The correct order of increasing bond angles in the following species is:

N O_{2}^{-} < N O_{2}^{+} < N O_{2}

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N O_{2}^{-} < N O_{2} < N O_{2}^{+}

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N O_{2}^{+} < N O_{2} < N O_{2}^{-}

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N O_{2}^{+} < N O_{2}^{-} < N O_{2}

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Solution

The correct option is B $N O_{2}^{-} < N O_{2} < N O_{2}^{+}$

$N O_{2}^{-}$ and $N O_{2}$ are $s p^{2}$ hybridized. Since $N O_{2}^{-}$ has lone pair-bond pair repulsion so the angle is smaller than $N O_{2}$ .
$N O_{2}^{+}$ is $s p$ hybridized.

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