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.

)sin(0 txx

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:

The frequency of the vibration is the number of waves per second and so is given by 1/T.

Thus,

2

1

T

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

)2sin(0 txx

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

So by simple substitution the time dependent displacement is:

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,

maF Using Hooke’s law and Newton’s second law

txt

mt

xmmakx 2sin

d

d

d

d02

2

2

2

txt

mkx 2cosd

d2 0Differentiating once:

xmtxmkx2

0

222sin2 And again:

22mk So:

A simple rearrangement leads to the relation

m

k

2

1

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

k

2

1

Where is the reduced mass

21

21

mm

mm

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 2 3

0% 0%0%

1. 300 – 400 Nm-1

2. 400 – 500 Nm-1

3. 500 – 600 Nm-1

12

27

27

227

21

21

13

9

18

9.5162Finally

10626.110673.1)351(

10673.1)351(mass reduced calc

10972.8108.3343

103frequency convert to

Nmk

kgkg

kg

mm

mm

Hzm

msc

Similar experiments were performed for the other hydrogen halides. The results are shown

below.

Halide

IR absorption / m k / Nm1 band wavenumber /

cm1

HF 2.416 966 4139

HCl 3.344 xxx 2990

HBr 3.774 412 2650

HI 4.329 314 2310

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.

C O

C O

bond typical band

wavenumber / cm1

1100

1700

CO 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 2

0%0%

1. True

2. False

114

8

14

14

2

27

2

44001044103

10320.1~

10320.1575

2

1

2

1

108365.022

diatomicr homonucleafor mass reduced

cmormc

Hzkg

kgsk

kgm

m

m

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?

2

~~2997

1099.810027.3

966

283.6

1

2

1Finally

10027.310673.1)192(

10673.1)192(mass reduced calc

966 so isotope oft independenconstant Force

1

113

21

27

27

27

227

21

21

1

HF

DFNotecmor

sk

kgkg

kg

mm

mm

Nmk

1 2 3 4

0% 0%0%0%

1. 2997cm-1

2. 2069cm-1

3. 4139cm-1

4. 8278cm-1

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: O=C=O should have 3(3) 5 = 4 modes. There are two stretching

modes:

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.

1500 2000 2500 3000 3500 4000-10

0

10

20

30

40

50

60

70

Infr

are

d (

Inte

nsity)

wavenumber (cm-1)

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:

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.

Tables and simulation results can help assign the vibrations!

(wavenumbers, cm-1)

GROUP

C-H Alkynes 3333-3267(s) stretch

700-610(b) bend

C=C Alkenes 1680-1640(m,w)) stretch

CC 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

O-H

Monomeric -- Alcohols, Phenols 3640-3160(s,br) stretch

Hydrogen-bonded -- Alcohols, Phenols 3600-3200(b) stretch

Carboxylic acids 3000-2500(b) stretch

N-H Amines 3500-3300(m) stretch

1650-1580 (m) bend

C-N Amines 1340-1020(m) stretch

CN Nitriles 2260-2220(v) stretch

NO2 Nitro Compounds 1660-1500(s) asymmetrical stretch

1390-1260(s) symmetrical stretch

Group Frequencies

v - variable, m - medium, s - strong, br - broad, w - weak

The Raman Effect

Polarization changes are

necessary to form the virtual

state and hence the Raman

effect

This figure depicts “normal”

(spontaneous) Raman effects

H. A. Strobel and W. R. Heineman, Chemical Instrumentation: A Systematic Approach, 3rd Ed. Wiley: 1989.

hv1

Scattering timescale ~10-14 sec

(fluorescence ~10-8 sec)

Virtual state

Virtual state

hv1

Ground state

(vibrational)

The incident radiation excites “virtual states” (distorted or polarized states)

that persist for the short timescale of the scattering process.

hv1 – hv2

Stokes line

hv1 – hv2

Anti-Stokes line

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):

)()()( ttt

P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3rd Ed. Oxford: 1997.

EM field Induced dipole moment

Expanding this product yields:

tttt )cos()cos(cos)( intint041

0

Rayleigh line Anti-Stokes line Stokes line

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:

ttt coscos)( 0int21

mean polarizability = max - min