The oxidation states of the most electronegative element in the products of the reaction between BaO2 and H2SO4 are:

Question

The oxidation states of the most electronegative element in the products of the reaction between BaO2 and H2SO4 are: (A)  0 and -1  (B)  -1 and -2  (C)  -2 and -3  (D)  -2 and 0

Tardigrade Admin · Accepted Answer

BaO 2+ H 2 SO 4 arrow BaSO -2 4+ H 2 O -1 2 Oxygen being the most electronegative element in the reaction has the oxidation states of -1 (in peroxide, ie , H 2 O 2 ) and -2 in BaSO 4.

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

$0$ and $- 1$

$- 1$ and $- 2$

$- 2$ and $- 3$

$- 2$ and $0$

$B a O_{2} + H_{2} S O_{4} \to B a S O^{- 2}_{4} + H_{2} O^{- 1}_{2}$
Oxygen being the most electronegative element in the reaction has the oxidation states of $- 1$ (in peroxide, $i e, H_{2} O_{2}$ ) and $- 2$ in $B a S O_{4}$ .

Q. The oxidation states of the most electronegative element in the products of the reaction between BaO2​ and H2​SO4​ are:

0 and −1

−1 and −2

−2 and −3

−2 and 0