The oxidation state of the most electronegative element in the products of the reaction. BaO2 with dil, H2SO4 are

Question

The oxidation state of the most electronegative element in the products of the reaction. BaO2 with dil, H2SO4 are (A) 0 and - 1 (B) - 1 and - 2 (C) - 2 and 0 (D) - 2 and - 1

Tardigrade Admin · Accepted Answer

Reaction between BaO 2 and H 2 SO 4 is as follows: beginarrayr BaO 2+ H 2 SO 4 arrow H 2 O 2+ BaSO 4 -1 -2 endarray 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 BaSO 4.

Q. The oxidation state of the most electronegative element in the products of the reaction. $B a O_{2}$ with dil, $H_{2} S O_{4}$ are

0 and - 1

- 1 and - 2

- 2 and 0

- 2 and - 1

Reaction between $B a O_{2}$ and $H_{2} S O_{4}$ is as follows:
$B a O_{2} + H_{2} S O_{4} \to H_{2} O_{2} + B a S O_{4} - 1 - 2$
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 $B a S O_{4}$ .

Q. The oxidation state of the most electronegative element in the products of the reaction. BaO2​ with dil, H2​SO4​ are