Question

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

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

Answer 1

Answer: (B)

$-1,-2$

The reaction between $Ba{O}_2$ and dil. $H_{2}S{O}_4$:

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

So, the products are $H_2{O}_2$ and $BaS{O}_4$. The most electronegative element in these two products is oxygen (in both cases).

Thus, the oxidation states are $-1$ in $H_2{O}_2$ and $-2$ in $BaS{O}_4$ respectively in each product.