# Sigma and Pi Bonds for Diatomic Oxygen Molecule

1. ### Conservation

37
According to the Molecular orbital theory, diatomic oxygen should have σ2px (internuclear axis) and $\pi$2py and $\pi$2pz orbitals filled with two unpaired electrons, one at antibonding $\pi$2py and the other at antibonding $\pi$2pz. And of course, the 2s bonding and antibonding orbitals as well.
According to the molecular orbital theory, does this imply that diatomic oxygen possesses three "bonds" and one set of unpaired electrons, opposed to the double bond (sigma pi) implied by the valence bond theory?

Thank you.

Last edited: Mar 20, 2014
2. ### DrDu

4,420
The unpaired electrons are in fact antibonding, so that the bond order is 3-2x1/2=2.
In contrast to folk expositions of VB theory, VB theory predicts the same ground state as MO theory and not two double bonds, which rather describes bonding in the excited singulet oxygen.
This was shown already in 1937 by Wheland and Lennard-Jones:
http://pubs.rsc.org/en/content/articlelanding/1937/tf/tf9373301499#!divAbstract

3. ### Conservation

37
Right, hence the paramagnetic properties displayed in oxygen to point out the flaws of VB theory.

However, if the two unpaired electrons are antibonding, does that make the π2py and π2pz bonds "half" bonds? So one "full" sigma bonds at x and two "half" bonds at y and z planes to add up to two?

4. ### DrDu

4,420
As I tried to explain, VB gives a correct description of bonding in oxygen, so it is not flawed.
Furthermore, also the singlet state of oxygen is about as paramagnetic as the triplet state, due to orbital momentum, so oxygen being paramagnetic does not help to decide whether the ground state is singlet or triplet.

Yes, the pi bonds both have a bond order of 1/2.