The key issue is not so much that they are in non-bonding orbitals, it is more that these non-bonding orbitals are singly-occupied (due to Hund’s rule).

This means that these compounds are very similar in behavior to free radicals – except because there are two free radicals, they are what are known as “diradicals”.

So imagine the instability of free radicals, times two.
(The barrier to recombination of free radicals is basically zero, so these types of compounds very readily combine with each other even if they manage to form – they also react rapidly with oxygen)

I also do not understand how I would know from the Frost diagram, why, if I take a 4n+2 system, and remove 2 electrons, the result would be the anti-aromatic system expected from a 4n compound. Because all filled orbitals are stabilising? I’m not sure how to see the aromatic stabilisation or lack thereof in this context.

The Frost circle is mostly useful for visualizing what the energy levels will look like.

Once you know that the energy levels look like, then you fill them up with the number of available pi electrons.

When you do this with cyclobutadiene you end up with one doubly-occupied orbital and two singly-occupied orbitals (unstable). But when you do this with benzene you end up with three doubly-filled orbitals (stable!)

Is this correct?

Fixed it, thank you!

Overall this is an excellent series of articles.

Very interesting! Also from the paper: “Comparison with all-trans octatetraene indicates that COT2− has a substantial aromatic stabilization energy (25 kcal mol−1) approaching that of benzene (33 kcal mol−1), but this favorable influence is swamped by Coulomb repulsion. “

The compound has been made as a disodium salt but the most recent theory is that it exists because it is stabilized by interactions with counterions: J. Chem. Theory Comput. 2013, 9, 4436−4443.

“Like many multiply charged anions, COT2− exists in isolation only as a short-lived resonance state lying above neutral COT.”
“….Charge-compensating complexation of COT2−with two sodium cations results in a thermodynamically stable Na2COT compound”

Cyclobutadienyl di-anion, yes, because it would have six pi electrons. Cyclobutadienyl anion, no, because it can only be formed from deprotonation of C-H and the negative charge (and ensuing lone pair) would be at right angles to the pi system.