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Spectrochemical series explanation

  1. Oct 31, 2013 #1
    After learning about crystal field theory, my main question was due to the arrangement of the molecules in the spectrochemical series. I can understand that the larger anions are placed towards the lower end since they do not cause high splitting in the d orbitals, but some of the arrangements of the molecules confuse me. My reasoning is largely due to the size of the molecules. For ions like I-, they have a large ionic radius and cannot pair well with the central metal well. This results in a low Δo. I can also see that CO is smaller and since it is not charged, the lone pair electrons on the oxygen atom can yield high splitting without being repelled by the outer electrons too much. My primary question is why OH- is not towards the end with higher Δo. Any clarification or corrections to what I have posted would be greatly appreciated. Also, any links or another explanation for the arrangement of the molecules on the spectrochemical series would be amazing!

    Thanks!
     
  2. jcsd
  3. Dec 14, 2013 #2
    Any help on how to distinguish b/w strong and weak field splitting ligands w/o the aid of a spectrochemical series?
     
  4. Jan 30, 2016 #3
    If splitting of d orbitals resulted simply from the effect of point charges (ions or dipoles), one should expect that anionic ligands would exert the greatest effect. To the contrary, most anionic ligands lie at the low end of the spectrochemical series. Furthermore, OH- lie below neutral H2O molecule and NH3 produces a greater splitting than H2O although the dipole moments are in reverse order.
    This is due to the assumption of purely electrostatic interactions between ligands and central metal ions. In fact, covalent interactions between metal ions and ligands have to be taken into account.
     
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