Why is the Enthalpy of the NO2 to N2O4 Reaction Exothermic?

[Qube]

Homework Statement

Determine the enthalpy of this reaction (below).

2NO_{2} \rightarrow N_{2}O_{4}

Homework Equations

Reaction is exothermic if product is more stable than reactants.

Reaction is endothermic if the reverse were true (product less stable than reactants).

The Attempt at a Solution

I'm trying to think of all the different ways I can explain why the forward reaction is exothermic.

1) Dimerization of nitrogen dioxide; a N-N bond is being formed using the radical on the nitrogen in nitrogen dioxide.

2) The radical on the nitrogen in nitrogen dioxide makes it inherently less stable than a molecule with no radicals.

3) The nitrogen dioxide has a positive formal charge on the nitrogen and a negative formal charge on its oxygen. Positive formal charges are also present on the nitrogens in nitrogen tetraoxide along with negative formal charges on two oxygens but the entire molecule is bigger, so the charge density is likely lower.

4) Hybridization? The nitrogen of the nitrogen dioxide is sp2 hybridized (do radicals count when determining hybridization)?

Anything else I can consider in explaining why the above reaction is exothermic?

 

[Redbelly98]

It's not really clear what you are asking.

The problem statement wants you to calculate the enthalpy value. Have you done that, and now you are basically looking for some intuitive reasoning to make sense of the result, to satisfy your own curiosity? Or is coming up with the reason actually part of the assignment, even though the problem statement does not explicitly ask for a reason?

(Sorry I can't actually help with coming up with the reason. But as nobody else has replied in 2 days, I just thought I'd try to get some clarification for the benefit of others who may be better able to help.)

 

[Qube]

Just really want to know.
It's the former. I know the enthalpies but why? Prof didn't give satisfying explanation.