The decreasing order of the bond angles in $N O_{2}, N O_{2}^{+}$ , and $N O_{2}^{-}$ is

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Solution

$N O_{2}^{+}$ has a linear structure and has a bond angle of $180^{\circ}$ . Both $N O_{2}$ and $N O_{2}^{-}$ are bent but $N O_{2}$ has an odd electron on N while $N O_{2}^{-}$ has the presence of a complete lone pair that repels the bond pairs further and hence the bond angle is lesser in it. Therefore, the order is $N O_{2}^{+} > N O_{2} > N O_{2}^{-}$

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