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Pairing Energy/LFSE (Inorganic)

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the LFSE for the following octahedral ions from first principles.

    d) Low Spin d6
    e) d9

    2. Relevant equations

    3. The attempt at a solution
    d) t2g6eg0

    e) t2g6eg3

    Actual solutions:
    d) 2.4[itex]\Delta[/itex]o-2P
    "However, since one pair of electrons would be paired in a spherical field the additional pairing energy is 2P."
    e) 0.6[itex]\Delta[/itex]o
    "However, since all four pairs of electrons would be paired in a spherical field there is no additional pairing energy."

    I'm really confused about the treatment of pairing energy. I'd really appreciate if someone could explain when I should/should not include pairing energy. Initially I thought any electrons that moved up and paired in eg would cause the addition of pairing energy but it seems it isn't so. I don't really understand what it means by "spherical field".

    Thank you!
  2. jcsd
  3. Feb 28, 2012 #2
    I realize that this question was asked a long time ago, but I figured I would answer to help out future readers.

    When calculating the Ligand Field Stabilization Energy you have to take into account the stability you gain from orbitals decreasing in energy and the amount of energy it takes to put two electrons in the same orbital (pairing energy)

    the LFSE gains 0.4Δo for every electron in a t2g orbital and loses 0.6Δ for every electron in an eg orbital. If you fill all orbitals (d^10) then you get no change in energy (zero stabilization). If you fill all but one electron in a eg orbital (d^9) then you gain 0.6Δo of stability (which would mean that the total energy would drop 0.6Δo)

    Now, as for pairing... it takes energy to put an electron into a spatial orbital that already has an electron. This is represented by the pairing energy P. Consider a d^6 complex. In its free ion state (when there are no ligands and all d orbitals are degenerate) it has two electrons in one orbital and one electron in each of the other four. If you bind it to several ligands (while keeping the same ionization state) you end up with three t2g orbital and two eg orbitals. If P is high it is energetically favorable to put electrons into the eg orbitals before you double up in the t2g orbital. The d^6 would have only two paired electrons, which is the same amount as it had in the free ion state, so the energetic contribution of the pairing energy to the LFSE is 0.

    If P is low then you double up the t2g orbitals first, which would mean that you would have three pairs of paired electrons: two pairs more than in free ion state. This destabilizes your structure, so you would have to subtract 2P from your LFSE.
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