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(B) $1$. In $NO_2^+$,the nitrogen atom is $sp$ hybridized with a linear geometry,so the $O-N-O$ bond angle is $180^{\circ}$.$2$. In $NO_2$,the nitroge

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Decrease of $O-N-O$ bond angle

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(B) $1$. In $NO_2^+$,the nitrogen atom is $sp$ hybridized with a linear geometry,so the $O-N-O$ bond angle is $180^{\circ}$.$2$. In $NO_2$,the nitrogen atom is $sp^2$ hybridized with one unpaired electron,resulting in a bent geometry with a bond angle of approximately $134^{\circ}$.$3$. In $NO_2^-$,the nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent geometry with a bond angle of approximately $115^{\circ}$.$4$. As we move from $NO_2^+ 	o NO_2 	o NO_2^-$,the $O-N-O$ bond angle decreases $(180^{\circ} > 134^{\circ} > 115^{\circ})$.$5$. The bond order decreases as electrons are added to antibonding orbitals or as the number of lone pairs increases,which leads to an increase in $N-O$ bond length.$6$. Therefore,the statement 'Decrease of $O-N-O$ bond angle' is correct.

Select the correct statement regarding the following change:
$NO_2^+ \to NO_2 \to NO_2^-$

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Select the correct statement regarding the following change:$NO_2^+ \to NO_2 \to NO_2^-$