The correct order of bond angles is

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(D) To determine the bond angle, we look at the hybridization and the number of lone pairs on the central nitrogen atom:$1$. $NO^{+}_{2}$: The central

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$NO^{+}_{2} > NO_{2} > NO^{-}_{2}$

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(D) To determine the bond angle, we look at the hybridization and the number of lone pairs on the central nitrogen atom:$1$. $NO^{+}_{2}$: The central $N$ atom is $sp$ hybridized with $0$ lone pairs. The geometry is linear, so the bond angle is $180^{\circ}$.$2$. $NO_{2}$: The central $N$ atom is $sp^{2}$ hybridized with $1$ odd electron (radical). Due to the presence of the odd electron, the bond angle is slightly less than $120^{\circ}$ (approximately $134^{\circ}$ in some contexts, but generally treated as $sp^{2}$ with repulsion).$3$. $NO^{-}_{2}$: The central $N$ atom is $sp^{2}$ hybridized with $1$ lone pair. The lone pair-bond pair repulsion decreases the bond angle to approximately $115^{\circ}$.Comparing these, the order is $NO^{+}_{2} (180^{\circ}) > NO_{2} (> 120^{\circ}) > NO^{-}_{2} (\,115^{\circ})$.Thus, the correct order is $NO^{+}_{2} > NO_{2} > NO^{-}_{2}$.

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