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So I was recently discussing the transitions in Egyptian Blue ($\ce{CaCu[Si4O10]}$) with some of my students, who had to prepare this compound. What I like in particular in this case is how, at least in a simplified view you can show, that the blue is not simply due to one single transition with the complementary color to blue but actually composed of three possible transitions in the visible region that cause an absorption by all colors but blue. I also looked for a paper so they could give a reference in their report and this paper summarizes the transitions pretty well.

But when I thought about it for a while some questions came up in my mind on the actual transitions. So there are three transitions, either from the $\mathrm{a_{1g}}$, the set of $\mathrm{e_{g}}$, or the $\mathrm{b_{2g}}$ into the $\mathrm{b_{1g}}$. As the square plane has an inversion center we are dealing with the same problem as in octahedral geometries, the odd / even parity rule upon excitation.

Then I found this line in a text on symmetry rules in electronic transitions

In cases where transitions coincide with vibrations of the initial or final state, the electronic transition moment R needs to be replaced by the transition moment Rv. [...] Here Ψv denotes a vibronic wave function. In complete analogy with electronic transitions, we could derive the following set of selection rules for vibronic transitions: $$\Gamma(\psi_v') \otimes \Gamma(\psi_v'') = \Gamma(T_x)$$

At first I misread that line and thought that the dipole moment would change to vibrational modes as well, so I could choose from a larger set of possible symmetry elements to do the transitions, but then going through the examples at the bottom of the page it seems more like the vibration would cause a lower symmetry and change the point group into one where the transition may be allowed. It further says:

The total symmetry of a system can be expressed as a conjunction of the symmetry of the electronic states Γ(Ψ) and the symmetry of the vibration Γ(Ψv). $$\Gamma(\psi_{ev}) = \Gamma(\psi) \otimes \Gamma(\psi_v)$$

And this is the line that I don't understand anymore. So I looked at the examples And for the one with the point group $\mathrm{c_{2v}}$ for the transition between $\mathrm{b_{1}}$ and $\mathrm{b_{2}}$ it says:

The picture changes if we account for vibrational modes too. An asymmetric stretching causes the molecule to get from $\mathrm{c_{2v}}$ to $\mathrm{c_{s}}$ and only the molecular plane remains as symmetry element. State $\mathrm{B_{2}}$ becomes $\mathrm{A'}$ and state $\mathrm{B_{1}}$ $\mathrm{A''}$. Consequently, the transition dipole moment is of symmetry $\mathrm{A''}$ and perpendicular to the plane of the molecule. Obviously, reduced symmetry increases the number of possible transitions.

I also found a correlation table for the point group $\mathrm{c_{2v}}$ that links it to it's sub-groups. For $\mathrm{c_{2v}}$ the irreducible representation for the vibrational modes should give: $$\Gamma_{vib} = 2 A_1 + B_2$$

So for $\mathrm{B_{2}}$ we have this case, mentioned in the link, where we have to change to a vibrational mode. But how do I determine what sub-group is created by this vibration? In their example they showed that $\mathrm{B_{2}}$ becomes $\mathrm{A'}$. In the correlation table this would be $\mathrm{c_{s} (σ_{yz})}$. And if I then go back to the character table for $\mathrm{c_{2v}}$ the only thing that I could find would be that for this entry, $\mathrm{c_{s} (σ_{yz})}$, $\mathrm{B_{1}}$ would return -1 while $\mathrm{B_{2}}$ remains unchanged.

So does this give me any hint on how to transform my point group to its sub-group by a vibrational mode? As I have never had any lecture on point groups and symmetry I can unfortunately only look for tables, texts or similarities since I have no clue about point groups at all. This means the above-stated ideas could be terribly wrong. I tried to understand the text I quoted above as well as possible and this would be my interpretation.

This means that in my Egyptian Blue case, I would determine the symmetry element of the transition dipole moment. If the transition is not allowed I consider the vibrational normal modes and if either the initial and or final state has the same symmetry element, then I need to find out (using a correlation table) how the point group can be changed to a sub-group, where the transition would be allowed.

Therefore, I finally remain with the question: How do I determine which sub-group is chosen by the actional of a vibrational mode in this specific case?

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    $\begingroup$ @ Tyberius has give you the answer, but for more detail have a look at chapter 7 of 'Modern Spectroscopy' by Hollas, (4th ed, publ Wiley.), in particular 7.3.2 to 7.3.5. $\endgroup$ – porphyrin Jan 1 at 18:58
  • $\begingroup$ Thank you, I will try to get a copy of it when I return to university tomorrow. $\endgroup$ – Justanotherchemist Jan 1 at 19:18
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I don't believe it is strictly necessary to determine the subgroup induced by a vibration in order to determine if a vibronic transition is allowed. You can simply use the equation from your 2nd quoted block of text.

For example, consider whether an electronic transition would be allowed between a $b_1$ and $b_2$ state of water. Without including vibrational contributions, $\langle b_1|\overset{a_1}{\underset{b_2}{b_1}}|b_2\rangle\neq a_1$, so the transition would seemingly not occur. However, if we instead consider the vibronic state $b_1 \otimes B_2=a_2$ (formed from the $b_1$ electronic state and the $B_2$ asymmetric stretch), the transition can occur, as $\langle a_2|b_1|b_2\rangle= a_1$.

The $b_2$ state is also a vibronic state, in this case the original electronic state paired with a totally symmetric vibrational state. We could also obtain other transitions by considering the $b_2$ electronic state paired with other vibrational states.

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  • $\begingroup$ Thank so, so that was what they meant by multiplying it with the vibrational state. I didn't do it explicitly so I just assumed it would not change anything if both symmetry elements were the same, to begin with. Now that I see the product it makes sense to me now. Thank you! $\endgroup$ – Justanotherchemist Jan 1 at 19:16
  • $\begingroup$ @Justanotherchemist yeah some of the notation of point groups and direct products can be confusing if it isn't clearly explained. Glad I could help! $\endgroup$ – Tyberius Jan 1 at 19:44
  • $\begingroup$ Can I combine it with any possible vibrational mode or does it have to be the one that has the same irreducible representation as the initial or final state? $\endgroup$ – Justanotherchemist Jan 2 at 10:43
  • $\begingroup$ @Justanotherchemist you can combine any of the molecules vibrational modes, not just those that match the electronic state. $\endgroup$ – Tyberius Jan 5 at 23:25
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I don't know if it's okay to give an additional question like that but I feared that editing the original post would lead to confusions in the continuity of the answers and it's probably not advisable to make it an extra question.

So I tried it with combining only one state with a vibrational mode and it worked. But can I really treat it that way? What I'm basically saying is that I need a distortion to make the transition happen. And as we know from Franck-Condon principle will the electronic transitions be much faster than the vibrational ones. So I guess this means that we need to have the vibration first, followed by an electronic transition. Which means I am starting from a distorted initial state, one where the geometry has to be different but there is apparently no change in the point group or overall symmetry and just this one initial state changes for some reason. But the change in electronic states should be almost vertical and therefore there wouldn't be any changes in the configuration. Sure the transition is into the excited vibrational state with the biggest overlap integral, but does that mean that it will automatically have this new vibrational quantum state? To my understanding, those are separated steps. The absorption at one molecular geometry, followed by relaxation and Stokes-shift into the lowest vibrational electronically excited state. So for the resonance condition of both initial and final state, don't we automatically have to assume the same vibrational distortion? Why would that vibronic coupling only act on one state and not on the other, because if it did to both it wouldn't change much about the transition conditions? So a description of it being a change in the point group as a distortion by some vibrational mode that lowers the overall symmetry and therefore allows more electronic transitions to happen seems probably a lot more plausible to me.

I'm okay if it can be solved by just having one of the two states being coupled to some vibration but what is the reason that the other state doesn't do the same thing?

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  • $\begingroup$ I think it may be better to post this as another question and link back to this one saying it is a continuation. I can post something on that new question, but I believe the issue you are having is that while the vibrational state and the geometry/configuration are related, they are not the same thing. $\endgroup$ – Tyberius Jan 6 at 16:07

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