Comparison of ONO Bond Angles in Nitrogen Oxide Species

To determine the correct order of the $\text{O-N-O}$ bond angles, we analyze the hybridization, structure, and electron-pair repulsions (VSEPR theory) for each species:

1. $\text{NO}_2^+$ (Nitronium ion):

   • Central nitrogen atom has $5 - 1 = 4$ valence electrons.
   • It forms two double bonds with two oxygen atoms ($2\ \sigma\text{-bonds} + 0\text{ lone pairs}$).
   • Hybridization of Nitrogen: $sp$
   • Molecular Geometry: Linear
   • $\text{O-N-O}$ bond angle: $180^\circ$

2. $\text{NO}_2$ (Nitrogen dioxide):

   • Central nitrogen atom has $5$ valence electrons. It forms two $\sigma$-bonds with oxygen atoms and retains $1$ unpaired electron (single electron).
   • Steric Number: $2\ \sigma\text{-bonds} + 1\text{ odd electron} \approx 3$
   • Hybridization of Nitrogen: $sp^2$
   • An unpaired electron exerts less repulsion on the bonding pairs than a full lone pair, causing the $\text{O-N-O}$ bond angle to be slightly larger than $120^\circ$.
   • $\text{O-N-O}$ bond angle: $\approx 134^\circ$

3. $\text{NO}_3^-$ (Nitrate ion):

   • Central nitrogen atom has $5 + 1 = 6$ valence electrons.
   • It forms three $\sigma$-bonds with three oxygen atoms ($3\ \sigma\text{-bonds} + 0\text{ lone pairs}$).
   • Hybridization of Nitrogen: $sp^2$
   • Molecular Geometry: Symmetrical trigonal planar
   • $\text{O-N-O}$ bond angle: $120^\circ$

4. $\text{NO}_2^-$ (Nitrite ion):

   • Central nitrogen atom has $5 + 1 = 6$ valence electrons.
   • It forms two $\sigma$-bonds with two oxygen atoms and holds $1$ complete lone pair ($2\ \sigma\text{-bonds} + 1\text{ lone pair}$).
   • Hybridization of Nitrogen: $sp^2$
   • The lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, compressing the $\text{O-N-O}$ bond angle to less than $120^\circ$.
   • $\text{O-N-O}$ bond angle: $\approx 115^\circ$

Comparing the calculated bond angles: $\text{NO}_2^- \, (115^\circ) < \text{NO}_3^- \, (120^\circ) < \text{NO}_2 \, (134^\circ) < \text{NO}_2^+ \, (180^\circ)$

Thus, the correct order of the $\text{O-N-O}$ bond angle is: $\text{NO}_2^- < \text{NO}_3^- < \text{NO}_2 < \text{NO}_2^+$

Correct Option: (B)