The correct order of ONO bond angle in the given species is

The correct answer is NO₂^- < NO₃^- < NO₂ < NO₂⁺.

To determine the correct order of the O-N-O bond angle in the given species, we analyze the hybridization, lone pairs, and odd electron repulsion around the central nitrogen atom for each species:

1. NO₂⁺ (Nitronium ion):
The central nitrogen atom has 5 valence electrons. Subtracting 1 electron for the positive charge gives 4 valence electrons. It forms two double bonds with two oxygen atoms and has zero lone pairs.
Steric number = 2 (hybridization is sp).
The geometry is linear, giving an O-N-O bond angle of exactly $180°$.

2. NO₂ (Nitrogen dioxide):
The central nitrogen atom has 5 valence electrons. It forms double/single bonds with two oxygen atoms and has 1 unpaired (odd) electron remaining.
The molecule has sp² hybridization. The single odd electron occupies an orbital on nitrogen and exerts less repulsion than a full lone pair. Thus, the bond angle is slightly reduced from $120°$ to approximately $134°$.

3. NO₃^- (Nitrate ion):
The central nitrogen atom is surrounded by 3 bonding regions with a negative charge distributed via resonance.
Steric number = 3 (hybridization is sp²) with 0 lone pairs.
The geometry is perfect trigonal planar, so the O-N-O bond angle is exactly $120°$.

4. NO₂^- (Nitrite ion):
The central nitrogen atom has 5 valence electrons + 1 extra electron from the negative charge = 6 electrons. It forms bonds with two oxygen atoms and retains 1 full lone pair.
Steric number = 3 (hybridization is sp²).
The strong repulsion from the full lone pair compresses the O-N-O bond angle to approximately $115°$.

Comparing the O-N-O bond angles:
NO₂^- ($\approx 115°$) < NO₃^- ($120°$) < NO₂ ($\approx 134°$) < NO₂⁺ ($180°$).