Chemistry 2
Lecture 6
Vibrational Spectroscopy
Learning outcomes from lecture 5
• Be able to draw the wavefunctions for the first few solutions to the Schrödinger
equation for the harmonic oscillator
• Be able to calculate the energy separation between the vibrational levels for the
harmonic oscillator
• Be able to explain the dependence of this separation on the masses of the atoms
and the strength of the bond
• Be aware of and be able to calculate the zero point energy of a vibrating
molecule
• Be able to draw the potential energy curve for a real molecule and explain why it
is different to the harmonic potential
Assumed knowledge
Light behaves like an oscillating electromagnetic field. The electric field interacts
with charges. Two charges separated in space represent a dipole moment which
can interact with an electric field. Energy can only be taken or added to the
electric field in units of hn (photons).
Revision: Light as a EM field
Revision:
The Electromagnetic Spectrum
Spectral
range
l (nm)
n (Hz)
Radio
~1 x 109
~108
~0.03
~10-8
NMR/ESR
Microwave
~100,000
~1012
~30
~10-2
Rotational
Infrared
~1000
~1014
~3,000
~103
Vibrational
Visible
400-750
4-6 x 1014
14,00025,000
1 – 3x105
Electronic
Ultraviolet
100-400
~1015
~40,000
~5x105
Electronic
<100
>1016
>100,000
>106
Core
electronic
X ray
n~
(cm-1)
Energy
(kJ/mol)
Spectroscopy
Revision: Fundamental equations
Quantity
E  hn
n  c
~
n  1
l
n
Energy
l
c
Symbol SI Unit Common
Unit
E
J
kJ/mol
Frequency
n
s-1 or
Hz
s-1 or Hz
Wavelength
l
m
nm or
mm
Wavenumber
n~
m-1
cm-1
Constant
Symbol
Value
Speed of light
c
3.00 x 108 m/s
Planck constant
h
6.626 x 10-34 Js
Classical absorption of light
Because light is an oscillating electromagnetic field, it can cause
charges to oscillate. If the charge can oscillate in resonance with
the field then energy can be absorbed.
Alternatively, an oscillating charge can emit radiation with
frequency in resonance with the original oscillation.
For example, in a TV antenna, the oscillating EM field broadcast by
the transmitter causes the electrons in your antenna to oscillate at
the same frequency.
Classical absorption of light
What about a molecule as an antenna?
How can we get oscillating charges in a molecule?
+
-
+
-
Classical absorption of light
+
-
+
Rotating a permanent
dipole causes an
oscillation of charge
Classical absorption of light
+
+
-
Vibrating a permanent
dipole causes an
oscillation of charge
The Quantum Harmonic Oscillator
Solution of Schrödinger equation:
n 
h
2
k 
1
n  
m
2
But: n 
1
k
2
m
n = 0,1,2,…
Oscillation (vibrational) frequency
 n  (n 
Need to know!
1
2
G (v)  (v 
)hn
1
2 )
S.I. units
G(v) &  in cm-1
G(v) is the “energy” from the bottom of the well and  is the “harmonic frequency”
“Anharmonic oscillator” (A.H.O.)
harmonic
anharmonic
Molecules dissociates
The Morse Anharmonic Oscillator
all in wavenumber (cm-1) units:
G (v) =
(
G (v) =
(
v + 12
) e
v + 12
) e
2
1
v + 2
(
)
e
2
4 De
2
1
v + 2  e xe
(
)
This one you DO have to know how to use
In spectroscopy, we tend to use the letter “vee” to indicate the quantum
number for vibration. The vibration frequency is indicated by e in cm-1.
When solving the general quantum mechanical problem we used the letter
n, to minimize confusion with “nu”, the vibrational frequency in s-1.
Harmonic
frequency

G (v)  v  1
Anharmonicity
constant


V (r)
The Morse energy levels

2
1
 - v  2  e xe
2 e
Harmonic
term
Anharmonic
term
wexe is usually positive, so vibrational
levels get closer together
The force constant for the AHO is the same
as for the harmonic case:
1
k
1
k
or  =
n 
2 c m
2 m
r
Different levels of approximation…
• General AHO:

G (v)  v  1






• Morse oscillator:

G (v)  v  1

2
3
1
1
 - v  2  e x e  v  2  e y e - ...
2 e

2
1
 - v  2  e xe
2 e
• Harmonic oscillator:
G ( v )  v  1 2 
The level of approximation to use depends on:
i) the information you have
ii) the information you need
The important equations!
• In wavenumber (cm-1) units:
G (v) =
(
G (v) =
(
v + 12
) e
v + 12
) e
2
1
v + 2
(
)
e2
4 De
2
1
v + 2  e xe
(
The ones you DO have to
know how to use
)
De 
e
2
4 e x e
De 
e
2
4 e x e
V (r)
Dissociation energy
D0
This is the energy from the bottom of the
well to the dissociation limit.
De
But we know about zero-point energy,
therefore slightly less energy is required to
break the bond.
D0 = De – G(0)
We can estimate the bond dissociation
energy from spectroscopic
measurements!
G(0)
(Zero-point energy)
r
Selection rules
• All forms of spectroscopy have a set of selection rules
that limit the number of allowed transitions.
• Selection rules arise from the resonance condition,
which may be expressed as a transition dipole moment:
μ 21    ( r ) μ ( r )  1 ( r ) dr  0
*
2
Upper
state
molecular lower vibrational
state coordinate
dipole
• Selection rules tell us when this integral is zero
Harmonic Oscillator Selection Rules
selection rule:
Dv=±1
Dv=+1: absorption
Dv=-1: emission
If an oscillator has only one
frequency associated with it, then
it can only interact with radiation
of that frequency.
Selection rules limit the number of allowed transitions
Thermal population
At normal temperatures, only
the lowest vibrational state
(v =0 ) is usually populated,
therefore, only the first
transition is typically seen.
Transitions arising from
v0 are called “hot bands”
(Their intensity is strongly
temp. dependent)
Much of IR spectroscopy can understood from just
these two results of the quantum harmonic oscillator:
E = (v+½)hn and Dv = ±1
For example:
v=10
More problems with harmonic model….
-1
CO
6352 cm
0 .8
(first overtone)
-1
4260 cm
1 .0
-1
A bsorbance
1 .2
2143 cm , (fundam ental)
CO
1 .4
What are these?
0 .6
(second overtone)
1 .6
0 .4
x 10
0 .2
x 100
0 .0
2000
4000
6000
-1
W a v e n u m b e r (c m )
What are the new peaks?
• Three peaks…
i. 2143 cm-1
ii. 4260 cm-1
iii. 6352 cm-1
These are almost 1 : 2 : 3
which suggests transitions
might be
01
02
03
v=4
v=3
v=2
v=1
v=0
Anharmonic oscillator (A.H.O.)
selection rules:
There are none!
But!...
Harmonic and anharmonic
models are very similar at low
energy, so selection rules of
AHO converge on HO as the
anharmonicity becomes less:
H.O.
A.H.O.
A.H.O. selection rule:
Dv=±1,±2, ±3
Intensity gets weaker and weaker (typically 10× weaker for each)
Anharmonic oscillator (A.H.O.)
A.H.O. selection rule:
Dv= ±1,±2, ±3
Dv = 1 : fundamental
Dv = 2 : first overtone
Dv = 3 : second overtone, etc
Typical Exam Question
• Consider the infrared absorption spectrum of CO below.
a) From the wavenumber measurements on the spectrum, assign the
spectrum, hence determine the harmonic frequency, e (in cm-1)
and the anharmonicity constant exe (in cm-1).
b) Estimate the bond dissociation, D0, for this molecule. There is no
absorption below 2000 cm-1.
Fundamental:
2143 = G(1) – G(0),
Overtone:
4260 = G(2) – G(0)
Using the spectra to get information…
G ( v )  v 
1
2
 e
- v 
  e xe
2
1
2
G(1)-G(0) = [(1.5)e – (1.5)2 exe] - [(0.5)e – (0.5)2 exe]
2143 = e – 2exe
…(1)
G(2)-G(0) = [(2.5)e – (2.5)2 exe] - [(0.5)we – (0.5)2 exe]
4260 = 2e – 6exe
…(2)
Two simultaneous equations (simple to solve)
→
e = 2169 cm-1, and exe = 13 cm-1
De 
e
2
De 
4 e x e
( 2169 )
4  13
2
 90 , 500 cm
-1
V (r)
Using the spectra to get information…
D0  De - G (0)
 90 ,500 - 1080  89 , 400 cm
D0
-1
De
89,400 cm-1 = 1069 kJ/mol
c.f. exp. value: 1080 kJ/mol
Why the difference?
Remember Morse is still an approx. to the
true intermolecular potential. Still 2%
error is pretty good for just 2
measurements!
G(0)
(Zero-point energy)
r
Equations to know how to use…
m1 × m2
mm +m
1
2
n 
1
k
2
m
G ( v )  ( v  1 2 )

G (v)  v  1



2
1
 - v  2  e xe
2 e
D0  De - G (0)
De 
e
2
4 e x e
Learning outcomes
• Be able to manipulate and use the key equations given in the green
box at the end of the lecture.
•Utilize the harmonic oscillator and anharmonic oscillator as a model
for the energy level structure of a vibrating diatomic molecule.
Next lecture
• The vibrational spectroscopy of polyatomic molecules.
Week 11 homework
• Work through the tutorial worksheet and the practice problems at
the end of the lectures and check your answers with those available
online
• 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
Practice Questions
Which of the following diatomic molecules will exhibit an infrared
spectrum?
a) HBr b) H2 c) CO d) I2
2. An unknown diatomic oxide has a harmonic vibrational frequency of ω =
1904 cm−1 and a force constant of 1607 N m −1. Identify the molecule.
a) CO b) BrO c) NO d) 13CO
3. As the energy increases, the vibrational level spacing for a harmonic
oscillator is:
a) increases b) decreases c) stays constant
4. As the energy increases, the vibrational level spacings for a Morse
oscillator usually:
a) increase b) decrease c) stay constant
5. As the energy increases, the vibrational level spacings for an anharmonic
oscillator usually:
a) increase b) decrease c) stay constant
1.
Practice Questions
6. For a Morse oscillator the observed dissociation energy, D0, is related to
the equilibrium vibrational frequency and the anharmonicity by the
following expression:
a) ωe2/4ωexe b) [ωe2/4ωexe]-G(0) c) (v+½)ωe+(v+½)2ωexe d) (v+½)ωe
7. Which of the following statements about the classical and quantum
harmonic oscillator (HO) are true (more than one possible answer here)?
a) The classical HO frequency is continuous, whereas the quantum
frequency is discrete.
b) The classical HO has continuous energy levels, whereas the quantum
HO levels are discrete.
c) The classical HO depends on the force constant, but the quantum HO
does not.
d) The classical HO may have zero energy, but the quantum HO may not.
e) The classical HO does allow the bond to break, whereas the quantum
HO does.
Practice Questions
8. Which of the following statements correctly describe features of the
quantum Morse oscillator and energy levels associated with it?
a) The vibrational energy levels get more closely spaced with
increasing v.
b) The vibrational energy levels approach a continuum as the
dissociation energy is approached.
c) The Morse oscillator and HO are nearly the same at very low v
d) The Morse potential is steeper than the HO for r < re
e) The Morse potential exactly describes the interatomic potential.