Why don't unstable odd electron species dimerize so that they become stable?

Question

Why doesn't unstable odd electron compound $\ce{NO}$ dimerize to $\ce{N2O2}$? Why doesn't this structure of ozonide dimerize? But why then does $\ce{BH3}$ dimerizes to $\ce{B2H6}$, and $\ce{AlCl3}$ to $\ce{Al2Cl6}$, both with something like bridged bonds?

Answer 1

Odd electron species do dimerize.

$\ce{NO}$

Nitric oxide dimerizes, but only at low temperature (and probably high pressure}.

$\ce{NO2}$

Nitrogen dioxide does dimerize. In fact, this is a well known property of $\ce{NO2}$. $\ce{NO2}$ (orange-brown) is in equilibrium with $\ce{N2O4}$ (colorless).

$$\ce{2NO2 <=> N2O4}$$ The position of this equilibrium is dependent on temperature and pressure (like all gas equilibria), and so it makes a nice demonstration of gas properties.

$\ce{O3}$

Ozone does not dimerize. Dimerization would lead to $\ce{O6}$, which likely is a 6-membered ring of all oxygen atoms. While this structure violates no rules, the oxygen-oxygen single bond is not stable. This instability of the peroxide bond makes it both useful and dangerous.

So why does $\ce{NO2}$ dimerize readily, $\ce{NO}$ dimerize under duress, and $\ce{O3}$ not dimerize at all?

The extra electron in the nitrogen oxides is found in an antibonding orbital. Conversion to the dimer gets that electron into a bonding orbital. There are no antibonding electrons in $\ce{O3}$.

$\ce{BH3}$

Borane dimerizes because the boron atom is electron deficient - it has an empty p orbital. This species is so electron deficient that it will form bonding interactions with nearly any electron pairs, including those already in sigma bonds. (More properly the B-H-B bonds are three-center-two-electron bonds). Aluminum compounds dimerize the same way.

So why does $\ce{BH3}$ dimerize and $\ce{O3}$ does not?

Ozone has no electron deficient atoms. Count the electrons. All atoms have an octet, and while one has a positive formal charge, the compound as a whole is not electron deficient.