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

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


[![enter image description here](https://i.stack.imgur.com/7PQki.jpg)](https://i.stack.imgur.com/7PQki.jpg)


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":


[![enter image description here](https://i.stack.imgur.com/dsRnq.jpg)](https://i.stack.imgur.com/dsRnq.jpg)


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




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]](https://doi.org/10.1021/ja046443a) 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>.


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