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I'm currently preparing a few slides for an upcoming talk in our group. I would like to mention some of the very basics regarding crystal field and ligand field theory as well, though this will not be the main topic of the talk.

I started from allowed transitions and how the (transition) dipole moment operator causes the transition between two states to be allowed or not (from just pointing out the area under an odd and even function and from group theory basics).

Then I'd like to continue to octahedral complexes. But here you've got two problems, one is the d-d transition that is forbidden and the other one is the Laporte forbidden transititon. When I need to compare this to the next slide with tetrahedral ligand fields I can always say that the Laporte rule doesn't work here since there is no inversion center.

So how is the Laporte rule still achieved for octahedral complexes? Due to something we call vibronic coupling. I always assumed this meant that a vibration, prior to the exitation deforms the octahedron to something that has less symmetry (which is probably happening, too) until I recently heard about transforming your set of p-orbital originating $T_\mathrm{1u}$ along the vibrational mode T1u that is also the dipole moment operator in this case, as x,y,z transforms as $T_\mathrm{1u}$ in this point group. Transforming $T_\mathrm{1u}$ along $T_\mathrm{1u}$ in an octahedral complex gives you, among others also a $T_\mathrm{2g}$, so the $T_\mathrm{2g}$ for the transition can also be described in terms of p-Orbital and thus you get p-orbital contribution. (Please correct me if I've understood anything wrong here).

Now this is a good explanation, what it doesn't explain however is how actual d-d transitions are allowed now? This means there is either another thing happening here or that there is also a vibronic coupling for the tetrahedral case.

Now, before I read into those group theory explanations I assumed that the d-d transitions happens because we are not talking about actual d-orbitals but we are talking about electrons in bonds. As the $T_\mathrm{2g}$ requires a π-type bonding and the $E_\mathrm{g}$ uses σ-bonding the actual transition is from a π-orbital into a σ-orbital, so p-s or if you take both sides pd-ps.

The other alternative would be that also the tetrahedral complex uses vibronic coupling. I can't find much on this though. There are calculations in combination with Jahn-Teller effects but no clear answer as with the octahedral case.

Hence my question: When talking about d-d forbidden transitions without taking the Laporte rule for odd even parity into account, is the often cited mixing of d-states with p- and s-states from something coming from the bonding situation or is the prohibition lifted due to the same vibronic coupling mechanism as used for partiy forbidden transitions?

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    $\begingroup$ For any transition to be allowed the transition moment integral $\int\psi_i\mu\psi_f \ne 0$ . In symmetry terms this means that $\Gamma_i\Gamma_\mu\Gamma_f$ contains the totally symmetric representation. Thus a vibration can distort, say $\psi_i$ to make this allowed, i.e. $\Gamma_i\Gamma_\mu\Gamma_f\Gamma_{vib}$ contains $A_1,\;A_{1g}$ etc, where $\Gamma$ means symmetry species. $\endgroup$
    – porphyrin
    Commented Dec 7, 2020 at 13:28
  • $\begingroup$ First of all thank you, I do understand that principle but I was curious if this was also the explanation for all d-d transitions in general or if, leaving group theory aside, the bonding situation in, say an octahedral complex, also adds this effect of what we call p-d and p-s mixing for the transition. Because we can still see that octahedral complexes show weaker absorptions due to parity. $\endgroup$
    – Justanotherchemist
    Commented Dec 7, 2020 at 14:44
  • $\begingroup$ I'm not sure I quite understand, but if you mix p-s or dp-ps or whatever you will change the shape of the wavefunction and this may make the integral non zero.In this case a vibration may not be needed. The bigger the overlap $\psi_i\mu\psi_f$ the more intense the transition will be, obviously. $\endgroup$
    – porphyrin
    Commented Dec 7, 2020 at 15:14
  • $\begingroup$ Yes, your mathematical and group theoretical approach proves it of course and I do understand the point of the irred. repr. in the end. My problem is I need something to destinguish the absorbance in tetrahedral vs. in octahedral complexes and something to lift the d-d problem as well. The former one is as you say vibronic coupling and can easily be shown using group theory. But I'm struggling a bit for the latter one. Like does this pd-ps mixing really happen through bonding situation or do we also consider mixing due to vibronic coupling when we refer to tet. complexes in this case? $\endgroup$
    – Justanotherchemist
    Commented Dec 7, 2020 at 16:27
  • $\begingroup$ Like literature will tell you d-d transitions are forbidden (ignoring parity here now). But, they happen due to p and s orbital mixing. Now if you search for this mixing you will find examples of vibronic coupling and parity included. So is there ever any other mechanism besides vibr. coupl. or Jahn-Teller involved here or are dd and parity rules lifted by the same mechanism? $\endgroup$
    – Justanotherchemist
    Commented Dec 7, 2020 at 16:30

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