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Chemistry 2 Lecture 10 Vibronic Spectroscopy

Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

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Page 1: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Chemistry 2

Lecture 10 Vibronic Spectroscopy

Page 2: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Learning outcomes from lecture 9

Excitations in the visible and ultraviolet correspond to excitations of electrons between orbitals. There are an infinite number of different electronic states of atoms and molecules.

Assumed knowledge

• Be able to qualitatively explain the origin of the Stokes and anti-Stokes line in the Raman experiment

• Be able to predict the Raman activity of normal modes by working out whether the polarizability changes along the vibration

• Be able to use the rule of mutual exclusion to identify molecules with a centre of inversion (centre of symmetry)

Page 3: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Which electronic transitions are allowed?

The allowed transitions are associated with electronic vibration giving rise to an oscillating dipole

Page 4: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Electronic spectroscopy of diatomics

• For the same reason that we started our examination of IR spectroscopy with diatomic molecules (for simplicity), so too will we start electronic spectroscopy with diatomics.

• Some revision:– there are an infinite number of different electronic states of atoms

and molecules– changing the electron distribution will change the forces on the

atoms, and therefore change the potential energy, including k, we, wexe, De, D0, etc

Page 5: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Depicting other electronic states

Ground Electronic State

Excited Electronic States1. Unbound

2. Bound

There is an infinite number of excited states, so we only draw the ones relevant to the problem at hand.

Notice the different shape potential energy curves including different bond lengths…

Page 6: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Ladders upon ladders…Each electronic state has its own set of vibrational states.

Note that each electronic state has its own set of vibrational parameters, including:- bond length, re

- dissociation energy, De

- vibrational frequency, we

re’re”

De”

De’

we”

we’

Notice: single prime (’) = upper statedouble prime (”) = lower state

Page 7: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

The Born-Oppenheimer ApproximationThe total wavefunction for a molecule is a function of both nuclear and electronic coordinates:

(r1…rn, R1…Rn)

where the electron coordinates are denoted, ri , and the nuclear coordinates, Ri.

The Born-Oppenheimer approximation uses the fact the nuclei, being much heavier than the electrons, move ~1000x more slowly than the electrons. This suggests that we can separate the wavefunction into two components:

(r1…rn, R1…Rn) = elec (r1…rn; Ri) x vib(R1…Rn)

Electronic wavefunction at ×each geometry

Total wavefunction = Nuclear wavefunction

Page 8: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

The Born-Oppenheimer Approximation

(r1…rn, R1…Rn) = elec (r1…rn; Ri) x vib(R1…Rn)

The B-O Approximation allows us to think about (and calculate) the motion of the electrons and nuclei separately. The total wavefunction is constructed by holding the nuclei at a fixed distance, then calculating the electronic wavefunction at that distance. Then we choose a new distance, recalculate the electronic part, and so on, until the whole potential energy surface is calculated.

While the B-O approximation does break down, particularly for some excited electronic states, the implications for the way that we interpret electronic spectroscopy are enormous!

Electronic wavefunction at ×each geometry

Total wavefunction = Nuclear wavefunction

Page 9: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Spectroscopic implications of the B-O approx.

1. The total energy of the molecule is the sum of electronic and vibrational energies:

Etot = Eelec + Evib

Eelec

Etot

Evib

Page 10: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Spectroscopic implications of the B-O approx.

• In the IR spectroscopy lectures we introduced the concept of a transition dipole moment:

|1

|2upper state

wavefunctionlower state

wavefunctiontransition

dipole moment

dipole moment operator

integrate over all coords.

0),( ˆ),( 1*221 drdRRrRr jjii μμ

dRRRR

drdRRrrR

jvib

ivib

jvib

jelec

ielec

ivib

)( )()(

)()( ˆ)()(

1*2

11*2

*221

μ

μμusing the B-O approximation:

2. The transition moment is a smooth function of the nuclear coordinates.

Page 11: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Spectroscopic implications of the B-O approx.

|1

|2

dRRR

dRRRR

drdRRrrR

jvib

ivib

jvib

ivib

jvib

jelec

ielec

ivib

)( )(

)( )()(

)()( ˆ)()(

1*20

1*2

11*2

*221

μ

μ

μμ

2. The transition moment is a smooth function of the nuclear coordinates. If it is constant then we may take it outside the integral and we are left with a vibrational overlap integral. This is known as the Franck-Condon approximation.

3. The transition moment is derived only from the electronic term. A consequence of this is that the vibrational quantum numbers, v, do not constrain the transition (no Dv selection rule).

Page 12: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Electronic AbsorptionThere are no vibrational selection rules, so any Dv is possible.

But, there is a distinct favouritism for certain Dv. Why is this?

Page 13: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Franck-Condon Principle (classical idea)

0 1 2 3 4 5

Ene

rgy

R

Classical interpretation:

“Most probable bond length for a molecule in the ground electronic state is at the equilibrium bond length, re.”

Page 14: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

0 1 2 3 4 5

Ene

rgy

R

Franck-Condon Principle (classical idea)

The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

In the ground state, the molecule is most likely in v=0.

Page 15: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

0 1 2 3 4 5

Ene

rgy

R

Franck-Condon Principle (classical idea)•The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The electron excitation is effectively instantaneous; the nuclei do not have a chance to move. The transition is represented by a VERTICAL ARROW on the diagram (R does not change).

Page 16: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

0 1 2 3 4 5

Ene

rgy

R

Franck-Condon Principle (classical idea)•The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The most likely place to find an oscillating object is at its turning point (where it slows down and reverses). So the most likely transition is to a turning point on the excited state.

Page 17: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Quantum (mathematical) description of FC principle

approximately constant with

geometry

Franck-Condon (FC) factor

μ21 = constant × FC factor

FC factors are not as restrictive as IR selection rules (v=1). As a result there are many more vibrational transitions in electronic spectroscopy.

FC factors, however, do determine the intensity.

dRRR jvib

ivib )( )( 1

*2021 μμ

Page 18: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Franck-Condon Principle (quantum idea)

In the ground state, what is the most likely position to find the nuclei?

0 1 2 3 4

0

1

2

3

(0)v=0

v

R

2Prob

Max. probability at Re

Page 19: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

0

1 2 3 4

Wave number

Franck-Condon Factors

If electronic excitation is much faster than nuclei move, then wavefunction cannot change. The most likely transition is the one that has most overlap with the excited state wavefunction.

v’ = 0

1

2

v” = 0

Page 20: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Look at this more closely…

• Excellent overlap everywhere

Negative overlap to left, postive overlap to right

overall zero overlap

Negative overlap in middlePositive overlap at edgesoverall very small overlap

Page 21: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Franck-Condon Factors

0

1

2

34

Wave number

Page 22: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Franck-Condon Factors

v=10

Note: analogy with classical picture of FC principle! v. poor v=0 overlap

Page 23: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Electronic Absorption

There are no vibrational selection rules, so any Dv is possible.

μ21 = constant × FC factor

Relative vibrational intensitiescome from the FC factor

Page 24: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Absorption spectrum of binaphthyl•Example of real spectra showing FC profile

30100 30200 30300 30400 30500 30600 30700 30800 30900

8

9

10

11

12

13

14

15

1617 18

19

2021

22

23

24

25 26 27

28

(3) (4) (5) 67

Wave number (cm-1)

Page 25: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Absorption spectrum of CFCl

17000 18000 19000 20000 21000

= CCl2 peaks

45

510

23

3

(0,0,1) hot bands}(1,n,0)(0,n,2)(0,n,1)(0,n,0)

(0,n,2)(0,n,1)(0,n,0)

Wave number (cm-1)

Page 26: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

If the excited state is dissociative, e.g. a p* state, then there are no vibrational states and the absorption spectrum is broad and diffuse.

Unbound states (1)

Page 27: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Even if the excited state is bound, it is possible to access a range of vibrations, right into the dissociative continuum. Then the spectrum is structured for low energy and diffuse at higher energy.

Unbound states (2)

Page 28: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Some real examples…A purely dissociative state

leads to a diffuse spectrum.HI

Page 29: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

The dissociation limit observed in the spectrum!I2

16000 18000 20000 220000.00

0.05

0.10

0.15

0.20

0.25

I2

Ab

sorb

an

ce

Wave number (cm-1)

Some real examples…

Page 30: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Analyzing the spectrum…All transitions are (in principle) possible. There is no Dv selection rule

Vibrational structure

16000 18000 20000 220000.00

0.05

0.10

0.15

0.20

0.25

I2

Ab

sorb

an

ce

Wave number (cm-1)

Page 31: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

cm-1

18327.818405.418480.918555.618626.818706.318780.018846.618911.518973.919037.5

v’2526272829303132333435

eee xG 22

12

1 vv)v( v”0

0

0

0

0

0

0

0

0

0

0

How would you solve this?(you have too much data!)

1. Take various combinations of v’ and solve for we and wexe simultaneously. Average the answers.2. Fit the equation to your data (using XL or some other program).

Analysing the spectrum…

Page 32: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

cm-1

18327.818405.418480.918555.618626.818706.318780.018846.618911.518973.919037.5

v’2526272829303132333435

eeeelecelectot xEGEE 22

12

1 vv)v(

v”00000000000 10 20 30 40 50 60 70

17000

17500

18000

18500

19000

19500

20000

Wa

ve n

um

be

r (c

m-1)

v'

Eelec = 15,667 cm-1

we = 129.30 cm-1

wexe = 0.976 cm-1

Dissociation energy = 19950 cm-1

vibelectot EEE

Analyzing the spectrum…

Page 33: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Learning outcomes• Be able to draw the potential energy curves for excited electronic

states in diatomics that are bound and unbound• Be able to explain the vibrational fine structure on the bands in

electronic spectroscopy for bound excited states in terms of the classical Franck-Condon model

• Be able to explain the appearance of the band in electronic spectroscopy for unbound excited states

The take home message from this lecture is to understand the (classical) Franck-Condon Principle

Page 34: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Next lecture

• The vibrational spectroscopy of polyatomic molecules.

Week 12 homework• Vibrational spectroscopy worksheet in tutorials• Practice problems at the end of lecture notes• Play with the “IR Tutor” in the 3rd floor computer lab and with the

online simulations:http://assign3.chem.usyd.edu.au/spectroscopy/index.php

Page 35: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of

Practice Questions1. Which of the following molecular parameters are likely to change when a molecule is

electronically excited?(a) ωe (b) ωexe (c) μ (d) De (e) k

2. Consider the four sketches below, each depicting an electronic transition in a diatomic molecule. Note that more than one answer may be possible(a) Which depicts a transition to a dissociative state?(b) Which depicts a transition in a molecule that has a larger bond length in the excited

state?(c) Which would show the largest intensity in the 0-0 transition?(d) Which represents molecules that can dissociate after electronic excitation?(e) Which represents the states of a molecule for which the v”=0 → v’=3 transition is

strongest?