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From my reading, I have understood examples a diatomic molecule to be $\rm N_2$ or $\rm O_2$, however, the below seems to suggest that $\rm N_2O_2$ is also diatomic. Is this correct and can someone please explain?

My answer for the below question was 6 degrees of freedom (3 translational, 3 rotational, not high enough temperature for vibrational).

Q. The main components of air are $\rm N_2$ and $\rm O_2$. Assume that we can approximate air as only made from these two gasses. At room temperature how many degrees of freedom does air have? State what type of movement (vibrational, rotational and translational) each of these degrees of freedom corresponds to.

Ans. This is a diatomic gas at approximately room temperature. It has 5 degrees of freedom. 3 of these degrees of freedom correspond to the translational movement of the particles (one in x, one in y and one in z direction) and 2 correspond to the rotational movement of the molecules (abut the two axes perpendicular to the line joining the two atoms).

Emilio Pisanty
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Jordan Lane
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The molecule $\rm N_2O_2$ does exist, for example as a dimer of nitric oxide, NO. Evidently it is not a diatomic molecule (it has 4 atoms, not 2). It is a nonlinear molecule, so if we treat it as a rigid molecule, your analysis of 3 translational and 3 rotational degrees of freedom would be correct. Most likely, the situation is not so simple, because the N-N bond is rather long (and hence probably quite weak), and there might be some internal conformational degrees of freedom to consider. But your counting of translations and rotations is correct.

However, none of this has anything to do with the question you are referring to. Air is a mixture, mostly, of $\rm N_2$ and $\rm O_2$, both diatomic molecules. The molecule $\rm N_2O_2$ is a completely insignificant component of air. The question asks you to treat air as a mixture of two gases which (at least at room temperature) behave in a similar way, as far as their degrees of freedom are concerned. If the temperature were much higher, we would need to take into account the fact that the bond in $\rm N_2$ is stronger than in $\rm O_2$, so the vibrational contributions would not start increasing with $T$ in the same way. But that's not relevant to the question.

Possibly you got confused by looking at a website that described the nitrogen-oxygen mixture as $\rm N_2O_2$, for example this one. But you shouldn't interpret this as meaning that air consists of $\rm N_2O_2$ molecules.