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Why do transition metal ions lose s electrons first?

  1. Dec 1, 2011 #1
    When transition metal start losing electrons they lose them from the s orbital before the d orbital. Why is this?

    The iron(II) ion has 24 electrons in this configuration:
    [Ar] 3d6
    The neutral chromium atom also has 24 electrons, but in this configuration:
    [Ar] 3d5 4s1

    I understand that empty, half, and full shells are preferred, but I don't under stand why two atoms with the same number of electrons would have different configurations. I would assume it must be due to the different number of protons, but I don't understand what the reason is.

    This also would seem to lead to the question of which orbital is higher energy for excited states? If 3d is higher energy than 4s then the chromium atom in an excited state would have its 4s electron jump up to 3d, and then match the iron ion. However, if 4s is higher than 3d I would also expect the iron ion to end up matching the chromium atom.

    I'm sure this all has perfectly logical explanations, but I evidently did not pick it up in my chemistry class. I'd appreciate it if anyone could explain this, or point me to a link where it is explained well.
  2. jcsd
  3. Dec 1, 2011 #2


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    Science Advisor

    Maybe you remember that in hydrogen all orbitals with the same principal quantum number are degenerate. In neutral atoms with more electrons this is no longer the case, as e.g. electrons in s orbitals are on the mean nearer to the nucleus than in d-orbitals and therefore see a higher effective charge on the mean, which means that they are energetically lower than d orbitals.
    In the transition metals this goes so far as 3d becoming energetically a tic higher than 4s.
    However when you consider positive ions, the electrons in the valence s and d orbitals see both a higher effective nuclear charge as the screening by the other electrons is reduced due to some electrons missing. Hence the orbitals in ions are more "hydrogen like" than in neutral atoms. As a consequence also 3d is lower in energy than 4s in the ion.
  4. Dec 1, 2011 #3
    Its about orbital geometry. When the ligands move in the d orbitals can squeeze in between them to get away from the charge they carry with them. The s orbital on the other hand is spherical so it can't go anywhere. This causes the s orbital to become higher in energy than the d orbitals. Thats my understanding of it anyway.
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