Oxidation state of oxygen in Super oxides is:

\frac{+ 1}{2}

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\frac{- 1}{2}

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+ 1

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Solution

The correct option is B $\frac{- 1}{2}$
Superoxide anions have the chemical formula $O_{2}^{-}$ with oxidation state of oxygen being $\frac{- 1}{2}$
For example: In $K O_{2},$ oxidation state of oxygen being $\frac{- 1}{2}$

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Oxidation number of oxygen in potassium super oxide $K O_{2}$ is

Y B a_{2} C u_{3} O_{7}

+ 3

F_{2} O

{K O}_{2}

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