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What is the correct structure for the $\ce{NO2}$ compound (not ion)? I always thought it was like:

enter image description here

since the negative charge would be residing the on the highly electronegative oxygen, but was reading this blogpost by the guys at WolframAlpha, who state that contrary to popular belief, the correct structure is actually:

enter image description here

They explain that it has to do with molecular orbital theory, but don't explain in depth.

Can someone please explain why this occurs?

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    $\begingroup$ Though it's hardly a rigorous argument, you can calculate the formal charges on each atom for both proposed structures. In the first, there is a +1 charge on the nitrogen atom, and a -1 charge on the singly-bonded oxygen atom (though really it's -0.5 on both oxygen atoms due to resonance). In the second structure, all atoms have zero formal charge, and so the second structure is expected to be lower in energy and hence to contribute more to the actual picture of bonding in $\ce{NO2}$. A higher level argument would still be desirable. $\endgroup$ – Nicolau Saker Neto Mar 24 '14 at 23:48
  • $\begingroup$ I would love another one of those research project answers by LordStryker on this one! $\endgroup$ – tschoppi Aug 10 '16 at 13:06
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Drawings of molecules are only representations to help us understand them, not the actual molecule itself. So both pictures you have there are correct; they both contribute to the actual structure of the $\ce{NO2}$ molecule. I'm guessing WolframAlpha chose the structure they did because it is the structure the most contributes to the actual structure of the molecule, with all formal charges being 0 like Nicolau said, it is the most 'favorable' of possible configurations that the molecule could be in.

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  • $\begingroup$ Does this mean electron clouds are more likely to stay near the N atom? $\endgroup$ – Soura Feb 27 '18 at 21:05
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According to Electronic Structure of NO2 Studied by Photoelectron and Vacuum-uv Spectroscopy and Gaussian Orbital Calculations J. Chem. Phys. 53, 705 (1970) :

The highest molecular orbital $4a_1$ is occupied by the 1 unpaired electron.

The electron population of this orbital is (see table VI):

0.53 on the N atom (0.16 2s, 0.37 2pz)

0.24 on each O atom (0.24 pz)

Experimentally, Oxides and Oxyions of the Non-metals. Part II. C02- and NO2 of the Chemical Society (1962): 2873-2880 collects values for the partition of unpaired electron density among N and O atomic orbitals.

The most purely experimental values are:

N 2s 0.094
N 2pz 0.364
N 2px 0.054
O 2pz 0.33

and

N 2s 0.103
N 2pz 0.471
O 2pz 0.33

(where the values for O are both O atoms added together)

There is extensive discussion of the various Lewis structures of NO2 from a dimerization point of view (that the opposite spin electrons should be on the atoms that form the dimer bond) starting on page 90 if the 2015 book Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increase-valence Structures , and also elsewhere in the book from a monomer point of view.

Overall, it is clear theoretically and experimentally that the unpaired electron is delocalized, having significant density on both N and the two O atoms. It is wrong to say that a particular one of the Lewis structures is "correct".

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  • $\begingroup$ How large of a basis set were they able to use? Being 1970, I suspect a rather small one, in which case the electron occupation pattern is likely suspect, due to basis set insufficiency. $\endgroup$ – hBy2Py Aug 10 '16 at 12:31
  • $\begingroup$ @hBy2Py I guess you wouldn't trust the values from this 1962 EPR paper either: pubs.rsc.org/en/content/articlepdf/1962/jr/jr9620002873 $\endgroup$ – DavePhD Aug 10 '16 at 12:31
  • $\begingroup$ @hBy2Py There is an experimental value for unpaired electron density at the N nucleus by microwave spectroscopy in scitation.aip.org/content/aip/journal/jcp/40/11/10.1063/… (1964) $\endgroup$ – DavePhD Aug 10 '16 at 12:34
  • $\begingroup$ @hBy2Py the 1970 article says "The best-atom-double-zeta (BADZ) basis set of Basch et al. [reference 26] was used throughout." Reference 26 is scitation.aip.org/content/aip/journal/jcp/47/4/10.1063/… J. Chem. Phys. 47, 1201 (1967) $\endgroup$ – DavePhD Aug 10 '16 at 12:41
  • $\begingroup$ I'll have to look around for a recent post-Hartree-Fock study. DFT gives results at odds with the ESR data. It looks likely that $\ce{NO2}$ is a system that DFT is no good for, or that one has to play some tricks in order to get to the correct result. $\endgroup$ – hBy2Py Aug 10 '16 at 13:20
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Nitrogen dioxide has 17 valence electrons, and is bent with an angle of 134 degrees and bond lengths 0.119 nm. Electron spin resonance experiments place the odd electron on the nitrogen in a $\sigma$ rather than a $\pi$ orbital. Thus formally we can consider the odd electron to be in a sigma bonding orbital (of $a_1$ symmetry species in $\ce{C_2v}$) of $sp^2$ hybrid orbital type. In so far as you want to use these types of diagrams experiment would indicate that the first structure is the most correct.

The ion $\ce{NO2^+}$ is linear, $\ce{NO2}$ is bent and so is $\ce{NO2^-}$ with an angle of 101 degrees. To understand these differences it is necessary to combine the p orbitals on each atom to form various $\pi$ orbitals and then consider how these orbitals change in energy as the molecule bends. To do this is quite easy but working out what happens to the energy on bending is more tricky. However, these have been worked out and we can use Walsh diagrams do do this. All that is necessary is to know the symmetry species of an orbital and fill up the orbitals with electrons.
(To do this is rather off topic for your question but I can add pictures if you are interested.)

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  • $\begingroup$ Can you link to literature describing the ESR results? A quick UPBE calculation shows all of the spin density divided on the oxygens. $\endgroup$ – hBy2Py Aug 10 '16 at 11:53
  • $\begingroup$ (That is, all three of the Bader, Loewdin, and Mulliken methods put all of the spin density on the oxygens.) $\endgroup$ – hBy2Py Aug 10 '16 at 12:30
  • $\begingroup$ I'm not an epr expert at all so i'm not familiar with the literature. I found the details in 'Stable Radicals: Fundamentals and Applied Aspects of Odd-Electron Compounds' edited by Robin Hicks' chapter 4 by D. Bohle, section 4.5. There are references there to original papers. $\endgroup$ – porphyrin Aug 10 '16 at 13:02
  • $\begingroup$ It's an interesting problem. Per my comment on DavePhD's answer, it may be that $\ce{NO2}$ is just a problematic system for DFT to handle. $\endgroup$ – hBy2Py Aug 10 '16 at 13:22
  • $\begingroup$ @ hBy2Py Yes $\ce{NO2}$ is a tough problem, I'm a bit out of date but I think that its electronic spectroscopy is not fully sorted out even after many years of study. There are lots of low lying and interacting states that are coupled to one another making life very messy. Presumably low lying states makes calculating things hard too. $\endgroup$ – porphyrin Aug 10 '16 at 13:36
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The $^{16}\ce{O}$ nucleus is spin-0, the $^{14}\ce{N}$ nucleus is spin-1. You can see the real world weighted fractional residence times of the unpaired spin with EPR. $\pi$-bonding will complicate the spectrum with spin cross-coupling.

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$\ce{NO2}$ molecules dimerize to form $\ce{N2O4}.$ The $\ce{N}$ adds to each other through the odd electron at $\ce{N}$ atoms. This may be a strong indication that the first structure of $\ce{NO2}$ has significant contribution.

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