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In a hypothetical (?) Og2+235 we would have a simple sigma bonding orbital occupied by one electron (leading to a bond order of 1/2). But how to take into account the giant repulsion of the two nuclei? Generally how are nucleis taken into account for energy calculations from molecular orbitals in MO theory? So far I've always read about the electrons but very few about the nuclei in the system.

Same in He23+, half bond order but very large repulsion. My intuition says that He21+ should be more stable, even if both have the same bond order.

Where in the framework of MO theory is exactly that taken into account? How does nuclei repulsion change the MO energy scheme?

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    Oh, no, not again ;> For diatomics you get +2, tops. Some absurd charge is out of question. – Mithoron Nov 29 '22 at 21:55
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    Apply the common sense of classical electrostatics and you would see that the monotonous steep slope of potential energy of nuclei cannot be disturbed by very tiny bump caused by interaction of electrons. – Poutnik Nov 29 '22 at 21:55
  • https://chemistry.stackexchange.com/questions/106826/does-the-hydrohelium-dication-exist – Mithoron Nov 29 '22 at 21:58
  • @Poutnik And why is it then so often taught that bonds are formed when one simply takes mo theory into consideration and only looks at electrons? Nuclei repulsion is hardly mentioned in mo theory –  Nov 29 '22 at 22:00
  • Well, gosh, it's not like you need to consider dications all that often. Even multiatomic break up pretty fast. – Mithoron Nov 29 '22 at 22:11
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    @FlawC https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation - calculating MOs first involves the assumption that that the nuclei are stationary (have fixed positions) and then optimising the electron wavefunction. The nuclear repulsion term is not ignored in the overall energy, but it is just a constant (for a given set of nuclear positions). – orthocresol Nov 29 '22 at 22:11
  • https://chemistry.stackexchange.com/questions/146481/why-is-oxygen-more-stable-than-oxygen-dication https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%e2%82%84o%c2%b2%e2%81%ba-exist – Mithoron Nov 29 '22 at 22:17
  • @orthocresol ok so considering a system just with the "Og2 +235", at what distance apart would they be stationary and form a bond? I mean what would be the hypothetical bond length of such a compound? –  Nov 30 '22 at 23:24
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    @FlawC Short answer: never. Long answer: what you need to do is to plot the energy of such a compound as a function of bond length. (You can google 'hydrogen bond energy curve' for an example of what this looks like for H2.) Now, your curve for $\ce{Og2^{235+}}$ would be so biased towards large distances, and the 'dip' in the curve would be so small, that it can hardly be considered a molecule. To be technical, it wouldn't admit a vibrational state which is required for something to be considered a molecule. https://goldbook.iupac.org/terms/view/M04002 So there is no bond at any length. – orthocresol Dec 01 '22 at 01:48
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    That said, the original question isn't silly at all (although I guess you did choose a rather extreme example). I would love to write a proper answer, but it's very late where I am now. Feel free to ping me again tomorrow and maybe I can get round to it. – orthocresol Dec 01 '22 at 01:50
  • @orthocresol that would be great :-) I've played a bit with a electrostatic system, two positive charges (p) x apart and one negative charge (e) x/2 apart, basically I've found out that the term: -pe/(x/2)^2 + p^2/x^2 describing the electrostatic force on one (and therefore the other) nuclei becomes zero at a "e to p ratio" of 1:4, so relative any more proton in the nucleis than 4*N(e) leads to a net repulsion at any distance. But of course we deal with quantum mechanics and I can't assume a electron being in the middle. I don't know enough about the quantum mechanical "forces" :-( –  Dec 01 '22 at 05:47
  • @orthocresol But I assume at a far enough distance apart (still repulsive, but it approaches the x axis asymptotic) the repulsion get overcompensatet by a type of quantum mechanical force which comes from the fact that the system would probably have an even lower energy than if the nuclei went to infinity apart and no longer felt any repulsion at all? (So can those electron interactions at some very very large distance overcome electrostatic forces in my example?) –  Dec 01 '22 at 11:02
  • @orthocresol OK so basically the only force here is actually electrostatic, then the only way I can explain it is that the atoms stay together when they are far apart and the fact that the electron interacts by staying in the area between the two nuclei, the moment where the electron would be @ x/2 doesn't contribute to the cohesion, only those where it's much closer to one or the other nuclei. Like a child that tries to hold two trucks together that rolling apart, but does not have the strength to hold them together in the middle, it always runs from left to right and pulls one back again? –  Dec 01 '22 at 11:16
  • I'm afraid that here you've gone off on a tangent which I don't really understand any more. There are no special quantum forces, it's all electrostatics. The problem is that the electrons are not point charges. They're charge distributions. So it's rather more involved than just doing Coulomb's law. – orthocresol Dec 01 '22 at 13:20
  • @orthocresol That's what I thought while writing, but unfortunately I know absolutely nothing about the nature of charge distributions, hence the analogy with the child. But if you have time to write an answer that would make me happy :) –  Dec 01 '22 at 15:53

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I would not bet on it. We apparently can't even get $\ce{He2^{3+}}$, let alone higher single-electron diatomic ions. From Wikipedia:

$\ce{He2^+}$ was predicted to exist by Linus Pauling in 1933. It was discovered when doing mass spectroscopy on ionised helium. The dihelium cation is formed by an ionised helium atom combining with a helium atom: $\ce{He^+ + He -> He2^+}$.[1]

The diionised dihelium $\ce{He2^{2+}} (1Σ_g^+)$ is in a singlet state. It breaks up $\ce{He2^{2+}->2 He^+}$ releasing 200 kcal/mol of energy. It has a barrier to decomposition of 35 kcal/mol and a bond length of 0.70 Å.[1]

Thus even with only two positive charges the helium dimer is prone to breaking up due to electrostatic repulsion. Although the dication above is metastable, we should expect the stability to only become worse with an additional positive charge and only one bonding electron instead of two.

A simple classical electrostatic calculation is instructive. Suppose you have a negative charge and two positive charges, the latter equidistant from the negative charge on opposite sides. The net force is attractive if all charges are one unit, but this net attraction is lost if we give two or more (and certainly 118!) units to both positive charges. The true quantum mechanical calculation is of course much more complicated, but can give only less favorable results because as the positive nuclear charges approach the electron cloud cannot stay fully between them. Thereby $\ce{H2^+}$ is predicted to be the only single-electron homonuclear diatomic ion that remains bound.

Cited Reference

  1. Grandinetti, Felice (October 2004). "Helium chemistry: a survey of the role of the ionic species". International Journal of Mass Spectrometry. 237 (2–3): 243–267. Bibcode:2004IJMSp.237..243G. https://doi.org/10.1016/j.ijms.2004.07.012.
Oscar Lanzi
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