How do you know if a bond according to VB theory consists of 2 "3 electron bonds" or one "fully paired" bond?

Question

First of all, I want to discuss the whole thing strictly according to the VB - theory and not use any concepts of the MO - theory.

According to Linus Pauling, the "3 electron Bond" is the resonance of one electron between 2 half filled orbitals:

A| •B <-> A• |B

I use his representation: "A•••B".

I applied that Idea to Triplett Oxygen, which has one sigma bond and 2 x ½ π Bonds or "2 x 3 electron bonds": "O÷÷÷O" so it forms one sigma Bond between the two half filled 2py orbitals and two ½ π Bonds between

1. one fully occupied 2px orbital and one half filled 2pz orbital.
2. one half filled 2pz orbital and one fully occupied 2px orbital

Then I moved on to Peroxid with O2--.

I now have two options:

1. Starting with the following electron configuration: O: 2s² 2px² 2py¹ 2pz¹, O--: 2s² 2px² 2py² 2pz²

That must lead to two "3 electron bonds": "O:::O".

2. Starting with the following electron configuration: O-: 2s² 2px² 2py² 2pz¹, O-: 2s² 2px² 2py² 2pz¹

That must lead to one sigma bond: "O-O".

I know that it can't be option one because Peroxide is diamagnetic and probably more energetically favourable to form a single sigma bond than two ½ π Bonds, but is there another explanation for that?

I also tried to apply that on Dioxygenyl, and endet up with one sigma bond, one π bond and one ½ π Bond: "O=•••O":

But then I didn't have to face this question because I could only start with O and O+ or O+ and O.

Answer 1

You can't, really.

Trihalide ions, $\ce{X3^-}$, are known for all halogens with stable isotopes ($\ce{F,Cl,Br,I}$). Conventionally they are regarded as having a two-elecron bond delocalized between two linkages, to wit:

$\ce{X-X\space\space :X^-\space\space <->\space\space X:^-\space\space X-X}$

But this apparently applies only for $\ce{Cl, Br, I}$. Braida and Hubertus[1] report that trifluoride ion, with the same formal valence structure as the others, has a stronger contribution from a delocalized three-electron bonding:

$\ce{F\therefore F^-\space\space \cdot F\space\space <->\space\space F\cdot\space\space F\therefore F^-}$

with the negative charge more delocalized here than in the "standard" structure that seems to apply to the heavier trihalide ions.

Both structures correspond to the same valence-bond arrangementcand thus simple valence bond theory cannot distinguish them. It's one area where molecular orbital theory is superior.

Reference

1. Benoit Braida and Phillipe C. Hubertus(2004). "What makes the F3- ion so special?" A breathing-orbital valence-bond ab initio study. _Journal of the American Chemical Society, 126(45), 14890-14898. https://doi.org/10.1021/ja046443a.