Vibrational Spectroscopy

The Comparison between a Classical Harmonic Oscillator and Its Quantum Counterpart

According to classical mechanics, the position and energy of a harmonic oscillator are:

tAx 2sin and 1

2k

(1) The description of the motion CM: Coordinate (x) and velocity (dx/dt) tell the exact location of the particle QM: Wave functions tell the probability at certain location.

(2) The dimension of the motion CM: With energy E, magnitude of a harmonic oscillator is:

and the particle can be only found in [-A, A]. QM: Anywhere

kEA 2

(3)  The frequency of the motion CM:

how many times the particle travels in between x = -A and x = A

QM: A simplification of

analogous to the CM term.

12

k

(4)  Nodes and Wave Property  CM: 2 nodes, no wave property QM: v nodes, wave-particle duality

The force constant is a measure of the curvature of the potential energy close to the equilibrium extension of the bond. A strongly confining well (one with steep sides, a stiff bond) corresponds to high values of k.

A molecular potential energy curve can be approximated by a parabola near the bottom of the well. The parabolic potential leads to harmonic oscillations. At high excitation energies the parabolic approximation is poor (the true potential is less confining), and is totally wrong near the dissociation limit.

The Morse potential energy curve reproduces the general shape of a molecular potential energy curve. The corresponding Schrödinger equation can be solved, and the values of the energies obtained. The number of bound levels is finite. The illustration also shows the relation between the dissociation energy, D0, and the minimum energy, De, of a molecular potential energy curve.

Pure Vibrational Spectra of Diatomics

• Harmonic Oscillator Potential– Selection Rule

– Wavelengths of Transitions

• The series of equally spaced energy levels gives a single line in the spectrum.

e

e

hchEhEEE

~~;~;1 2/12/1

The anharmonic oscillatorReal bonds, although elastic for small compressions and extensions, do not strictly obey Hooke’s Law for amplitudes > 10% of bond length.Empirical expression derived by P.M. Morse Morse Function for the potential energy:

2( )

12 2

1

2

ea R Re

e

V hcD e

ahcD

Where a is a constant for a particular molecule

When this equation is used in the Schrödinger equation, the pattern of allowed vibrational energy levels is found to be:

2

1

1

1 1 cm 0, 1, 2, .... 2 2

is an oscillator frequency expressed in cm is the corresponding anharmonicity constant

e e e

e

e

G

e is always small and positive (~ +0.01) for bond stretching vibrations, so that the vibrational levels crowd more closely together with increasing .

This is only an approximation more precise expressions for the energy levels require cubic, quadratic, etc. terms in (+1/2) with anharmonicity constants ye, ze etc. rapidly diminishing in magnitude.these values are important only at very large values of and we shall ignore them.

1 1as 12 2e eG

Rewriting term values:

And compare with the energy levels of the harmonic oscillator

11 G cm 0, 1, 2, .... 2

we can write

1 12

osc

osc e e

Ehc

anharmonic oscillator behaves like the harmonic oscillator but with an oscillation frequency which decreases steadily with increasing This is the case for any real state with v specified as a positive integer.

Ground state ( = 0)

10

1

10

1 1 0 cm2

1since cm v 0, 1, 2, .... 2

1 cm2

e e

osc

osc

EGhc

G

Unlike the harmonic oscillator, e cannot be measured directly.

Wavenumbers for the (+1)transitions are given by:

1/ 2 ( 1) ( ) (2 2)e e eG G G

Measuring anharmonicity

• We ignore the higher order anharmonicity constants. The dissociation energy is then given approximately by:

• In order to determine the equilibrium frequency and anharmonicity constant, at least two transition wavenumbers must be obtained.

2

4e

ee e

D

Example problem• From the following separations of vibrational levels in the

ground electronic state of CO, obtain values of e, ee , and De.

• Note that in practice ee is used as a single constant. It is written as a product of a constant and frequency for historical reasons.

''' 1 2 3 6

G/cm1 2143.1 2116.1 2088.9 2061.3 2033.5 2005.5

Solution

• Plotting G vs. (2+2) gives a straight line of slope ee and intercept e. A least squares fit gives

e = 2171.1 ± 0.4 cm-1

e e = 13.8 ± 0.1 cm-1

21 185393 cm 1022 kJ mol

4e

ee e

D

Anharmonic Oscillator Spectra

• Selection Rule

• For most molecules only the first three energy transitions are important.

• Transitions

~

~2/1~2/1 2

hcEEE

hchcE eee

Important Transitions

• Fundamental (first harmonic)

– This transition will have strong intensity.

• First overtone (second harmonic)

– This is a weak intensity transition.

)21(~~10

ee

)31(~2~20

ee

Not So Important Transitions

• Second overtone (third harmonic)

– This transition is normally too weak to be observed.

• Only absorption from = 0 need to be considered below 200 C because virtually all of the molecules are in the ground vibrational energy state.

)41(~3~30

ee

Vibrational-Rotational Spectroscopy for Diatomics

• Harmonic Oscillator Approximation

• Anharmonic Oscillator

1;1

~)2/1()1(~~

J

hJJB ee

1;3,2,1

~2/1~)2/1()1(~~ 2

J

hhJJB eeee

Spectrum

• The vibration-rotation spectrum appears as series of lines called bands or branches, classified according to the J of the transition.

J= -2 -1 0 1 2– Branch O P Q R S

• Line Spacing– The spacing between lines is . eB~2

Lines

• Lines in the R branch become closer together at higher values of J.

• Lines in the P branch becomes farther apart at higher

values of J.

– The lines have different intensities due to different rotational populations.

Example: HCl

• There are double headed peaks due to H35Cl vs. H37Cl with an isotopic ratio of ~ 1:3

Consider P and R branches in more detail

• The higher energy state is for =1, the lower energy for =0 (this is the transition for the fundamental frequency).

• To a first approximation, the transitions within the branches are separated by .

• However, the rotational constant is inversely dependent on r2, and this increases with .

• Thus, .

,,~

JJ EEhcE

B~2

Why Unequal Spacings?

• R branch

– The second term shifts lines to higher frequency with increasing J.

– The third term is negative, and dependent on J2, so the lines become closer together in progressing along the series.

2)~~()~~()~21(~~1,1

JBBJBB

JJJ

eeR

P branch

• Here the second term is being subtracted, so the series is shifted to lower frequencies with increasing J.

• The third term now spreads the lines out in progressing along the series.

2)~~()~~()~21(~~1,1

JBBJBB

JJJ

eeP

- very intense absorption at 2886 cm-1

- weaker absorption at 5668 cm-1

- very weak absorption at 8347 cm-1

Example: pure vibration spectrum of HCl

To determine equilibrium frequency of the molecule – must solve any two of the 3 equations:

e (12e ) = 2886 cm-1

2e (13e ) = 5668 cm-1

3e (14e ) = 8347 cm-1

e = 2990 cm-1, e = 0.0174

for the ideal harmonic oscillator the spectral absorption occurs exactly at the classical vibration frequency, but for real anharmonic molecules the observed fundamental absorption frequency and the equilibrium frequency may differ considerably.The force constant of the bond in HCl may be calculated directly from the equation

1-5

222

cm dynes10 x 5.16

4

ck e

when the fundamental constants and the reduced mass are inserted

The dissociation energy is the sum of the separations of the vibrational energy levels up to the dissociation limit just as the length of a ladder is the sum of the separations of its rungs.

G is the transition wavenumber

The Diatomic Vibrating – non-rigid Rotor

A typical diatomic molecule has rotational energy separations of 1-10 cm1, while vibrational energy level separations of HCl were nearly 3000 cm1. Since the energies of the two motions are so different, as a first approximation consider that a diatomic molecule can execute rotations and vibrations quite independently(e.g., Born – Oppenheimer approximation

Etotal = Eelectronic + Evibration

+ Erotation)then Etotal = Evibration

+ Erotation

or total = vibration + rotation cm1

(later will see where this approximation does not apply)

Initially ignore the small centrifugal distortion constants D, H etc., and hence:

2

1, e e e

1 11 cm 2 2J vS BJ J v v

Note: retention of D has only very minor effect on the spectrum, however, it is not logical to ignore D as this implies that the molecule is rigid yet vibrating.

,

2 32 3

21

e e e

1 1 H 1 ......

1 1 - cm 2 2

J v J vS F G

BJ J DJ J J J

v v

Selection rules for the combined motions are the same as those for each separately.v = 1, 2, ….. J = 1(Actually may also have v = 0, but this corresponds to the purely rotational transitions.

Note, however, that a diatomic molecule, except under very special and rare circumstances, may not have J = 0, in other words a vibrational change must be accompanied by a simultaneous rotational change.)Designate rotational quantum numbers in the v = 0 state as J” and in the v = 1 state as J’

use of single prime for the upper stateand double prime for the lower state is conventional in

all branches of spectroscopy.

Rotational levels J” are filled to varying degrees in any molecular population, so the transitions will occur with varying intensities.

If consider only the v = 0 v = 1 transition, then:

, ', 1 ", 0

e e e e e e

10

0 e e

1 1 1 1 ' ' 1 1 2 " " 12 4 2 4

' " ' " 1 cm

where 1 2

J v J v J vS S S

BJ J BJ J

B J J J J

Note: taking B to be identical in the upper and lower vibrational states is a direct consequence of the Born-Oppenheimer approximation – rotation is unaffected by vibrational changes.

Now we can have:(i) J = + 1 i.e. J’ = J”+1 or J’ – J” = +1

1, 0 2 " 1 cm " 0,1, 2,.....J vS B J J

(ii) J = 1 i.e. J” = J’+1 or J’ – J” = 1

1, 0 2 ' 1 cm ' 0,1,2,.....J vS B J J

These two expressions may conveniently be combined into:1

, 0 2 cm J v specS Bm

Where m replaces J”+ 1 in eqn. (i) and J’+ 1 in eqn. (ii)m has a positive value for J = + 1m has a negative value for J = 1

Note: m can not be 0, since this would imply values of J’ or J” to be 1.The frequency is usually called the band origin or band center.

0

This equation represents the combined vibration-rotation spectrumIt will consist of equally spaced lines (spacing = 2B) on each side of the band originBut since m 0, the line at itself will not appear.

0 0

- Lines to the low frequency side of correspond to negative m (i.e., J = 1) are referred to as the P branch.

0

- Lines to the high frequency side of correspond to positive m (i.e., J = + 1) are referred to as the R branch.

0

Inclusion of the centrifugal distortion constant D leads to the following expression for the spectrum:

3 1, 0 2 2 cm 1, 2, 3.... J v spec Bm Dm m

Note: B is 10 cm1 or lessD is 0.01% of B

Since a good IR spectrometer has a resolving power of ~ 0.5 cm-1

D is negligible to a very high degree of accuracy

But, the anharmonicity factor is not negligibleIt affects not only the position of the band origin, since

0 e e 1 2 But by extending the selection rules to include v = 2, 3 etc. also allows the appearance of overtone bands having identical rotational structure.From the band centers can calculate the equilibrium frequency and the anharmonicity constant

Breakdown of the Born-Oppenheimer Approximation: the interaction of rotations and vibrations for diatomic vibrating -rotatorValue of B will depend on the v quantum numbere.g., for v = 0 v = 1, take respective B values as B0 and B1 with B0 > B1 This gives:

2 1, 0 1 0 1 0( ) 2( ) cm 1, 2, 3.... P R B B m B B m m

Where positive m values refer to the R branch and negative to PSince B0 > B1, the last term is always negative irrespective of the sign of m.The effect on the spectrum of a diatomic molecule is to crowd the rotational lines more closely together with increasing m on the R branch side, while the P branch lines become more widely spaced as (negative) m increases.Normally B1 and B0 differ only slightly and the effect is marked only for high m values.

Vibrating-Rotor for Diatomics

2, vib

2 2rot

3 2

2 2 2 2 2

1/ 2 1/ 2 E

( 1) ( 1) E1/ 2 ( 1) interaction

4 3 1 1;16

J e e e

e J

e

e eJ e

e e e e e

E h h

hB J J hD J Jh J J

B hDr D ar a r

As a molecule vibrates the r changes and hence I changes. The rotation vibration interaction term accounts for this effect.

Combination Differences

• This is a method used to determine the two rotational constants,

• It involves setting up expressions for the difference in the wavenumbers of transitions to a common state; the resulting expression then depends solely on properties of the other state.

10~ and ~ BB

Common Upper States

• The transitions

have a common upper state, and hence depend on

• From the equations above, it can be shown that

)1(~ and )1(~ JJ PR

1~B

)2/1(~4)1(~)1(~0 JBJJ PR

Obtaining Rotational Constants• A plot of the combination difference against J+1/2 should be

linear with a slope of , so the rotational constant of the molecule in the state =0 can be determined.

• Similarly, have a common lower state, so their combination difference

can be used to determine the rotational constant of the molecule in the state =1.

0~4B

)(~ and )(~ JJ PR

)2/1(~4)(~)(~1 JBJJ PR

Example Problem

• The fundamental infrared band of H35Cl has lines at the following wavenumbers: 2775.77, 2799.00, 2821.59, 2843.63, 2865.14, 2906.25, 2925.92, 2944.99, 2963.35, 2981.05, 2998.05 cm1. Use the method of combination differences to determine the rotational constants B0 and B1.

Procedure

1. Identify P and R branches, and assign J values to each line. Note that R branch (J = +1) starts at J = 0, but the P branch (J = 1) starts at J = 1.

2. Calculate the two series of combination differences.

3. Make plots and do linear regression to obtain slope = 4B.

Vibrational Spectra of Polyatomics

• Selection Rules

• In the harmonic approximation the vibrational quantum number may change for only one normal mode at once. – The motion must modulate the dipole moment of the

molecule, i.e., it must cause the dipole moment to oscillate in value.

1,0,2,1,0

J

Not all vibrational modes are IR active

• Example: CO2

• Both bends and the asymmetric stretch are active, but symmetric stretch is IR inactive. Why not?

• The normal mode must modulate the molecule's dipole moment. This is not the case for a symmetric stretch of any Dh molecule.

Dipole Moments

• If a molecule has a permanent dipole moment, all or most of its modes will modulate the dipole and thus be IR active.

• If there is not a permanent dipole, some modes can produce a fluctuating dipole.

• Determining whether a given normal mode will give rise to a vibrational band can often be done by inspection, i.e., look at the mode to see whether the motion leads to modulation of the dipole moment.

• Group theory can also be used, and is sometimes necessary for complicated cases.

Each normal mode may show Q, P, and R branches

• In CO2 the two bending modes can produce a motion that is the same as a rotation of a bent molecule.

• This results in J = 0 (Q branch) being allowed.

Combination bands are possible

• Two (or more) normal modes can change their quantum numbers at once. For a combination band each change is positive.

• E.g., = +1 for one mode = +1 for another mode

• The resultant wavenumber is about equal to the sum of the wavenumbers for the fundamental bands.

Difference bands

• Concept is similar to combination bands.

• One is positive, the other is negative.

• The resultant wavenumber is about equal to the difference between the fundamental bands.

Overtones

= +2, +3 for one band. Higher overtones are possible but will be extremely weak.

• The resultant wavenumber is approximately a multiple of the fundamental frequency.

• As with combination and difference bands, the difference from predicted values is due to anharmonicity effects.

Example spectrum: H2O

• Fundamental frequencies occur at 3756 cm-1 (asymmetric stretch), 3652 cm-1 (symmetric stretch) and 1595 cm-1 (bend).

Predict other observed bands.

Self-test

• Which of the following molecules have infrared active vibrations?Do not use group theory for this, only the modulating dipole rule.

• H2, NO, N2O (bent), OCS (linear), CH4

Coordinates• For a molecule with n atoms, 3n coordinates (aka

degrees of freedom) are needed to locate all the nuclei in three-dimensional space.

• Three coordinates can be used to specify the center of mass of the molecule.

• For a linear molecule two angular coordinates specify the orientation of the molecule. For a non-linear molecule three angular coordinates are required.

Normal Modes of Vibration• The remaining 3n-5 or 3n-6 coordinates may be chosen in a

certain manner to give normal coordinates.

• The vibrational motions of the normal coordinates are called normal modes of motion. Each normal mode is an independent, synchronous motion of atoms or groups of atoms that may be excited without leading to excitation of any other normal mode.

• Since each normal coordinate is a linear combination of cartesian coordinates of the nuclei, a normal mode is a concerted motion of some or all of the nuclei.

Normal modes for triatomics• For each triatomic there is a symmetric stretch, in

which both bonds lengthen or shorten together.

• There is also an asymmetric stretch, in which one bond lengthens while the other shortens.

• A linear triatomic can bend in two perpendicular directions, so there are two bending modes with the same frequency.

But…• A non-linear triatomic can only bend in the plane of

the molecule. Other motions which appear to be bends are in fact rotations.

• Only in special cases, such as CO2, are the normal modes purely stretches or purely bends. In general, a normal mode is a composite motion of simultaneous stretching and bending of bonds.

• Heavy atoms generally move less than light atoms in normal modes.

Normal modes in larger molecules

• In many cases in larger molecules some of the normal modes correspond to large oscillations of one bond length or angle while the remaining are nearly constant.

• E.g., ethane has a C—C and C—H stretches. The frequencies of such normal modes are nearly the same for the same pair of elements in different compounds. E.g., most hydrocarbons have C—H stretches in the range 2850 - 3000 cm-1.

• As is the case for triatomics, in general asymmetric stretches usually have the highest frequencies, symmetric stretches slightly lower, and bends have much lower frequencies.

Influence of Rotation on the Spectra of Polyatomic Molecules

Selection rules for the simultaneous rotation and vibration of a diatomic molecule was

= 1 J = 1 J 0Which gives rise to a spectrum consisting of ~ equally spaced line series on each side of a central minimum designated as the band centre.

Vibrations of complex molecules can be subdivided into those causing a dipole change either(i) parallel(ii) Perpendicular to the major axis of rotational symmetry.

The selection rules and the energies depend on the shape of the molecule also.

Consider first the linear molecule as the simplest and then consider other types of molecules.

The purpose of this distinction is that the selection rules for the rotational transitions of complex molecules depend on the type of vibration, parallel or perpendicular, which the molecule is undergoing.

Linear Molecules• Parallel Vibrations• - the selection rule is identical with that for

diatomic molecules• i.e., J = 1 = 1 for simple harmonic motion• J = 1 = 1, 2, 3, … for anharmonic

motion • (this is expected since a diatomic molecule is linear and

can undergo only parallel vibrations)• The spectra will thus be similar in appearance consisting

of P and R branches with lines about equally spaced on each side, no line occurring at the band center.

In general, as compared to a diatomic, for a polyatomic molecule the moment of inertia (dependent on mass) may be considerably larger, the B value (inversely dependent on I) correspondingly smaller, and the P or R line spacing (dependent on B) will be less.

e.g., compare the spacings between lines for HCN and CO HCN 2.8 – 3.9 cm1 CO 4.0 cm1

• For still larger molecules the value of B may be so small that separate lines can no longer be resolved in the P and R branches- can get a rough value of B from the separation between the maxima of the P and R envelopes.

• NOTE: a nonlinear molecule cannot give rise to this type of band shape, so its observation somewhere within a spectrum is sufficient proof that a linear or nearly linear molecule is being studied.

- the selection rule is

for simple harmonic motion

Which implies that now, for the first time, a vibrational change can take place with no simultaneous rotational transition.

Perpendicular vibrations eliminate the linearity of the molecule, so unlike parallel vibrations, a Q branch is observed.

(ii) Perpendicular Vibrations

P and R branch lines

1, 0 2 cm 1, 2, 3.... J specS Bm m

3 1, 0 2 2 cm 1, 2, 3.... J specS Bm Dm m Or

The P and R lines are given as before by equations:

-If the oscillation is taken as simple harmonic the energy levels are identical with those of

2

1, e e e

1 11 cm 2 2JS BJ J

Transitions with J = 0 , however, correspond to a Q branch whose lines may be derived from the equations

, 1 ,

e e e e e e

10

1 1 1 1 1 2 ( 1) 12 4 2 4

cm for all

J JS S S

BJ J BJ J

J

Thus the Q branch consists of lines superimposed upon each other at the band center one contribution arising for each of the populated J values.- The resultant line is usually very intense

e

Q Branch

, 1 ,

e e e e e e

10

1 1 1 1 1 2 ' ( 1) " 12 4 2 4

( 1)( ' ") cm

J JS S S

B J J B J J

J J B B

Further, if B’<B”, we see that the Q branch line would become split into a series of lines on the low frequency side of (since B’B” is negative)0

If we take into account the fact that the B values differ slightly in the upper and lower vibrational states (breakdown of the Born-Oppenheimer approximation), we would write instead:

• Normally, however, B’-B” is so small that the lines cannot be resolved and all the Q branch appears as a somewhat broad resonance centered around 0

The formation of P, Q, and R branches in a vibration-rotation spectrum. The intensities reflect the populations of the initial rotational levels.

Note: polyatomic molecules with zero dipole moment do not give rise to pure rotation spectra in the microwave region (e.g., CO2, HCCH, CH4).

Such molecules do however show vibrational spectra in the infrared region (or Raman) and if these spectra exhibit resolved fine structure, the moment of inertia of the molecule can be obtained.

Symmetric Top MoleculesFollowing the Born-Oppenheimer approximation we can take the vibrational-rotational energy levels for this type of molecule to be the sum of the vibrational levels

21

e e e1 1 cm 0, 1, 2, 3... 2 2vibG

And the rotational levels

2 11 ( ) cm 0,1,2...

, ( 1), ( 2)......rotF BJ J A B K J

K J J J J

2

, e e e

2 1

1 1 2 2

( 1) ( ) cm

J v vib rotS G F

BJ J A B K

Again it is necessary to divide the vibrations into those which change the dipole

(i) Parallel(ii) Perpendicular to the main symmetry axes – which is nearly always the

axis about which the “top” rotatesThe rotational selection rules differ for the two types

This equation assumes, of course, that centrifugal distortion is negligible.

Parallel Vibrations• Selection rules = 1, J = 0, 1, K =

0• Exception: For K=0, J = 1, thus no Q branch is

allowed.• Since here K = 0, terms in K will be identical in

the upper and lower state and so the spectral frequencies will be independent of K

• Therefore the situation will be identical to that discussed for the perpendicular vibrations of a linear molecule

• The spectrum will contain P, Q, R branches with a P, R line spacing of 2B (which is unlikely to be resolved) and a strong central Q branch.

• The intensity of the Q branch (relative to lines in the P and R branches) varies with the ratio Ia/Ib

• In the limit when Ia 0, the symmetric top becomes a linear molecule and the Q branch has 0 intensity.

10 2 1 ( ) 1 2 cm spectS B J A B K

(a) (R branch lines)

2

, e e e

2 1

1 1 2 2

( 1) ( ) cm

J v vib rotS G F

BJ J A B K

Selection rule is = 1, J = 0, 1, K = 1Each of the following expressions are readily derivable for the spectral lines taking the energy levels of the equation

(ii) Perpendicular vibrations

(b) J = 1 K = 1 (P branch lines)

10 2 1 ( ) 1 2 cm spect B J A B K

(c) J = 0 K = 1 (Q branch lines)

10 ( ) 1 2 cm spect A B K

Therefore this type of vibration gives rise to many sets of P and R branch lines since for each J value there are many allowed values of K (K = J, J1,…. –J)The wings of the spectrum will therefore be quite complicated and will not normally be resolvable into separate lines.

This latter term may not be small (and is equal to zero only for spherical top molecules which have all their moments of inertia equal)For A>>B (e.g., CH3I) the Q branch lines will be well separated and will appear as a series of maxima above the P, R envelope.In the IR spectra of even the simple diatomic molecule may contain a great many lines- That of a polyatomic molecule may be extraordinarily complex even though some of the details of fine structure are blurred by insufficient resolving power.

The Q branch is also complex since it too will consist of a series of lines on both sides of separated by 2(AB)0

Assignments are based mainly on experience with related molecules, on the band contour (from which the type of vibration (|| or) can usually be deduced), and on the use of Raman spectra

Consideration of the symmetry of the molecule is also important because this determines which vibrations are likely to be IR active.

Although in favourable cases much information may be obtained about bond lengths and angles or at least the general shape of a moleculeIn others even the assignment of observed bands to particular molecular vibrations is not trivial

Spherical tops• For tetrahedral and octahedral molecules, the

vibrational selection rules become more restrictive. Symmetry limits the number of allowed vibrational transitions (more later).

• Neglecting centrifugal distortion, rotational term values are given by

and selection rules are

• The spectrum will be similar to the perpendicular band of a linear molecule.

( ) ( 1)F J B J J

Vibrational Raman Spectra of Diatomic Molecules

Gross selection rule for a vibrational Raman transition: polarizability should change as the molecule vibrates.

Both homonuclear and heteronuclear diatomic molecules are vibrationally Raman active.

As they swell and contract during a vibration, the control of the nuclei over the electron varies, and hence the molecular polarizability changes.

Vibrational selection rule = 1harmonic approximation

Lines to the high frequency side of incident light • – anti-Stokes lines from = 1• -usually weak because very few molecules are in excited

vibrational state initially Lines to the low frequency side of incident light • – Stokes lines from = + 1• gas-phase spectra show branch structure arising from

simultaneous rotational transitions that accompany the vibrational excitation

• Selection rules are J =0, 2 as in pure rotational Raman spectroscopy O branch (J = 2), Q branch (J = 0), S branch (J = + 2)

The formation of O, Q, and S branches in a vibration-rotation Raman spectrum of a linear rotor.

2 4

6 4

O i

Q i

S i

J B BJ

J

J B BJ

Q branch arises from differences in rotational constants of upper and lower rotational states.

Note that the frequency scale runs in the opposite direction to that in IR, because the higher energy transitions (on the right) extract more energy from the incident beam and leave it at lower frequency.

The structure of a line in the vibrational Raman spectrum of carbon monoxide, showing the O, Q, and S branches.

Note: unlike IR spectroscopy, a Q branch is obtained for all linear molecules

Vibrational Raman Spectra of Polyatomic Molecules

• A transition will be Raman active if one of the six different components of the polarizability tensor (which is symmetric) changes during the vibration.

• For our purposes, this is equivalent to the symmetry species of the normal mode being the same as the symmetry species of a quadratic form (see character table).

xx xy xz

yx yy yz

zx zy zz

Depolarization

• Assignment of Raman lines to particular vibrational modes aided by noting the state of polarization of scattered light.

• The depolarization ratio, , of a line is the ratio of the intensities of the scattered light with polarizations perpendicular and parallel to the plane of polarization of the incident radiation:

Depolarization planes

• The fat arrows represent the electric vector of the incident (green) and scattered (orange) radiation. There is also a perpendicular scattered component, as indicated by the simple wave-like line.

Measuring

• Intensity of a Raman line is measured with a polarizing filter, first parallel and then perpendicular to the polarization of the incident beam.

• If the emergent light is not polarized, then both intensities are the same and is close to 1. If the light retains its initial polarization, then 0 so = 0

Classification

• Line classified as depolarized if it has equal to or greater than 0.75, and as polarized if < 0.75.

• Only totally symmetric lines give rise to polarized lines in which the incident radiation is largely preserved. Vibrations which are not totally symmetrical give rise to depolarized lines because the incident radiation can result in radiation in the perpendicular direction as well.

Assignment• Bands which are polarized can be immediately

assigned to the totally symmetric representation and the depolarized bands can be assigned to modes of any other symmetry.

• The more highly symmetric the molecule, the closer to zero will be for a totally symmetric vibration.

• These vibrations usually give rise to the strongest Raman bands.

Example- CCl4 (Td)

CCl4 spectrum

• The strong band at 459 cm1 almost disappears when the polarizers are perpendicular.

• The bands at 314 and 218 cm1 have depolarization ratios of 0.75.

• The 459 cm1 band must have A1 symmetry, and the ones at 314 and 218 cm1 cannot have A1 symmetry.