The oxidation states of the most electronegative element in the products of the reaction of BaO2 with dilute H2SO4 are

Question

The oxidation states of the most electronegative element in the products of the reaction of BaO2 with dilute H2SO4 are (A) 0 and - 1 (B) - 1 and - 2 (C) - 2 and 0 (D) - 2 and + 1

Tardigrade Admin · Accepted Answer

In H2O2 oxygen shows = -1 (peroxide) oxidation state and in BaSO4 oxygen shows = - 2 oxidation state.

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

0 and - 1

- 1 and - 2

- 2 and 0

- 2 and + 1

In $H_{2} O_{2}$ oxygen shows $= - 1$ (peroxide) oxidation state and in $B a S O_{4}$ oxygen shows $= - 2$ oxidation state.

Q. The oxidation states of the most electronegative element in the products of the reaction of BaO2​ with dilute H2​SO4​ are