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

For $\text{NO}_2^+$, the central nitrogen atom is $sp$ hybridized with no lone pairs, resulting in a linear geometry and a bond angle of $180^\circ$.

For $\text{NO}_2$, the central nitrogen atom is $sp^2$ hybridized with one unpaired electron. The repulsion from a single electron is less than that from a bond pair, so the bond angle opens up to be greater than $120^\circ$ (approximately $134^\circ$).

For $\text{NO}_3^-$, the central nitrogen atom is $sp^2$ hybridized with no lone pairs, resulting in a trigonal planar geometry and a bond angle of exactly $120^\circ$.

For $\text{NO}_2^-$, the central nitrogen atom is $sp^2$ hybridized with one lone pair. The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing the bond angle to less than $120^\circ$ (approximately $115^\circ$).

Therefore, the correct order of O-N-O bond angles is $\text{NO}_2^- < \text{NO}_3^- < \text{NO}_2 < \text{NO}_2^+$.

Answer: $\text{NO}_2^- < \text{NO}_3^- < \text{NO}_2 < \text{NO}_2^+$