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QM- Tiny question on sigma & pi bonding states

  1. Nov 27, 2005 #1
    Could somebody please explain the 2-fold degenerate sigma 2p is lower in energy than that of 4-fold degenerate the pi 2p bondings state in the O2 molecule? Does it have anything to do with resonance energies perhaps?
     
  2. jcsd
  3. Nov 27, 2005 #2

    Gokul43201

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    A simple way to think about this is to keep in mind that it is wavefunction overlap that causes a reduction or increase in orbital energy. To have a large change in energy (with respect to the parent atomic orbital) you must have large spatial overlap. Besides the extent of spatial overlap, there is the matter of whether the wavefunctions add or subtract, to form the corresponding molecular orbitals. In the first case, there is a decrease in energy, and in the second, an increase.

    Now, we know that the [itex]\sigma _z [/itex] orbitals form by overlap of the [itex]2p_z[/itex] orbitals in the internuclear direction. Clearly, the wavefunction overlap in this case is much greater than with the formation of [itex]\pi_x[/itex] or [itex]\pi_y[/itex], where the overlap is only very small. Hence, the changes in energy of the [itex]\sigma _z [/itex] orbitals with respect to the 2p energy will be much greater. As a result, the bonding [itex]\sigma _z [/itex] has a lower energy than the bonding [itex]\pi_x[/itex] or [itex]\pi_y[/itex], while the antibonding [itex]\sigma^* _z [/itex] has a greater energy than the antibonding [itex]\pi^*_x[/itex] or [itex]\pi^*_y[/itex].

    The moral : more spatial overlap, more change in energy.

    (In terms of stationary perturbation theory you can think of as resulting from larger off-diagonal terms in the hamiltonian)
     
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