Why is the ABMO more destabilized then BMO is stabilized

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SUMMARY

The antibonding molecular orbital (ABMO) is more destabilized than the bonding molecular orbital (BMO) is stabilized due to the normalization of molecular orbitals when mixing atomic orbitals from different atoms. The bonding and non-bonding orbitals are expressed as \(\psi_\pm=\frac{1}{\sqrt{2}(1\pm S)}(\phi_A\pm \phi_B)\), where \(S=\langle B|A \rangle\). The bonding energies are approximated by \(E_\pm\approx \pm H_{AB}/(1\pm S)\). Since \(1-S\) is significantly smaller than \(1+S\), the destabilization of the antibonding orbital exceeds the stabilization of the bonding orbital.

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When we mix the atomic orbitals of two atoms, why is it that the antibonding molecular orbital is more destabilized than the bonding molecular orbital is stabilized. Note that the two atoms are different atoms.
 
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That's mainly due to the normalization of the molecular orbitals.
The bonding/non-bonding orbitals are normalized as \psi_\pm=\frac{1}{\sqrt{2}(1\pm S)}(\phi_A\pm \phi_B) where S=\langle B|A \rangle where the plus sign refers to bonding and the minus sign to the anti-bonding orbital.
With the exchange energy H_{AB}=\langle B|H|A\rangle the bonding energies are approximately E_\pm\approx \pm H_{AB}/(1\pm S). Now with 1-S being much smaller than 1+S, the anti-bond is more destabilzed than the bond is stabilized.
 

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