http://chemwiki.ucdavis.edu/Theoret...rbital_Theory/How_to_Build_Molecular_Orbitals Hi, I'm currently trying to understand MO theory and how diagrams are made and interpreted. I stumbled across this website and it shows the general MO diagrams for different diatomic molecules. It also shows the 2_{p∏} of B_{2}, C_{2} and N_{2} as being higher in energy than compared to O_{2} and F_{2}. I was trying to reason it out and considered that since there will be more ∏ antibonding orbitals for the O_{2} and F_{2} that this is the reason why. But aren't their structural constraints to this reasoning? Aren't ∏ orbitals of a specific energy due to their perpendicular orientation and their sideways overlapping? Why exactly is their energy lower in the MO diagrams? I'm assuming my explanation for this is incorrect so so any clarification on the above matters would be great! Thank you! :)
These pictures are only correct as far as the ordering of the sigma and pi orbitals are concerned however, the absolute energies of these orbitals is probably incorrect. The point is that the s-p splitting increases in a period (that is from B to F, in our case). The anti-bonding sigma formed from the s-orbitals and the bonding sigma from the p orbitals repell, that is why the bonding p-sigma is shifted above the bonding pi orbitals in case of B, C and N. In O2 and F2, the orbitals are too far appart, so that this repulsion won't change the order of the sigma and pi orbitals derived from p.