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Electron Configuration of Platinum

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Why is the electron configuration of neutral Platinum:

    [itex]Pt = [Xe]4f ^{14} 5d ^{9} 6s ^{1}[/itex]

    and not?:

    [itex]Pt = [Xe]4f ^{14} 5d ^{10} 6s ^{0}[/itex]

    I thought that if the two electrons from the 6s sub-shell were to move to the 5d sub-shell, then 5d sub-shell would be fully paired, hence stability? However, this is not the case as the actual configuration calls for only one electron to be moved from the 6s sub-shell, rather than two. Can someone explain to me why this is the case?
     
  2. jcsd
  3. Sep 29, 2017 #2

    TeethWhitener

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    Science Advisor
    Gold Member

    This is a pretty good resource on advanced concepts regarding electron configurations in atoms:
    http://pubs.acs.org/doi/full/10.1021/ed8001286?src=recsys
    In case it's paywalled, the main point is that electron-electron interactions matter. In the case of platinum specifically, there is quite a lot going on. First, the 6s and 5d orbitals are very close in energy in the platinum system. In addition, there is a (generic) difference between electron repulsion in the d orbitals and in the s orbitals. S orbitals tend to be more diffuse and d orbitals tend to be more compact. Therefore, electron repulsion tends to drive d-orbital filling to higher energies than s-orbital filling. There are also spin-orbit interactions to consider (spin-orbit terms are zero in the s-orbital case but nonzero in the d-orbital case). These all combine and balance out in a certain way to give you the experimentally observed electron configuration. In the case of platinum, this means roughly that the repulsion felt by adding the final electron to the d-subshell is larger than the energy gained by completing that subshell, so the electron ends up in the s-orbital.
     
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