What is oxidisation of "O" in potassium superoxide:

-1

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-2

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\frac{- 1}{2}

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Solution

The correct option is C $\frac{- 1}{2}$
As its a stable molecule,

(Oxidation number of $K$ ) + 2(Oxidation number of $O$ ) = $0$ .

So, oxidation state of Oxygen in $K O_{2}$ is $\frac{- 1}{2} .$

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Similar questions

$K O_{2}$ (potassium superoxide) is used in oxygen cylinders in space and submarines because it:

{K O}_{2}

{K O}_{2} (s) + H_{2} O (l) ⟶ K O H (s) + O_{2} (g)

{K O}_{2}

H_{2} O

O_{2}

K_{2} O

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