Among ${K O}_{2}, {A I O}_{2}^{-}, {B a O}_{2} a n d {N O}_{2}^{+}$ , unpaired electron is present in :

{N O}_{2}^{+} a n d {B a O}_{2}

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{K O}_{2} a n d A I O_{2}^{-}

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{K O}_{2}

only

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B a O_{2}

only

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Solution

The correct option is D ${K O}_{2}$ only
This can be done by counting number of valence electrons

$K O_{2} = 1 + 2 \times 6 = 17$ -Odd number of electrons and hence unpaired.

$A l O_{2}^{2 -} = 3 + 2 \times 6 + 1 = 14$ -Even number of electrons and hence paired.

$B a O_{2} = 2 + 2 \times 6 = 14$ -Even number of electrons and hence paired.

$N O_{2}^{+} = 5 + 2 \times 6 - 1 = 16$ -Even number of electrons and hence paired.

Hence option C is the correct answer.

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