Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the...
-
Upload
brian-stephens -
Category
Documents
-
view
226 -
download
0
Transcript of Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the...
![Page 1: Chemistry 2 Lecture 10 Vibronic Spectroscopy. Learning outcomes from lecture 9 Excitations in the visible and ultraviolet correspond to excitations of.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/1.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/2.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/3.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/4.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/5.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/6.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/7.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/8.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/9.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/10.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/11.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/12.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/13.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/14.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/15.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/16.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/17.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/18.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/19.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/20.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/21.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/22.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/23.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/24.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/25.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/26.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/27.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/28.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/29.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/30.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/31.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/32.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/33.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/34.jpg)
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.](https://reader036.fdocuments.us/reader036/viewer/2022062314/56649d095503460f949db549/html5/thumbnails/35.jpg)
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?