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Among $KO_2,AlO_2,BaO_2$ and $NO_2^+$,unpaired electron present in

$\begin{array}{1 1}(a)\;KO_2\;only&(b)\; KO_2\;and\;AlO_2^-\\(c)\;NO_2^+\;and\;BaO_2&(d)\;BaO_2\;only\end{array}$

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$KO_2\Rightarrow K^+O_2^-$
$O_2$ has unpaired electron.
Hence (a) is the correct answer.
answered Mar 7, 2014 by sreemathi.v
 

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