}_2$
Oxidation state of $O$ in $H_2{O}_2=-1$
Oxidation state of $O$ in $BaS{O}_4=-2$