Finding band origin of electronic transition and wavenumber of vibrational mode

[alphabetagamm]

Homework Statement

The first vibrational band in the absorption spectrum of H₂CO is at 28871 cm^-1. The first two bands in the emission spectrum (from v'=0 level) lie at 27021 cm^-1 and 24687 cm^-1.

Explain these observations. Determine the band origin of the electronic transition and the wavenumber of the vibrational mode in the excited electronic state that is responsible for the vibronically induced intensity.

The vibrational numbers of H₂CO in its ground electronic state are:
v₁=2766 cm^-1 (A₁)
v₂=1746 cm^-1 (A₁)
v₃=1501 cm^-1 (A₁)
v₄=1167 cm^-1 (B₁)
v₅=2843 cm^-1 (B₂)
v₆=1251 cm^-1 (B₂)

Homework Equations

G(v)=w_e (v +1/2) - w_ex_e (v + 1/2)²

The Attempt at a Solution

From ground state A₁ the molecule is excited to A₂ (absorption) it then rapidly relaxes to v=0 of A₂. The emission spectrum is from this level.

I do not know where to begin with the band origin or how to find the vibrational mode wavenumber.

Any help greatly appreciated.

 

[JohnRC]

(1) This is a fairly tricky problem. I am presuming that you are in an honours stream. (It is not an exercise that I would have felt free to set for my undergraduate students). However, although tricky, it is not hard.

(2) There are several quite different and quite valid ways to approach it: here is a start on just 3 of them:

a) Think about the electronic structure of formaldehyde. What electronic structure, and more particularly, which of the symmetry representations, would you expect to be associated with its first excited state? What implications does that have for the observed vibrational structure of the spectrum?

b) The two observed emission bands relate to a single excited vibronic state going to two different vibrational levels of the ground electronic state. Can you relate the energy difference between them to any of the fundamental vibrational frequencies for the ground state that you were given?

c)Although there are six different vibrational modes in the molecule, and transitions in every one of the six would be symmetry allowed in the pure vibrational (infrared) spectrum, the electronic spectrum has particularly few bands, and no visible band at the band origin. What are the implications of this -- in symmetry terms? -- in vibrational terms?

None of my sets of questions will provide the answers for you, but I think that providing answers to anyone of the three will get you to a position where you can "follow your nose" through the rest of the problem.

 

[alphabetagamm]

John.
(1) I am indeed an undergraduate

2a) C_2v point group with ¹A₁ ground state and ¹A₂ excited state resulting from promotion of b₂ oxygen lone pair electron to \pi* of C=O bond.

The reducible representation on the vibrations of both states are \tau_vib= 3A₁ + 2B₂ + B₁

As fro what this does to the spectrum it would mean 6 vibrational wavenumbers (the ones given).

b) The difference between the 2 emission wavenumbers is equal to 2B₁ transitions.

c) Electronically forbidden due to the irrep for n to \pi* being electronic dipole forbidden.

Further question-
Probably simple but is the absorption line exciting to the v'=0 level or is it exciting to v'=n then rapidly relaxing to the v'=0 state?

Is the band origin for the electronic transition the v=0 for the ¹A₁ state?

Thanks so much.

 

[JohnRC]

Further question-
Probably simple but is the absorption line exciting to the v'=0 level or is it exciting to v'=n then rapidly relaxing to the v'=0 state?

Is the band origin for the electronic transition the v=0 for the 1A1 state?

Absorption line is exciting from v"=0 to v_n' = 1 level, which is then rapidly relaxing to v'=0

"Band origin" means the energy gap between a v'=0 and a v"=0 level of the electronically excited and ground states respectively. In this particular case there will be no spectral line at the band origin because that transition is symmetry forbidden.

So if a direct transition between the two electronic states is symmetry forbidden, it means that the band origin will be inactive -- the upper electronic state is only accessible to vibrationally excited states that will combine with the electronic symmetry to make an overall transition that is dipole allowed. That will obviously rule out totally symmetric A₁ vibrational modes.

It is likely, but not necessarily the case, that the hidden mode will also be the one that is showing in progressions.

That is, the two transitions we see in emission might be
(¹A₂)v'=0 → (¹A₁)v₄"=1,3
or

(¹A₂)v'=0 → (¹A₁)v₄"=0,2; v₅"=1
or
(¹A₂)v'=0 → (¹A₁)v₄"=0,2; v₆"=1

One of these three possibilities can be ruled out immediately -- which, and why?

I have said that the first is "more likely" a priori, but you should be able to find a better reason for distinguishing between the other two possibilities.

A final point I will help you with is the "mirror image" principle for absorption and emission vibronic spectra. The important point about the way it applies here is that

If the first strong emission line is from zero vibrational level of electronically excited state to v_n" = 1 of ground state, then the first strong absorption line will involve the same mode: a transition from zero vibrational level of ground state to v_n' = 1, same n, of electronically excited state.