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Molecular Orbitals

  1. Dec 12, 2013 #1
    http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Molecular_Orbital_Theory/How_to_Build_Molecular_Orbitals [Broken]


    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 2p∏ of B2, C2 and N2 as being higher in energy than compared to O2 and F2. I was trying to reason it out and considered that since there will be more ∏ antibonding orbitals for the O2 and F2 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! :)
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 12, 2013 #2


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    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.
  4. Dec 12, 2013 #3
    That was a great explanation. Thank you!
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