Why is the ABMO more destabilized then BMO is stabilized

  • Thread starter Thread starter hivesaeed4
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
hivesaeed4
Messages
217
Reaction score
0
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.
 
Chemistry news on Phys.org
That's mainly due to the normalization of the molecular orbitals.
The bonding/non-bonding orbitals are normalized as [itex]\psi_\pm=\frac{1}{\sqrt{2}(1\pm S)}(\phi_A\pm \phi_B)[/itex] where [itex]S=\langle B|A \rangle[/itex] where the plus sign refers to bonding and the minus sign to the anti-bonding orbital.
With the exchange energy [itex]H_{AB}=\langle B|H|A\rangle[/itex] the bonding energies are approximately [itex]E_\pm\approx \pm H_{AB}/(1\pm S)[/itex]. Now with 1-S being much smaller than 1+S, the anti-bond is more destabilzed than the bond is stabilized.