In which of the following has the oxidation number of oxygen been arranged in increasing order?

$O F_{2}$ < $K O_{2}$ < $B a O_{2}$ < $O_{3}$

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$B a O_{2}$ < $O_{3}$ < $O F_{2}$ < $K O_{2}$

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$B a O_{2}$ < $K O_{2}$ < $O_{3}$ < $O F_{2}$

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$K O_{2}$ < $O F_{2}$ < $O_{3}$ < $B a O_{2}$

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Solution

The correct option is C
$B a O_{2}$ < $K O_{2}$ < $O_{3}$ < $O F_{2}$

Compound NO of Oxygen

$B a O_{2}$ -1

$K O_{2}$ $\frac{- 1}{2}$

$O_{3}$ O

$O F_{2}$ +2

$O F_{2}$ > $O_{3}$ > $K O_{2}$ > $B a o_{2}$

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