Chemistry - Structure of NO2 compound?
Solution 1:
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".
Solution 2:
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.)
Solution 3:
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.
Solution 4:
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.