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Oxidation (Charge) State and Screening

  1. Jul 22, 2013 #1
    Why exactly does an increase in the oxidation state of an element (i.e., higher positive states) reduce the screening of core electrons? For example, why would fewer electrons in a d state alter (increase) the effective charge experienced by a p state?
     
  2. jcsd
  3. Jul 24, 2013 #2
    citw,

    Your question as it stands is incomplete. Maybe you could expand it, and indicate the context of the screening phenomenon that you are interested in (e.g. XPS spectra, interatomic potentials, electrostatics or whatever).
     
  4. Jul 25, 2013 #3
    I'm not entirely sure how to expand upon it, other than that I'm interested for XPS. What I've read in texts is that chemical shifts change the effective charge of core states. Specifically, if the oxidation state is increased (more positive), that core electrons experience reduced screening. I don't understand the effect of valence electrons on core electrons, with respect to screening. This is a general phenomenon, but I don't have an explanation for it.

    If you look at p. 25 here: http://mmrc.caltech.edu/SS_XPS/XPS_PPT/XPS_Slides.pdf, they mention that "the electrostatic shielding of the nuclear charge from all other electrons in the atom (including valence electrons)." I don't understand why this is the case. I can understand why core electrons shield valence electrons, but not vice-versa.
     
  5. Jul 25, 2013 #4

    DrDu

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    The orbitals of the valence electrons have always also smaller maxima of electron density inside the region where the core electrons are located. So reducing their occupancy will result in a lowering of screening.
     
  6. Jul 25, 2013 #5
    Bancroft and McIntyre are two of Canada's most experienced surface scientists, but nevertheless the reference you give is highly misleading in places, and I think they would cringe to read it.

    Page 28 starts off by saying that the chemical shift is "usually thought of as an initial state effect (i.e. relaxation processes are similar magnitude in all cases)". Nothing could be further from the truth. The energetics of photoemission depend equally on the initial and final (core-ionized) states, and the observed chemical shift is the sum of energy changes in both states. This principle is absolutely fundamental in XPS, and has been firmly established since 1980 or so. Nevertheless, much of the published literature fails to understand or acknowledge it.

    The same page states that spin-orbit splitting is "largely an initial state effect". In fact, it is purely a final state effect (i.e. the energy difference between two final state spin orientations).

    With respect to the (meaningless) statement about the "electrostatic shielding of the nuclear charge" - your instinct is basically correct. This can be demonstrated very easily and elegantly for metals using the so-called (Z+1) approximation. In this model, binding energy shifts can be calculated to a precision of about 0.1 eV by replacing the final state with a Z+1 atom (this mimics the effect of the core hole on the valence electrons). The point to note is that the model only calculates contributions to the chemical shift that arise from changes in the valence electron states. Its success shows that these are the dominant contribution.

    In summary, core-level shifts in XPS must be rationalised in terms of total energy changes between the initial and final states.

    Refs. for further reading.

    B. Johansson. N. Martensson: Phys. Rev. B 21 (1980) 4427.

    D. Tomanek, Surf. Sci. 126 (1983) 112.
     
  7. Jul 25, 2013 #6
    DrDu makes an interesting point about the valence electron density... does this make sense with what you're trying to explain?
     
  8. Jul 25, 2013 #7
    I am referring only to the chemical shift, not to electronic properties of the initial state alone. If, for example, you postulate or learn that there is some change in the valence states due to chemical bonding (quite possible), you only have half the information you need. The chemical shift can go in either direction, depending on how the same effect (e.g. changes in electron density) manifests itself in the final state.

    The XPS experiment is symmetric between the initial and final states, so any explanation of chemical shifts (e.g. electron density modifications) must consider how both states are affected.
     
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