Vibrational Spectroscopy
In this part of the course we will look at the kind of spectroscopy which uses light to
excite the motion of atoms. The forces required to move atoms are smaller than
those required to move electrons (as nucleus and electron are bound by strong
electrostatic forces). Thus the photon energies involved are somewhat lower;
instead of UV/Vis radiation infra-red radiation is sufficient, hence the alternative title,
infra-red spectroscopy.
In the first section we will see that we can use classical arguments (springs again) to
describe resonances in vibrational spectroscopy. We will learn how to relate the
resonance frequency to the strength of a chemical bond and the mass of the nuclei.
In the next section we will see how analysis of vibrational spectra can lead to
information about molecular shapes.
In the third section we will see that the infra-red spectrum can be used as a powerful
analytical tool, because the spectrum yields a ‘fingerprint’ characteristic of a
particular molecule.
Finally some quantum aspects of molecular vibrations will be investigated.
Absorption of IR radiation is sufficient to cause molecules to vibrate (the electronic
state stays the same). The vibrational motions may be usefully sub-divided into
stretching and bending. In general stretching involves separating nuclei whereas
bending does not, thus stretching modes require higher energy photons than
bending modes. In complex molecules vibrations are characteristic of functional
groups, so IR spectra give an indication of what groups are inside a molecule, and
are thus a useful analytical tool.
As before we start with the simplest cases and progress to more complicated
molecules.
Diatomic molecules — “molecular springs”
For diatomics, there are no bending modes only a single vibrational mode. The
vibrating molecule is often likened to an oscillating mass on the end of a spring:
Typically the mass
undergoes simple harmonic
motion – see physics
courses.
For small oscillations the restoring force is thus proportional to the extension, so the
oscillation and the displacement (x) follows a simple sinusoidal motion. Thus we can write
generally:
x  x0 sin(t )
The frequency of the vibration is the number of waves per second and so is given by 1/T.
Thus,
1

 
T
2
The quantity  is now seen to be related to the frequency of the motion. We call it
the angular frequency and rearranging this equation we have
  2
So by simple substitution the time dependent displacement is:
x  x0 sin(2t )
Now we can examine the motion of the mass (m) according to Newton’s laws. Provided
displacement is not too large Hooke’s law applies
The constant here k is called the spring constant. A big k means that a large restoring force
is exerted for a particular extension of the spring, i.e. we have a ‘strong’ spring. Now, the
equation of motion is given by Newton’s second law,
F  ma
Using Hooke’s law and Newton’s second law
d2 x
d2
 kx  ma  m 2  m 2  x0 sin 2t 
dt
dt
Differentiating once:
And again:
So:
d
x0 cos2t 
 kx  m2 
dt
2
2
 kx  m2  x0 sin2t   m2  x
k  m2 
2
A simple rearrangement leads to the relation
1

2
k
m
Simply, this tells us that a ‘strong’ spring (big k) vibrates at a higher frequency than
a weak ‘springy’ one. Also, if the mass is light the frequency will be high, so heavy
masses vibrate at low frequencies. k has units kgs-2, or Nm-1
Application to bonds — the ‘clamped’ diatomic molecule
The idea is exactly the same as before — a small extension of the bond
leads to a proportional restoring force
where k is now the force constant of the bond. In the same way as k
relates to how strong the spring is, for molecules it relates to how strong
the bond is. This makes it a useful quantity to know.
Problem: we can see that the vibrational frequency of HCl will depend on which
end of the molecule we clamp because of m in our equation. In reality molecules
are not clamped and will possess one unique vibrational frequency. Clearly the
equation has to be modified for real molecules and the modification must be
related to the mass. A detailed derivation gives us

1
2
k

Where  is the reduced mass

m1m2
m1  m2
Thus if the frequency of the diatomic vibrational mode is measured and the
masses are known then the force constant (related to bond strength) can be
determined. This is an important result.
Estimate or calculate exactly the force constant of the
H1Cl35 bond if the molecule absorbs at 3343.8 nm? [c = 3 x
108 ms-1, proton mass = 1.673 x 10-27 kg]
1. 300 – 400 Nm-1
2. 400 – 500 Nm-1
3. 500 – 600 Nm-1
3  108 ms 1
13
 

8
.
972

10
Hz
9
 3343.8  10 m
c
convert to frequency
m1m2
(1  35)  1.673  1027 kg 
 27



1
.
626

10
kg
 27
m1  m2
(1  35)  1.673  10 kg
2
calc reduced mass
Finally
k   2   516.9 Nm 1
2
0%
1
0%
2
0%
3
Similar experiments were performed for the other hydrogen halides. The results are shown
below.
IR absorption / m
k / Nm1
band wavenumber /
cm1
HF
2.416
966
4139
HCl
3.344
xxx
2990
HBr
3.774
412
2650
HI
4.329
314
2310
Halide
The table shows that photons of greater energy are required to stretch the HF bond
compared to the HCl bond. This makes sense as we know that F atoms are
smaller that Cl atoms and can therefore make shorter, stronger bonds. This trend
continues down the table. We can also see this trend by looking at the force
constants. Thus the HF bond is seen to be very difficult to stretch (966 Nm-1) being
almost twice as difficult as the HCl bond (517 Nm-1).
These ideas are easily extended to give some clue as to the nature of chemical
bonds in a molecule. Consider the following series of carbon – oxygen bonds.
bond
C O
C O
CO
typical band
wavenumber / cm1
1100
1700
2170
We see that more energy is required to stretch a double bond than a single bond — this
makes sense as more electrons are found between the nuclei in a double bond and so the
bond is stronger. The result for carbon monoxide suggests that the bond is even stronger
than a double bond — perhaps a triple bond? Such a structure can be envisaged:
Thus IR spectroscopy has proved useful in elucidating a chemical structure.
As hydrogen is a small atom a strong bond is expected in the H2 molecule. An IR
spectrum was recorded over the range 400 - 5000 cm-1 but no absorption was
observed. Independent measurements suggest the H—H bond has a force
constant of 575 Nm-1.
The HH bond is strong and the atoms light. The reason the
vibration was not observed was that the 400 – 5000 cm-1
wavenumber range was not wide enough.
1. True
2. False
reduced mass for homonuclear diatomic
1

2
k
1

 2
575kgs2 
 1.320  1014 Hz
 kg 
14

1
.
320

10
4
1
~  

44

10
m
8
c
3  10
m2 m

  0.8365  1027 kg
2m 2
or
4400cm 1
0%
1
0%
2
Thus the spectrometer should/should not have observed a transition.
We recall that just because absorption is in principle possible absorption may still be
absent. This is another example of a selection rule; the vibrational transition in H2
is forbidden . It turns out that the selection rule associated with IR spectroscopy
that stops H2 from absorbing IR radiation is associated with the symmetry of the
molecule and the change in dipole moment during the vibration.
There is another method, Raman Spectroscopy, which allows frequencies of H2,
and generally X2 molecules, to be determined. The selection rules for Raman are
different because the interaction with the radiation is due to the polarisabiltiy, not
the dipole moment.
Selection Rule
“Vibrations may only show up in the IR spectrum if the vibration concerned causes
a change in the molecular dipole moment”
There is an equivalent rule for Raman, requiring a change in the (shape of) the
polarizability during a vibration.
It is easy to see how this selection rule arises. Recall that light has an oscillating
electric field associated with it. If a vibration causes a change in the molecule’s
dipole moment then the vibrating molecule is really an oscillating electric field and
can interact with light — if there is no oscillating dipole then it cannot. HCl and H2
illustrate this nicely:
This means that in general we can say for diatomics:
homonuclear diatomic molecules — IR inactive
heteronuclear diatomic molecule — IR active
For the HF molecule the vibration at 4139 cm-1 shows k =
966 Nm-1. At what wavenumber does the isotope DF
absorb?
1.
2.
3.
4.
2997cm-1
2069cm-1
4139cm-1
8278cm-1
Force constant independen t of isotope so k  966 Nm 1
m1m2
(2  19)  1.673  1027 kg 
 27
calc reduced mass  


3
.
027

10
kg
 27
m1  m2
(2  19)  1.673  10 kg
12
0%
0% 13 10%
0%
1 k
1 
966

Finally  

 8.99  10 s

 27 
2  6.283  3.027  10 1
2
3
4
~

1
or
2997cm
Note ~DF  HF
2
2
Polyatomic Molecules
In general a molecule will contain N atoms (N = 2, 3, …). How many vibrations can
such a molecule possess? Each atom requires 3 coordinates to specify its location
so that 3N numbers determine the entire molecule. For the whole molecule, three
coordinates may be used to specify where the centre of mass is (during vibrations
c.o.m. does not move) and three coordinates are required to specify the orientation
of the molecule as a whole. This means that there are 3N - 6 coordinates left over
and this is the number of vibrations that the molecule possesses. This is generally
true for a non-linear molecule: if the molecule is linear then we only need two
coordinates to specify its orientation.
Overall:
non-linear: 3N - 6 vib modes
linear:
3N - 5 vib modes (e.g. diatomics: 3(2) - 5 = 1)
Example: N = 3
Linear, e.g. CO2:
modes:
O=C=O should have 3(3)  5 = 4 modes. There are two stretching
and two bending modes:
Overall, two peaks appear in the IR spectrum.
Another Example
Non-linear Molecule, e.g. water: should have 3(3)  6 = 3 modes. There are two stretching
modes:
and one bending mode:
Thus three peaks appear in the IR spectrum of water.
These examples illustrate one way in which IR spectroscopy can be used to
determine the structure of a molecule — the IR spectra of carbon dioxide and
water alone will tell us that CO2 is linear and water is bent.
70
Infrared (Intensity)
60
50
40
30
20
10
0
-10
1500
2000
2500
3000
3500
-1
wavenumber (cm )
4000
Calculate the number of vibrational modes in
acetylene (C2H2) then draw them. Label
each for IR/Raman activity.
Larger Molecules - Fingerprinting
Obviously as N gets bigger the number of vibrational modes becomes large.
Spectra of such molecules will be complex. However, to a good approximation the
vibrational frequency of a particular bond or functional group is independent of the
surrounding bonds. This means that we can create a table of stretching and
bending modes associated with different bonds — these are called correlation
tables, an example of which is shown in the attached chart and table
IR Spectra: Formaldehyde
 Certain types of vibrations have distinct IR frequencies – hence the

chemical usefulness of the spectra
The gas-phase IR spectrum of formaldehyde:
(wavenumbers, cm-1)
 Tables and simulation results can help assign the vibrations!
Formaldehyde spectrum from: http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/InfraRed/infrared.htm#ir2
Results generated using B3LYP//6-31G(d) in Gaussian 03W.
Group Frequencies
GROUP
3333-3267(s) stretch
C-H
Alkynes
C=C
Alkenes
1680-1640(m,w)) stretch
C C
Alkynes
2260-2100(w,sh) stretch
C=C
Aromatic Rings
1600, 1500(w) stretch
C-O
Alcohols, Ethers, Carboxylic acids, Esters
1260-1000(s) stretch
C=O
Aldehydes, Ketones, Carboxylic acids, Esters
1760-1670(s) stretch
Monomeric -- Alcohols, Phenols
3640-3160(s,br) stretch
Hydrogen-bonded -- Alcohols, Phenols
3600-3200(b) stretch
Carboxylic acids
3000-2500(b) stretch
O-H
700-610(b) bend
3500-3300(m) stretch
N-H
Amines
C-N
Amines
1340-1020(m) stretch
C N
Nitriles
2260-2220(v) stretch
NO2
Nitro Compounds
v - variable, m - medium, s - strong, br - broad, w - weak
1650-1580 (m) bend
1660-1500(s) asymmetrical stretch
1390-1260(s) symmetrical stretch
The Raman Effect



The incident radiation excites “virtual states” (distorted or polarized states)
that persist for the short timescale of the scattering process.
Polarization changes are
necessary to form the virtual
state and hence the Raman
effect
This figure depicts “normal”
(spontaneous) Raman effects
Virtual state
Virtual state
hv1
hv1
hv1 – hv2
Anti-Stokes line
hv1 – hv2
Stokes line
Scattering timescale ~10-14 sec
(fluorescence ~10-8 sec)
H. A. Strobel and W. R. Heineman, Chemical Instrumentation: A Systematic Approach, 3rd Ed. Wiley: 1989.
Ground state
(vibrational)
More on Raman Processes

The Raman process: inelastic scattering of a photon when it is incident on
the electrons in a molecule
– When inelastically-scattered, the photon loses some of its energy
to the molecule (Stokes process). It can then be experimentally
detected as a lower-energy scattered photon
– The photon can also gain energy from the molecule (anti-Stokes
process)


Raman selection rules are based on the polarizability of the molecule
Polarizability: the “deformability” of a bond or a molecule in response to an
applied electric field. Closely related to the concept of “hardness” in
acid/base chemistry.
P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3rd Ed. Oxford: 1997.
More on Raman Processes

Consider the time variation of the dipole moment induced by incident
radiation (an EM field):
Induced dipole moment
 (t )   (t ) (t )
EM field
polarizability

If the incident radiation has frequency  and the polarizability of the molecule
changes between min and max at a frequency int as a result of this
rotation/vibration:
 (t )    12  cos int t  0 cos t
mean polarizability

 = max - min
Expanding this product yields:
 (t )   0 cos t  14  0 cos(  int )t  cos(  int )t 
Rayleigh line
Anti-Stokes line
P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3rd Ed. Oxford: 1997.
Stokes line