# Hybridization of orbitals and forming of bonds in the nitrogen dioxide molecule [closed]

I considered the nitrogen atom hybridizing its orbitals and then ending up having 3 $$\mathrm{sp^2}$$ orbitals and the oxygen with 2 $$\mathrm{sp}$$ orbitals. The thing which bothers me is when people draw a radical (single electron) over the Nitrogen atom after there is the bond formation between nitrogen and 2 oxygen atoms.

Let us consider the Nitrogen, it has one full $$\mathrm{p}$$ orbital (AKA lone pair) and 3 single electrons in 3 $$\mathrm{sp^2}$$ orbitals, so that means that it can form 3 bonds – first it bonds with one oxygen forming a σ bond, leaving the nitrogen with 2 electrons, now the nitrogen forms one π bond with the same oxygen and the remaining single electron forms another σ bond with another oxygen atom. Where is the single radical on the nitrogen after all the bonding has taken place?

Moreover, as in the above picture, people draw the other oxygen (the one that did not form a π bond with nitrogen) as having 3 lone pairs, whereas, after the hybridization of the oxygen atom, I can only see it having 2 lone pairs in 2 $$\mathrm{p}$$ orbitals.

What is actually correct? Am I going wrong somewhere?

In your simple model, you start off with:

• nitrogen: $$\mathrm{3(sp^2)^1 + p^2}$$
• left oxygen: $$\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$$
• right oxygen: $$\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$$

The σ bonds are easy. Use $$\mathrm{sp(O) + sp^2(N)}$$. That leaves you with:

• left oxygen: $$\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{O-N})^2}$$
• nitrogen: $$\mathrm{p^2 + (sp^2)^1 + (\sigma_\ce{N-O(1)})^2 + (\sigma_\ce{N-O(2)})^2}$$
• right oxygen: $$\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{N-O})^2}$$

Now the question is how to do a π system (obviously there can only be one σ bond between two atoms). The orbitals used for the π system must be orthogonal to the bond axis, since that axis must be a node. Both oxygens have two p-type orbitals which fulfill this condition. Nitrogen also must use its p-type orbital to form the π bond. The $$\mathrm{sp^2}$$ orbital cannot because it is in the same plane as the other two $$\mathrm{sp^2}$$ orbitals, angled at $$120^\circ$$ rather than $$90^\circ$$. This gives us a π system composed of two oxygen p orbitals (one on each oxygen) and one nitrogen p orbital; choosing our orbitals well will result in this π system having four electrons. Essentially, this is theme and variation of Your Average 4-electron-3-centre bonding; but you can also consider it akin to the allyl anion.

The one remaining unpaired orbital would be in $$\mathrm{sp^2}$$ on nitrogen, pointing away from both $$\ce{N-O}$$ bonds but being in the same plane.

The above are the considerations and reasonings that you should use from the starting point you had. It is not the only way. You could assume that nitrogens five electrons are actually distributed as $$\mathrm{2(sp^2)^1 + (sp^2)^2 + p^1}$$. In this case, the argument outlined above is essentially the same except that the π system now contains three electrons rather than four; it is an allyl radical system, if you wish. Since this ab initio construction will lead to all orbitals with s contribution ($$\mathrm{sp^2}$$ on nitrogen, $$\mathrm{sp}$$ on oxygen) being fully occupied, it is more consistent with Bent’s rule and likely to be a better proposition.

If you do it like this, you can – in the Lewis 2-electron-2-centre formalism – choose either oxygen and have the radical form an $$\ce{N-O}$$ π bond on that side; the other will remain a radical. By resonance, you can draw the mirrored molecule.

$$\ce{O=N-\overset{.}{O} <-> \overset{.}{O}-N=O}$$

In these structures, the radical is never centred on nitrogen although we find that dimerisation at low temperatures leads to $$\ce{O2N-NO2}$$.

Actual measurement shows, that the radical is centred in an orbital that transforms as $$\mathrm{a_1}$$. As $$\mathrm{a_1}$$ is symmetric with regards to all symmetry transformations, it cannot be a π type orbital which can transform either as $$\mathrm{a_2}$$ (node through nitrogen) or $$\mathrm{b_1/b_2}$$ (no node through nitrogen; depending on your choice of coordinates). You see, it gets more and more complicated the more you rely on experimental data.

• Very nice answer! Interesting that CO2 has one electron less (is isoelectronic with NO2+), and nitrite ion has one electron more. Both are easier to imagine, but halfway in between is a monster! – James Gaidis Sep 29 '20 at 13:42

$$\ce{NO2}$$ , $$\ce{NO}$$, $$\ce{ClO2}$$ and a few other are deceptively simple molecules with surprisingly sophisticated bonding. Their reasonable understanding requires use of molecular orbitals theory. Below I try to invoke resonance to provide a simpler understanding for $$\ce{NO2}$$

$$\ce{NO2}$$ is a 'hypervalent' compound, i.e. to describe it adequately, one should invoke tricks involving dative bonds. I think, it is easiest to think about it as a resonance hybrid of $$\ce{O = N^{.+}-O^{-}}$$ and $$\ce{O^{-}- N^{.+}= O}$$. Indeed, $$\ce{NO2}$$ fragment bound via nitrogen is fairly common in chemistry and $$\ce{NO2}$$ readily interacts with, for example, chlorine:

$$\ce{2NO2 +Cl2 = 2 ClNO2}$$

and direct ESR spectroscopy data * suggest strong spin density on nitrogen. However! $$\ce{NO2}$$ radical is too stable for this simple interpretation, and same ESR data suggest that considerable spin density in NO2 is localized on oxygen atoms. This means that structures $$\ce{O^. - N = O}$$ and $$\ce{O = N - O^.}$$ also strongly participate.

Since no structure with unpaired electron in pi-system allows for a considerable spin density on nitrogen, it leads to conclusion that spin density is delocalized with participation of $$sp^2$$ hybridized lone electron pairs of oxygen, in particular ones opposed to the in-plane $$sp^2$$ orbital with the lone electron. This description is apparently in agreement wit MO treatment of $$\ce{NO2}$$. This IS counter-intuitive, since common organic sense is to put unpaired electron into $$\pi$$-system

• Thanks for your reply, but I am asking as to why the hybridization won't work here. The thing which you told about resonance, I am in the first place confused as to why does formal charges appear in the molecule, and how does the radical appear on the nitrogen, as it would not make sense wrt my diagram. – FinalBOSS Sep 28 '20 at 17:52
• @FinalBOSS You can make it work if you use the above treatment, it implies every atom is $sp^2$ hybridized. But in general you shouldn't ask 'why hybridization doesn't work here', but 'why hybridization would work here'. – permeakra Sep 29 '20 at 5:03
• How is $\ce{NO2}$ hypervalent? – Jan Sep 29 '20 at 12:17
• @Jan It has significant spin density on the nitrogen atom and the nitrogen forms three bonds, i.e. it has nitrogen with a 'dangling bond' and three normal bonds. But it cannot form more than three bonds when neutral. – permeakra Sep 29 '20 at 14:38