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Ligand field stabilization energies, is it correct to say

  1. May 22, 2012 #1
    the following?:

    low-spin d5 metal. Energy = -2Δ -4K (where K is exchange energy. i,e. minimization of energy due to parallel spins)

    low-spin d6 metal. Energy = -1.6Δ -3K

    Therefore: energy needed to add an electron = ΔE= 0.4Δ + K

    Is that correct?

    maybe I should be talking about Spin Pairing Energy (P) ....

    HEEEEEEEEEEEEEEEEEEEEEEEEEEEELP!!!!:cry:
     
  2. jcsd
  3. May 22, 2012 #2

    DrDu

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    The energy to add an extra electron (i.e. electron affinity) depends much more on other factors than ligand field stabilization. This should already be clear from the fact that electron affinity is not zero for atoms in the gas phase where there is no ligand field stabilization.
    It mainly depends on the increased intraatomic repulsion of the electrons.
     
  4. May 22, 2012 #3
    that would be the pairing energy right?
     
  5. May 23, 2012 #4

    DrDu

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    The pairing energy would also contribute, but this contribution would be rather small.
    Your question is not very specific. What is the concrete situation you have in mind? It would also be helpful to specify the coordination of the complex ...

    I would consider some Born cycle, i.e. desolvation of the complex ion, decomplexation of the ion with charge i+1, electron affinity of the ion, formation of the complex with charge i, solvation of the complex. Similar cycle for the ion or atom delivering the electron.
    The ligand field statilization only contribute to the second and before last step.
    The ligand field stabilization is only a relative stabilization of some set of d orbitals relative to others. Even in a complex where there is no ligand field effect, like going from s^0 to s^1, the energy of complex formation is generally non-zero, e.g. due to the increasing (or decreasing) electrostatic attraction between the central atom and the ligands.
     
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