On non-existence of molecular helium (and or why that's equivalent to bond order zero) 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):

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*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

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?
 A: 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.
A: 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).
