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Why does molecular orbital theory (MOT) imply that molecular helium does not exist? All answers I found in the web use following two standard arguments, but I not see why these are sufficient to answer my concern (see below for more details):

  • the energies of electrons in bounding MO and antibounding MO neutralize each other

and/or

  • the bond order (half the difference in the numbers of bonding and antibonding electrons) is zero

But why do these two observations imply that molecular helium not exist? Consider the energy diagram

enter image description here

By ? (certain universal priciple, I forgot it's name; somebody knows?) the system tends to stay in a state with lowest energy. In this example the electrons can have three energy states: the $1s$-energy $E_{1s}$, the bounding MO energy $E_b$ and the antibounding MO energy $E_{ab}$. We have $E_{ab} > E_{1s} > E_b$.

Naively I wound say that the electrons prefer the state where their common energy is minimal. So my question reduces to:

Why $2E_{ab} +2E_b > 4E_{1s}$

More generally I would like to understand: Why bond order zero (same number of electons in bounding und antibounding MO's) implies that electons prefer to stay in thier atomic orbits instead of the MO's? Is there also any energetic reason behind?

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  • $\begingroup$ The # of molecular orbitals is equal to the # of atomic orbitals of the individual atoms. $\endgroup$
    – Jun Seo-He
    Feb 17, 2022 at 0:51
  • $\begingroup$ That's the naive consequence. But reality shows that the molecular helium configuration not exist. And my question is why MOT implies this. $\endgroup$
    – user267839
    Feb 17, 2022 at 15:08

2 Answers 2

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Yes, $E_{ab} - E_{1s} > E_{1s} - E_{b}$, that is, the energy difference between the anti-bonding molecular orbital and the individual atomic orbital is larger than the energy difference between the bonding molecular orbital and the individual atomic orbital.

The stabilization energy of the bonding molecular orbital, $\Delta E_+ = E_{1s} - E_{b}$, and the destabilization energy of the anti-bonding molecular orbital, $\Delta E_- = E_{ab} - E_{1s}$ are given by

$$\Delta E_{\pm} = \frac{e^2}{4 \pi \epsilon_0 R} + \frac{J \pm K}{1 \pm S},$$

where $J$ is the Coulomb integral (always negative), $K$ is the exchange integral (also always negative), and $S$ is overlap integral (a value between $0$ and $1$).

So, $\Delta E_- - \Delta E_+$ is always a positive value:

$$\Delta E_- - \Delta E_+ = \frac{|J| + |K|}{1+S} - \frac{|K|-|J|}{1-S},$$

since $|J| + |K|$ is always greater than $|J| - |K|$.

Check here, and here. This results in a bond energy of $4.6 \times 10^{-5} \, \textrm{kJ}/\textrm{mol}$, far smaller than that of H$_2$, $436 \, \textrm{kJ}/\textrm{mol}$, and a bond length of $5500 \, \textrm{pm}$, far larger than that of H$_2$, $74 \, \textrm{pm}$ (see table in the reference).

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  • $\begingroup$ yes, a result of that kind I was looking for, thank you! a nitpick: in the other answer Jun Seo-He argued that that's the consequence of the observation that the 'overall energy' of the He molecule is bigger than the combined energy of the individual He atoms. I'm not 100% sure, is the 'overall energy' exactly the $2E_{ab} +2E_b$? $\endgroup$
    – user267839
    Feb 21, 2022 at 20:25
  • $\begingroup$ also, do you know what is the name of this 'fundamental priciple' involved crucialy here that the system always prefers to stay in the configuration state of lowest energy? $\endgroup$
    – user267839
    Feb 21, 2022 at 20:28
  • $\begingroup$ Yes, that's the overall energy of Coulomb and exchange terms, but does not includes other stuff as center of mass motion, spin-orbit coupling, vibrational and rotational states, etc. And yes, He molecule has higher energy than the individual He atoms, as atomic orbitals lie below the arithmetic average of bonding and anti-bonding molecular ones. As to the principle of minimal energy, perhaps you would like to call it second law of thermodynamics?. $\endgroup$
    – Arc
    Feb 22, 2022 at 1:27
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The electrons in the antibonding molecular orbitals are very close to one of the nuclei so the repulsion between the nuclei becomes bigger than the attraction due to the electrons in the bonding molecular orbital.

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  • $\begingroup$ so you mean consequenty the difference $dE':= E_{ab}-E_{1s}$ is finally bigger than $dE:=E_{1s}-E_b$? $\endgroup$
    – user267839
    Feb 17, 2022 at 15:38
  • $\begingroup$ I say that the overall energy of the He molecule is bigger than the combined energy of the individual He atoms thats why He2 isnt formed ,the electrons in the molecular antibonding orbital raise the energy of the system more than the electrons in the bonding orbital reduce it. $\endgroup$
    – Jun Seo-He
    Feb 17, 2022 at 15:41
  • $\begingroup$ by 'overall energy' of He molecule you mean $2E_{ab} +2E_b$ ? $\endgroup$
    – user267839
    Feb 18, 2022 at 22:32

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