Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct order of ONO bond angle in the given species is :

Welcome to a classic chemical bonding problem! We are tasked with arranging four nitrogen-oxygen species—${\text{NO}}_2^{+}$, ${\text{NO}}_3^{-}$, ${\text{NO}}_2^{-}$, and ${\text{NO}}_2$—in order of their $\text{O-N-O}$ bond angles. To conquer this, we must rely on Valence Shell Electron Pair Repulsion (VSEPR) theory, which allows us to predict the 3D geometry of molecules based on the electrostatic repulsion between electron pairs.

Let's break down each species one by one by calculating their steric numbers and analyzing the repulsive forces at play.

Let's start with the nitronium ion, ${\text{NO}}_2^{+}$. If we draw its Lewis structure, we see the central nitrogen atom forms two double bonds with the oxygen atoms and has zero lone pairs.

This gives it a steric number of $2$ (two $\sigma$-bonding domains). A steric number of $2$ corresponds to $sp$ hybridization. Naturally, to minimize repulsion, the two bonding pairs will position themselves as far apart as physically possible—in a straight line. Because it is perfectly linear, the $\text{O-N-O}$ bond angle is exactly $180^{\circ }$. This will be our maximum possible angle among the given species.

Next, look at the nitrate ion, ${\text{NO}}_3^{-}$. The central nitrogen atom is bonded to three oxygen atoms (one double bond, two single bonds) and has no lone pairs left.

A steric number of $3$ means $s{p}^2$ hybridization, which gives us a beautiful, flat, trigonal planar geometry. In a perfect trigonal planar setup, the $360^{\circ }$ space is divided equally among the three bonds. Therefore, the $\text{O-N-O}$ bond angle is exactly $120^{\circ }$.

Now, things get interesting with the nitrite ion, ${\text{NO}}_2^{-}$. It also has a steric number of $3$ (two $\sigma$-bonds and one lone pair), meaning its base geometry is trigonal planar. However, one of those positions is occupied by a lone pair of electrons, giving the molecule a bent shape.

According to VSEPR theory, a lone pair is bulkier and exerts more repulsion than a bonding pair. This strong lone pair-bond pair repulsion pushes the two oxygen atoms closer together, compressing the bond angle from the ideal $120^{\circ }$ down to approximately $115^{\circ }$.

Finally, let's examine nitrogen dioxide, ${\text{NO}}_2$. It is a fascinating odd-electron molecule. It has two $\sigma$-bonds and a single, unpaired electron residing on the nitrogen atom. Like the nitrite ion, it has a bent shape, but the repulsion dynamics are entirely different.

A single electron exerts significantly less repulsion than a full lone pair, and even less than the bulky double bonds. Because the downward push from the top single electron is weak, the oxygen bonding pairs push each other apart, expanding the angle beyond the ideal $120^{\circ }$ to about $134^{\circ }$.

Let's put it all together. By comparing the angles we've deduced:

The nitrite ion has the smallest angle, followed by the nitrate ion, then nitrogen dioxide, and finally the nitronium ion with a perfect $180^{\circ }$. This matches option (B) perfectly.

The golden rule here is understanding the hierarchy of repulsion: $\text{Lone Pair}>\text{Bond Pair}>\text{Odd Electron}$. A lone pair pushes the hardest, compressing angles. An odd electron pushes the weakest, allowing angles to open up. Keep this in mind, and you will master any VSEPR question!