[Compaq]

Hi!

I am preparing for an exam in a spectroscopy course, and here is a question I am having trouble answering. I was hoping someone here could help me out. The question is attached.

I really appreciate any help I can get. Symmetry of molecules and orbitals is new to me, and it is difficult to grasp!

[Corribus]

Well, start with the first part. It's not entirely clear what "irrespective of the substituents R" means - but let's assume they are small and identical. What is the point group of, say, acetone?

[Compaq]

Acetone has C_2v symmetry.

I guess I am not sure about what kind of symmetry information I am supposed to be able to extract form the text.

If the dipole transition is symmetry forbidden, then what can I say about the symmetry label of S1 compared to the S0 state? Also, the symmetric stretch of the C=O bond will not change the symmetry of the molecule, so only the asymmetric bending would lead to a vibrationally allowed transition, yes?

[Corribus]

Ok, let's take it one step at a time. For now: Ignore the vibronic component and just consider the pure electronic transition of a carbonyl.

1. Identify it. (What is the nature of the ground and first excited singlet state? What orbitals are involved?)
2. Assign symmetry labels to the participating states.
3. Show that the transition is forbidden.

Can you do these three steps?

[Compaq]

1. The pure electronic transition from the ground state would be the HOMO-LUMO transition. I am trying to construct a MO diagram, but I am a little unsure how to do it. The C atom is sp2 hybridized and has 4 valence electrons: one in each sp2 orbital and one in a remaining p orbital. O is sp hybridized and has six  valence electrons: two full sp orbitals and two in the remaining p orbital. Four of O's electrons are lone pairs, and contribute not to the bonding or anti-bonding orbitals.

An sp2 and sp orbital form sigma and sigma* MO. The remaining p orbitals form pi and pi* MO. There must be two non-bonding MO (n1/n2). Are these of same energy? The remaining sp2 orbitals from C are involved with the bonding to the methyl groups of the acetone molecule.

Are non-bonding MOs always in-between the energy of the pi MO and pi* MO? It seems the HOMO-LUMO transition is n(1/2?) -> pi*.

That is about as far as I can go. I don't know how to go from a MO to a state or term symbol, and assign symmetry labels to them.

[Corribus]

Alright, building an MO diagram is a fine place to start, but may be a little overkill.  I think you have most of the pieces there to do it, but we can simplify things. We don't need to get into hybridization and all that stuff for this problem. All we want to know are the orbitals involved, and what the symmetry of those orbitals are in the C_2v point group - and only those involved with the carbonyl group.

Considering just the C and the O, there are basically five molecular orbitals to worry about:

1. A σ orbital between C and O.
2. A π orbital between C and O.
3. Two non-bonding atomic orbitals (n) on oxygen
4. A π* orbital between C and O.
5. A σ* orbital between C and O.

I've already ordered these from low to high energy. This is all in line with what you wrote in your last post, more or less. The nonbonding orbitals are intemediate in energy between the π and π* orbitals. This should be intuitive, since the π and π* orbitals are stabilized and destabilized, respectively, compared to nonbonding p-orbitals, even if one is carbon and the other oxygen. So, your HOMO is going to be one of the nonbonding orbitals on oxygen and the LUMO is going to be the π* orbital. The lowest energy transition for carbonyl is π* n. Note that you have higher energy transitions (e.g., between the π and π* orbitals), and it wouldn't be a bad idea to go through the exercise of showing whether these transitions are symmetry allowed or forbidden.

On to step 2: assign symmetry labels to the participating states. The participating orbitals have been defined, but not the states. We can go whole hog here but again some simplification is in order. Any filled orbital has complete symmetry (A₁), because any symmetry designation times itself is completely symmetric, so electrons in the low lying π and σ orbitals don't really contribute to the state symmetry. So we can just think about what happens to the electrons in the nonbonding and π* orbitals. If our transition is π* n, then the states we have to worry about are n¹π*¹ and n². Can we determine what symmetry representations these states conform to?

To get you started, call the axis of the molecule the z-axis and the zy plane contains the R groups. In this scheme, using C_2v symmetry group, assign a representation for each of the molecular orbitals defined above. Then, it is straightforward to assign the symmetry labels to the participating states simply by doing a little multiplication of the orbital representations. If you need some help with that, let me know. Once you have symmetry designations for the states, we can look at why the transition is forbidden.

One other thing, the two nonbonding orbitals typically transform together. For simplicity, I think it is easier just to visualize the participating nonbonding electrons to be located in a straight-up (i.e. unhybridized) p_y orbital. Use this to determine the symmetry of the nonbonding electrons. It's not strictly correct but the symmetry designation will be.

NB: The π (pi) symbol and letter n symbol look practically identical on this forum! I've tried to help that out by italicizing the "n" where possible.

[Compaq]

I found that the transition is symmetry allowed, which of course is incorrect. Here's what I did (see attachment for my notes).

For a transition to be symmetry allowed, the multiplication of the two states and each of the x,y,z directions must equal the A1 characters. If none of the directions meet this criteria, the transition is forbidden. If the criterium is met for one direction, then the transition is allowed for that particular polarization of the light.

I already know that n2 is an a1 orbital. Both the n and pi* I found to be a2 orbitals, which leads to the state n1pi*1 to be of A1 symmetry. Therefore it is easy to see that for the transition to be allowed, one of the directions must also be A1, which for the C2v point group z is. Hence I found the transition to be symmetry allowed if the radiation is z polarized. I am not sure what I did wrong, but it should have to do with my calculation of the characters, I guess.

Thank you for helping me!

[Corribus]

If the criterium is met for one direction, then the transition is allowed for that particular polarization of the light.

In principle, yes. But since the molecular frame is seldom fixed wrt to the lab frame, any polarization of light can usually access the molecular transition.

I already know that n2 is an a1 orbital. Both the n and pi* I found to be a2 orbitals, which leads to the state n1pi*1 to be of A1 symmetry.

Well here is your problem. Neither one of these orbitals have A₂ symmetry. Here's why. If the C₂ axis is the z axis and the yz plane is the plane of the molecule, then the π molecular bonding orbital of the carbonyl sits above and below the yz plane. Importantly, the portions above and below the plane have opposite polarity. An identity operation obviously does not change anything, so the associated character is 1. A C₂ rotation (around the z axis) basically inverts the polarity of the π orbital, so the associated character for this operation is -1. (This rules out an A₂ symmetry representation, because the character for A₂ is 1 for the C₂ operation.) Mirror reflection in the xz plane does not change the orientation of the π orbital (χ = 1) and the reflection through yz plane inverts it (χ = -1). Therefore the symmetry representation of the π orbital is Γ = 1 -1 1 -1, which is B₁. The π* orbital has the same symmetry.

A p_y atomic orbital on oxygen (which we've identified as a simplification of the nonbonding orbitals) does not belong to A₂ either for the same reason that the π orbitals do not: C₂ rotation inverts its polarity, so the associated character is -1. This means it must belong either to B₁ or B₂, the only two representations in C_2v that have a -1 character for the C₂ operation. Can you determine which it is?

(EDIT: Here's a picture if it's helpful illustrating the π bond concept. Sorry for the bad scan. Hope you can make it out.)

[Compaq]

Oh, man. I totally messed up the rotation around the z axis (which is the easiest symmetry operation to visualize!). And as you say, the σ_v character for π* is +1, not -1. Looking at my figure, it seems obvious now! But cannot I not draw the pi* orbital, with non-matching faces of the orbitals (no overlap?) the way I did? Why did you use the pi orbital and then say that pi* has the same symmetry?

Regarding the non-bonding orbital, the σ_v is +1 (no change of polarity), giving the orbital B1 symmetry. But this leads to the same issue as earlier; When both n and pi are of the same symmetry, the combined state of the two will be of A1 symmetry, again leading to a symmetry allowed transition.

[Corribus]

Oh, man. I totally messed up the rotation around the z axis (which is the easiest symmetry operation to visualize!). And as you say, the σ_v character for π* is +1, not -1. Looking at my figure, it seems obvious now! But cannot I not draw the pi* orbital, with non-matching faces of the orbitals (no overlap?) the way I did? Why did you use the pi orbital and then say that pi* has the same symmetry?

Because it was easier to describe without a picture, which I included in the end anyway. There's no problem with your drawing.

Regarding the non-bonding orbital, the σ_v is +1 (no change of polarity), giving the orbital B1 symmetry. But this leads to the same issue as earlier; When both n and pi are of the same symmetry, the combined state of the two will be of A1 symmetry, again leading to a symmetry allowed transition.

Nonbonding orbital is oriented along the y axis (for simplicity - in reality it has both y and z projections, but it is the y projection that matters here). We've already discussed the effect of C₂ rotation (χ=-1). What about the mirror plane reflections? There are two of them here - one in the yz plane and the other in the xy plane. One of them does nothing to the y-axis. The other one flips the polarity. Which one flips the polarity? This determines whether the symmetry designation is B₁ or B₂.

(Here's a little hint, by the way. The π orbital we've already assigned to B₁ and it is oriented more or less in the x-direction. Obviously the p_y orbital is oriented in the y direction. First, this should be a strong clue that they may belong to different symmetry classifications, at least for a point group that has all 1-dimensional representations. Additionally, do you notice that the x-oriented π orbitals belong to B₁, which has an x-basis according to the character table? What does this tell you to expect of the symmetry assignment for the p_y orbital?)

[Compaq]

Okay, for the n orbital, χ_σxz and χ_σxz is -1 and and +1, respectively. Then the excited state has A2 symmetry. For f1f2f3 to equal A1, the directions need to have A2 symmetry too, which none have. Therefore symmetry forbidden.

If I only pay more attention performing the symmetry operations, I would have gotten the correct answer the first time.

Somehow one of the vibrations help the transition. I speculated that the symmetry needs to change for one of the vibrations if it is to make a difference to the symmetry assessment above. Is that correct?

[Corribus]

Right: basically you know that the product of the reducible representation of the initial state and the reducible representation of the final state, and at least one of the representations with a linear basis, must be or contain a component with A₁ symmetry to be allowed. (This is because, if this condition is not met, the transition moment integral vanishes.)

Using only pure electronic states you have now shown that the transition moment integral vanishes.

The goal of the problem is to show that if you use vibronic states, this is no longer the case.

We can express the symmetry of the vibronic states as follows:

[tex]\Gamma_{el} \otimes \Gamma_{vib} = \Gamma_{vibronic} [/tex]

And therefore, for example:

[tex]\Gamma_{i, vibronic} = \Gamma_{i,el} \otimes \Gamma_{i,vib} [/tex]

And

[tex]\Gamma_{f, vibronic} = \Gamma_{f,el} \otimes \Gamma_{f,vib} [/tex]

It remains, then to identify the symmetry assignments of potential vibrational states.  The ground or initial vibrational state is A₁, so the ground vibronic state has the same symmetry as the ground electronic state. What about the excited vibrational states? Conveniently the problem gave you structures of what the normal modes are that you should consider. Can you assign symmetry classes to them, and show that the transition becomes partially allowed?