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Molecular Oribtal theory, anyone?

  1. Sep 24, 2004 #1
    I figured that since the bonding of elements is involved, this should be regarded as chemistry...but honestly, I am not sure which forum this should go.

    Anyway, my question is simple: Can anyone eplain the bonding and anti-bonding in MO theory to me? My textbook mentioned that the bonding depended on the symmetry of the electron orbitals and such...but that doesn't really say much.

    Anyone here to offer some help please?
     
  2. jcsd
  3. Sep 25, 2004 #2

    chem_tr

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    Hello,

    I'll try to be as simple as possible, since MO theory has lost much in terms of simplicity. Barnard Shaw has said, "Real, but not beautiful; beautiful, but not real"; this sentence can be adapted to MO theory compared with crystal field theory.

    Now, for simple biatomic species of second period (2p electrons are present), MO theory is still easy to understand. Please view the attachments to see the molecular orbital diagrams (simplificated) for N2, and any other heterodinuclear molecule.

    Here, for symmetric molecule, you'll see that only atomic orbitals of similar symmetry will couple with each other to make a molecular orbital. There are 2s and 2p orbitals, and two hybrid orbitals with different s/p ratios are produced, designated as "h1" and "h2". Let's say, h1 orbital has 70%s/30%p; and the reverse for h2, of course.

    For energetic reasons, h1 sees h1 of the other atomic orbital; and pi orbitals do the same. But there is another factor; only symmetric ones can couple with the other, so pi orbitals only couple with the other pi orbital, and sp hybrid orbitals couple another sp hybrid orbital.

    Now let me discuss bonding and anti-bonding concepts. If a molecular orbital occurs with involvements of two atomic orbitals, then there must be a anti-bonding orbital, because every orbital carries two electrons; and two orbitals combine. The remaining two electrons should indicate another orbital, i.e., anti-bonding. But anti-bonding orbitals are of very high energy, so these two electrons go somewhere else (of similar symmetry); according to the well-known electronic configuration rules.

    If you're stuck somewhere, please let me know, and we'll try to solve it.

    Regards, chem_tr
     
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