Question

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

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

Answer 1

Answer: (B)

$-1$ and $-2$

$Ba{O}_2+H_{2}S{O}_4\to {BaSO}_4+H_2{O}_2$

The most electronegative element in the product is oxygen.

The oxidation state of oxygen in $BaS{O}_4$ is -2 and in $H_2{O}_2$ is -1.

$Note$: Oxidation state of hydrogen is always +1 so in $H_2{O}_2$ oxidation of oxygen is -1 instead of -2.

Hence, the correct option is $\text{B}$