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  4. The oxidation number of oxygen in KO3, Na2O2 is

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Q. The oxidation number of oxygen in $KO_3, Na_2O_2$ is

Redox Reactions

Solution:

O.N. of O in $KO_3 = 1 + 3 x = 0$
$\therefore \, x = - 1/3 = - 0.33$
O.N. of O in $Na_2O_2 = 2 \times 1 + 2x = 0$
$\therefore \, x = - 1 $