First slide

Hydrogen Peroxide

Question

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

Easy

A

0 and –1
B

–1 and –2
C

– 2 and 0
D

–2 and +1

Solution

$\large Ba{O_2}\; + \;\mathop {{H_2}S{O_4}\;}\limits_{(dil)} \to \;BaSO_4^{ - 2}\; + \;{H_2}O_2^{ - 1}$

In the above reaction the products formed are BaSO₄ & H₂O₂
In BaSO₄ the oxidation state of oxygen [more electronegative atom] is "-2"
In H₂O₂ the oxidation state of "O" [More electro-negative atom] is -1
Therefore. Option -2 is correct.

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