The compound $K O_{2}$ is a

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peroxide

normal oxide

super oxide

sub oxide

answer is B.

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Detailed Solution

${KO}_{2}$ is a superoxide.Becuase in ${KO}_{2}$ the oxidation sate of oxide is $- 1 / 2$ .potassium loses one electron to two oxigens so each oxygen gains half electron and forms ${KO}_{2}$ .
Hence option B is correct.

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