Laser Molecular Spectroscopy CHE466 Fall 2009

Laser Molecular SpectroscopyCHE466Fall 2009

David L. Cedeño, Ph.D.Illinois State University

Department of Chemistry

Vibrational Spectroscopy

Diatomic Molecules: The Harmonic Oscillator Model

Vibrational motion is that related to the displacement of nuclei relative to each other along a given axis (called a stretching) or a common plane (bending). Vibrational motions represent most of the nuclear motion in a molecule. Recall that there are (3N – 6) or (3N – 5) vibrations for non-linear and linear molecules respectively, with N = # atoms. In order to describe vibrational motions we assume a classical model in which nuclei are represented by spheres of various masses joined by springs of different thickness (force constants). For a diatomic molecule:

If the motion is described using Newtonian physics (classical) then the motion of one sphere relative to the other along the dimension x (molecular axis) would be harmonic, i.e. the motion will be oscillatory around an equilibrium position (re when atoms are at rest).

Vibrational Motion

re

x axis


The kinetic (T) and potential energies (V) are:

k is the force constant of the spring and is the reduced mass. The potential energy is a function of the displacement relative to equilibriumx = r – re

Vibrational Motion

2)(

22

2

kxxV

pT x

Potential Energy for k = 70 N/m, m = 6.0 x 10-27 kg

0.00E+00

1.00E-20

2.00E-20

3.00E-20

4.00E-20

5.00E-20

6.00E-20

-5.00E-11

-4.00E-11

-3.00E-11

-2.00E-11

-1.00E-11

0.00E+00

1.00E-11

2.00E-11

3.00E-11

4.00E-11

5.00E-11

x (m)

V (x

) (in

J)


Of course, molecular vibrational motion cannot be described using classical physics. Thus a quantum mechanical solution is required. The Hamiltonian is given by

The energy is obtained using the Schrodinger equation and appropriate wavefunctions. These have the form:

and Hv are the Hermite polynomials:

Vibrational Motion

22ˆ

2

2

22 kxx

H

2

2/2/1v

4/1

vv

with

)(!v2

1)(2

k

exHx x

v Hv(1/2x) 0 1 1 21/2x 2 4x2 – 2 3 83/2x3 –121/2x 4 162x4 –48x2 + 12


The solution of the Schrodinger equation for the energy of the vibrational motion is quantized. In other words, there are vibrational states (with v = 0, 1, 2, …) that have an specific energy Ev:

Ev = he(v + ½)

Where

is the so-called harmonic frequency.

Vibrational Motion

k

e 21

Energy Diagram for a harmonic oscillator with e = 1.72 x 1013 Hz

PROBABILITY DENSITY

0.00E+00

1.00E-20

2.00E-20

3.00E-20

4.00E-20

5.00E-20

6.00E-20

-7.00E-11 -5.00E-11 -3.00E-11 -1.00E-11 1.00E-11 3.00E-11 5.00E-11 7.00E-11

x (m)

Ener

gy (J

)

v = 0

v = 1

v = 2

v = 3

v = 4


Note that a consequence of the quantum nature of vibrational motion is that a molecule will always have a vibrational energy (v = 0). This is called the zero point energy.For a diatomic molecule using the harmonic oscillator model:

E0 = ½he

Note that the harmonic oscillator model predicts that the spacing between adjacent vibrational levels is the same and equal to he.

Example: Find the zero point energy and spacing between vibrational states of the HI molecule. Use k = 4.0277 x 104 N/m and assume a harmonic oscillator model.

Vibrational Motion

Diatomic Molecules: The Transition dipole integral

We learned that the transitions probability for a rotational transition is extremely small (for practical reasons zero) if the molecule does not have a permanent dipole moment. Is that also the case for vibrational motion?Since in a vibration there is a displacement of nuclei realtive to each other, the dipole moment may not be constant, it may vary with the displacement (x). We can represent that variation using the following equation:

Where e is the permanent dipole moment and the other term is the first order variation of the dipole moment with displacement. It turns out that based on this, a molecule that does not have a permanent dipole moment (i.e. e = 0) could be excited vibrationally (i.e. the probability of transitions is not negligible) as long as the vibrational motion induces a transitory dipole moment and the transition rule related to changes in v number does not vanish the transition dipole integral:

Vibrational Motion

....μμ)(μe

e

x

dxdx

dxxdxd

dxxdxddx

dxxdxd

v"*v'

ev'v"

v"*v'

ev"

*v'ev'v"

v"e

e*v'v'v"

μR

μμR

μμR

Diatomic Molecules: Transition Rules under the Harmonic Oscillator Model

Based on the previous equation, homonuclear diatomic molecules cannot be vibrationally excited with light, because they do not have a permanent dipole moment, and its vibrational motion cannot create a transitory one.On the contrary, heteronuclear diatomic molecule are able to absorb infrared radiation (strongly). The transition rule is:

v = ± 1

Most molecules at room temperature are in the lowest vibrational state (v = 0), thus the transition that is more likely to be observed is the one corresponding to v” = 0 to v’ = 1. This is called the fundamental transition.An increase in temperature may populate other vibrational states (say v = 1 and 2), thus in adition to v = 0, thus the transitions v” = 1 to v’ = 2, and v” = 2 to v’ = 3 are also possible. Bands associated to this transitions are called hot bands.

Vibrational Motion

Diatomic Molecules: Anharmonicity

It is obvious that a harmonic potential energy does not represent vibrational motion accurately, especially at large values of v (i.e. highly excited vibrational states). Pictorially, putting a large amount of energy in a vibration does not translate into infinite elongation (x) of the bond. Instead, the coulombic and covalent attraction of the nuclei will be overcome and the molecule will dissociate. Also, compression of the bond at a very large displacement will induce repulsive forces that increase the potential energy at a larger rate than predicted by a harmonic oscillator model.

Vibrational Motion

Potential Energy Curve for H2

-10

90

190

290

390

490

590

-0.5 0.5 1.5 2.5x (Angstrom)

V(x)

(kJ/

mol

)

Morse

Harmonic


Morse proposed a potential that represents diatomic molecules very well:

De is the dissociation energy of the molecule (relative to V(x) = 0) and is the Morse parameter for the molecule (it defines the width of the potential curve).

A solution to the Schrodinger equation is still possible upon using wavefunctions that are slightly different from the ones obtained using the harmonic oscillator model. The vibrational energy is given by:

Ev = he(v + ½) – hexe(v + ½)2 + heye(v + ½)3 + …

The parameters xe and ye are the first and second order anharmonicity constants. In most cases (for routine work) the energy is expressed as a two term equation (i.e. depending up to (v + ½)2), but more terms may be necessary.

Vibrational Motion

21)( xe eDxV


A direct results of anharmonicity is that the vibrational states of a given molecule are not evenly spaced. Indeed the energy difference between adjacent levels decreases as v increases.

Vibrational Motion


The energy spacing between adjacent levels is given by:

E = Ev’ – Ev” = he – 2heexe (v”+ 1)

Which can be written in wavenumbers as:

= e – 2exe (v”+ 1)

Note then that if have the wavenumbers for v = 1 transitions, a linear plot of w vs. v’ would give e and exe from the intercept and the slope.

Vibrational Motion

Diatomic Molecules: Anharmonicity, zero point energy and dissociation energy

The zero point energy corrected for anharmonicity is equal to:

E0 = ½ he – ¼ heexe

The dissociation energy can be defined either in terms of the bottom of the potential curve (De) or

more realistically from the zero point energy (D0

from v = 0)

Vibrational Motion

-10

90

190

290

390

490

590

-0.5 0.5 1.5 2.5 3.5x (Angstrom)

V(x)

(kJ/

mol

)

D eD 0

Diatomic Molecules: Anharmonicity, zero point energy and dissociation energy

D0 can be obtained form the E = Ev’ – Ev” = he – 2heexe (v”+ 1) plot as an integral or summation

of all the spaces between adjacent levels (i.e. E) extrapolated to the maximum v possible, assuming that the extrapolation is valid. De can be calculated as: De = D0 + E0

Vibrational Motion

Birge-Sponer Plot for HI

y = -79.363x + 2309.5

-500

0

500

1000

1500

2000

2500

0 5 10 15 20 25 30 35

v"+1

(c

m-1

)

Diatomic Molecules: Transition rules

The anharmonicity of the potential perturbs the wavefunctions and breaks the harmonic transition rule. Due to anharmonicity, it is possible to reach higher vibrational states via light absorption because allowed transitions correspond to:

v = ±1, ±2, ±3, …

Most common transitions are those originating from v” = 0, thus it is possible to observe bands for:

v” = 0 → v’ = 1 (Fundamental band)v” = 0 → v’ = 2 (first overtone band)v” = 0 → v’ = 3 (second overtone band) et cetera.

Usually the intensities of the overtone bands are reduced dramatically (about 10v’-fold as v’ increases)

Vibrational Motion

Diatomic Molecules: Transition rules

Example: CO FTIR spectra of fundamental, and first and second overtone transitions. Mina-Camilde et al., J. Chem. Ed., 1996, 73, 804-807.

Vibrational Motion

Diatomic Molecules

So far, vibrational motion has been described ignoring the effect of rotational motion on the spectra. Since the photons exciting a vibration contain enough energy to excite the molecule rotationally, one cannot ignore rotational motion. The figure shows a typical high resolution spectrum of a fundamental vibrational transition for a diatomic molecule

Vibrational-Rotational Spectroscopy

Diatomic Molecules

The main band consists of two groups of narrow bands, separated by a gap at about the center of the main band


(cm-1)

Diatomic Molecules

Each band corresponds to a rotational transition occurring simultaneously with the vibrational transition. The rotational transitions correspond to J = ± 1 (transition rule)


v” = 0J” = 0

v’ = 1J’ = 0

J’ = 1

2

3

4

J” = 1

2

3

4

Transitions with J = 1 occur at larger frequencies than those with J = –1 (see red arrows vs. blue arrows). Thus they group apart from each other in the spectrum. The set of bands at higher frequency is called the R branch (J = 1), while the other set is the P branch (J = -1). The gap in between is a consequence of the absence of J = 0 transitions in diatomic molecules.The lowest frequency R band corresponds to J” = 0 → J’= 1, while the highest energy P band corresponds to J” = 1 → J’= 0. They are denoted R(0) and P(1) respectively (in wavenumbers).

Diatomic Molecules: Energy and Spectral Bands

The energy of a molecule is then the sum of rotational and vibrational energies:

Ev,J = Ev + EJ = he(v + ½) – hexe(v + ½)2 + BvJ(J +1) – DJJ2(J +1)2

With Bv = Be – (v + ½), where a is the vibration-rotation coupling constant.

Therefore the spectral bands correspond to E = Ev’,J’ – Ev”,,J” , which can be separated into R

and P bands (in wavenumbers). For the fundamental vibrational transition (v” = 0 to v’ = 1):

Where 0,1 = e – 2exe


3J

201010,1R 1)(J"4D1))(J"B(B1))(J"B(Bω)(J"ω

3J

201010,1P J"4D)J"B(B)J"B(Bω)(J"ω

Diatomic Molecules: Energy and Spectral Bands

If centrifugal distortion is neglected (i.e. DJ ≈ 0), then:

Where 0,1 = e – 2exe. This is the so-called band center of the fundamental vibrational transition.It is possible to obtain all the spectroscopical constants B1, B0, e, exe, Be, and from the analysis of vibration-rotational spectra.This implies using a combination of the equations above in a way that other equations are created in a way that the dependence on some f the parameters is cancelled out. This methodology is called a Combination of Differences approach.


2010110,1R )J"B(B)J"B(3BB2ω)(J"ω

201010,1P )J"B(B)J"B(Bω)(J"ω

Diatomic Molecules: Spectroscopical Constants and the Combination of Differences Method

•Obtaining B1 and B0 and DJ

If one subtracts P(J”) fromR(J”) the following is obtained:

Which can be written as:

Thus a linear plot can be obtained, in which the slope is -4DJ and the intercept is 2B1.

Similarly, if one subtracts P(J”+1) fromR(J”-1) the following is obtained:

Which can be written as:

Thus a linear plot can be obtained, in which the slope is -4DJ and the intercept is 2B0.


1)J"1)(J"(2J"4D1)(2J"2B1)(J"ω)1(J"ω)ω(J"Δ 2J0PR

"2

1)J"1)(J"(2J"4D1)(2J"2B)(J"ω)(J"ω)ω(J"Δ 2J1PR

'2

1)J"(J"4D2B1)(2J"

)ω(J"Δ 2J1

'2

1)J"(J"4D2B1)(2J"

)ω(J"Δ 2J0

"2

Diatomic Molecules: Spectroscopical Constants and the Combination of Differences Method

•Obtaining Be and

Given that Bv = Be – (v + ½), then at least two values of Bv are needed to obtain Be and .

•Obtaining e and exe

If P(J”) fromR(J”+1) are added the following is obtained:

Thus a linear plot can be obtained with an intercept equal to 0,1 = e – 2exe.

Knowledge of other band centers, for instance the one for the first overtone band (v” = 0 to v’ = 2, 0,2) allows one to obtain e and exe. It can be shown that any overtone, the band

center has a wavenumber equal to:


eeev'0, x1)ω(v'v'ωv'ω

2010,1PR 1))(J"B2(B2ω1)(J"ω)(J"ω

Diatomic Molecules: Intensities

The intensities of the P and R bands are dictated by Boltzmann’s statistics, as is the case for the population of rotational levels of the initial state:

Where EJ” is the rotational energy.


kTE

JJ

eJNN "

)1"2(0

"

Diatomic Molecules:transitions

It is possible to obtain the Raman vibrational and rotational spectra of diatomic molecules. The vibrational transition rule still holds (i.e. v = ±1, ±2, ±3,…) as well as the rotational transition rule (J = 0, ±2). The interesting event is that a vibrational-rotational band in the Raman has three branches: S (for J = +2), Q (for J = 0) and O (for J = -2)

Vibrational-Rotational Raman Spectra

v” = 0J” = 0

v’ = 1J’ = 0

J’ = 1

2

3

4

J” = 1

2

3

4

Diatomic Molecules:

Below is the Raman spectrum of the fundamental band of 12C16O. Notice that most Q bands overlap.


Also, the S band with the lowest energy correspond to J” = 0 and denoted S(0), while the O band

with the highest energy correspond to J” = 2 and denoted O(2) in wavenubers.

Diatomic Molecules:

Neglecting the effect of centrifugal distortion:

A combination of differences method can be used to determine B0, and B1:

0,1 is obtained from the origin of the Q band (i..e Q(0)):

Q(0) = 0,1


2010110,1S )J"B(B)J"B(5BB6ω)(J"ω

1)(J")J"B(Bω)(J"ω 010,1Q

2010110,1O )J"B(B)J"B(3BB2ω)(J"ω

1)(2J"4B2)(J"ω2)(J"ω)ω(J"Δ

1)(2J"4B)(J"ω)(J"ω)ω(J"Δ

0OS"4

1OS'4

Polyatomic Molecules

Normal ModesAs seen previously, the number of vibrational modes of motion increases proportionally to the number of atoms in the molecule (as 3N – 5 in linear molecules, or 3N – 6 in non-linear ones). Each vibrational motion may be, however, involve the combination of the motions of various atoms in terms of both stretching and bending.It is possible to break down these complicated motion into a set of harmonic motions that move at the same frequency, and in phase. Such a vibrational mode is called a normal mode.For example in water, the vibrational motion of the molecule is expressed in terms of 3 normal modes, which all of you already are familiar with:


A1 B2 A1



1 3 2

It is possible to solve the Schrodinger equation for the molecules in terms of the energy of each individual normal mode. It turns out that each mode has a vibrational quantum number associated to it (vi). Thus the vibrational energy of each mode assuming a harmonic motion is given by:

Ei = hi(vi + ½)

If a normal mode is degenerate, then:

Ei = hi(vi + di/2)

Where di is the degeneragy.The total vibrational energy of the molecule is the summation of the normal mode energies:

63N

iivib EE



In the water molecule, the 3 normal modes have associated quantum numbers v1, v2, and v3. In triatomic molecules (like water), the bending mode is denoted with the label 2, while the symmetric stretch is label 1. The transition rules are the same as for diatomic molecules: v = ±1, ±2, ±3…, which means that there is a possibility to have as many as 3N – 6 fundamental, and overtone bands. Fortunately, there are some restrictions based on symmetry to the probability of a transitions.Besides the fundamental and overtone bands, there is the possibility of having combination bands, which originate from the excitation of two normal modes simultaneously. Thus for water, the following is possible:

Three fundamental bands: vi” = 0 → vi’ = 1 (i = 1, 2 or 3)

Various first overtones: vi” = 0 → vi’ = 2, 3, 4, etc., (i = 1, 2, or 3)Various combination bands, see some examples:

v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 1, v3’ = 1

v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1

v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 1, v3’ = 0

Let us introduce some notation to account for the transitions. There are two notations, which are used by spectroscopists to distinguish a particular transition. Let us see some examples:

Sym. Stretch of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 0. Notation:

Notation: |0,0,0> → |1,0,0>

1st overtone of the bending mode of water:v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 2, v3’ = 0. Notation:

Notation: |0,0,0> → |0,2,0>Combination of sym. stretch and asym. stretchv1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1. Notation:

Notation: |0,0,0> → |1,0,1>

Polyatomic Molecules: Notation


101

202

10

1031

Let us introduce some notation to account for the transitions. There are two notations, which are used by spectroscopists to distinguish a particular transition. Let us see some examples:

Sym. Stretch of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 0. Notation:

Notation: |0,0,0> → |1,0,0>

1st overtone of the bending mode of water:v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 2, v3’ = 0. Notation:

Notation: |0,0,0> → |0,2,0>Combination of sym. stretch and asym. stretchv1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1. Notation:

Notation: |0,0,0> → |1,0,1>

Polyatomic Molecules: Notation


101

202

10

1031

Polyatomic Molecules: An IR spectrum of CO2 (gas phase)


Bands:

201 : 667.5 cm-1

301 : 2349 cm-1

2

3

Polyatomic Molecules: An IR spectrum of N2O (gas phase)


2

2

1

3

2+ 1

1

2+ 3

1+ 3

Bands:

201 : 588.8 cm-1

202 : 1167.0 cm-1

101 : 1285.0 cm-1

301 : 2223.5 cm-1

101 20

2 : 2461.5 cm-1

102 : 2563.5 cm-1

201 30

1 : 2798.5 cm-1

101 30

1 : 3498.1 cm-1

Polyatomic Molecules: Selection Rules and Symmetry

Why is that all fundamental bands of N2O, and H2O (both bent molecules) are observed in the IR spectrum, but for CO2 only two of them are observed?Because of the symmetry of the molecule. Recall that IR intensities are proportional to the strength of the dipole moment that is induced by the vibrational motion. In other words the transition dipole integral Rv’v” must be valued (i.e. ≠ 0). There is a symmetry requirement that implies that the quantity to be integrated must be totally symmetric or contain a term that is totally symmetric:

Г(v’) x Г() x Г(v”) ⊇ A

Since the dipole moment follow the symmetry of translational motion then the equation above can be expressed as:

Г(v’) x Г(Ti) x Г(v”) ⊇ A (with i = x, y, or z)

If the lower state is the ground state (i.e. vi” = 0), then the wavefunction is totally symmetric and Г(v’) = A. Thus the equation reduces to:

Г(v’) x Г(Ti) ⊇ A (with i = x, y, or z)

Meaning that allowed transitions will be those with wavefunctions that include or follow the symmetry representation of the dipole moment (i.e. translational motion) in any direction:

Г(v’) ⊇ Г(Ti) (with i = x, y, or z)



EXAMPLE 1: Fundamental Transitions of Water

1. Determine symmetry group: C2v

2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For water these are A1, B1 and B2.

3. Allowed transitions that originate from the ground state are those that correspond to normal

modes that correspond to A1, B1 and B2 symmetry. In the case of water: Гvib = 2A1 + B2, so all

fundamental bands are IR active.


A1 B2 A1

Sym Stretch Asym Stretch Bending


EXAMPLE 2: Fundamental Transitions of CO2

1. Determine symmetry group: D∞h

2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For CO2 these are A2u and E1u.

3. Allowed transitions that originate from the ground state are those that correspond to normal

modes that correspond to A2u and E1u symmetry. In the case of CO2: Гvib = A1g + A2u + E1u, so

only A2u (asym. stretch) and E1u (two degenerate bending) are IR active.


A1g A2u

+ +

E1u

Sym Stretch Asym StretchBending

Note: For D∞h: A1g is also notated as g+, A2u as u

+ and E1u as u.


Cross Product Symmetry Rules: This are required to find the symmetry representation of overtones and combination transitions.

1. Single degeneracy: A x A = A A x B = B B x B = A1 x 1 = 1 1 x 2 = 2 2 x 2 = 1g x g = g g x u = u u x u = g

Example: A1g x A2u = A2u A1u x A2u = A2g

2. Degenerate: Depends on symmetry group, but overall must multiply characters and then reduce the result.

Example: Determine cross product of E x E for group C3v:

3. Tables of cross products rules are available in the course website. Some worth noting are the one for the linear molecules. For instance the notation is different


C3v I 2C3 3v A1 1 1 1 Tz A2 1 1 -1 Rz E 2 -1 0 (Tx,Ty), (Rx, Ry)

The reducible representation is: 4 in E, 1 in 2C3, and 0 in 3v.Reduction yields:

E x E = A1 + A2 + E


EXAMPLE 3: Overtone Transitions of CO2


2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For CO2 these are A2u and E1u.

3. 1st overtone of asym. stretch (i.e. transition 302): In this case the cross product of the

symmetry representation is required. The fundamental (301)corresponds to symmetry A2u, thus

the overtone corresponds to symmetry A2u x A2u = A1g. Therefore, the overtone 302 has a

symmetry representation that DOES NOT contain one of the translational representations, thus the band is IR inactive.

4. 1st overtone of bending (i.e. transition 202): In this case the cross product of the symmetry

representation is required. The fundamental (201)corresponds to symmetry E1u, thus the

overtone corresponds to symmetry E1u x E1u = A1g + A2g + E2g (Also notated as: u x u = g+

+ g– + g). Then, this overtone (20

2) has a symmetry representation that DOES NOT contain

one of the translational representations, thus the band is IR inactive.



EXAMPLE 4: Overtone Transitions of Water


2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A1, B1 and B2.

3. 1st overtone of asym. stretch (i.e. transition 302): In this case the cross product of the

symmetry representation is required. The fundamental (301)corresponds to symmetry B2, thus

the overtone corresponds to symmetry B2 x B2 = A1. In other words, the first overtone has a

symmetry representation that contains one of the translational representations, thus the band is IR active.

4. 2nd overtone of asym. stretch (i.e. transition 303): In this case the cross product of the

symmetry representation is required. The fundamental (301)corresponds to symmetry B2, thus

the overtone corresponds to symmetry B2 x B2 x B2 = B2. Then, the second overtone has a

symmetry representation that contains one of the translational representations, thus the band is IR active.



EXAMPLE 5: Combination Transitions of NH3


2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A1, E.

3. The representation for vibrational motions is Гvib = 2A1 + 2E. Sym. Stretch (v1) and a bending

(v2) are A1, Asym. stretch (v3) and another bending (v4) are E.

4. Combination (10130

1): In this case the cross product of the symmetry representation is

required. Two fundamental are combined. They (101 and 30

1)correspond to symmetry

representations A1 and E respectively. Thus the combination corresponds to symmetry A1 x E

= E. Thus, the combination has a symmetry representation that contains one of the translational representations, thus the band is IR active.



required. Two fundamental are combined. Both (101 and 20

1) correspond to symmetry

representation A1. Thus the combination corresponds to symmetry A1 x A1 = A1 . Thus, the

combination has a symmetry representation that contains one of the translational representations, thus the band is IR active.



EXAMPLE 6: Combination Transitions of acetylene (C2H2)


2. Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A2u and E1u.

3. The representation for vibrational motions is Гvib = 2A1g + A2u + E1g + E1u. Sym C-H and C≡C

stretchs (v1 and v2) are A1, Asym. C-H stretch (v3) is A2u, the trans bending (v4) is E1g and the

cis bending is E1u


v1, A1g v2, A1g v3, A2u

+ + + +- - - -

v4, E1g v5, E1u

Note that out of this 5 fundamental transitions, only 30

1 and 501 are IR active.


EXAMPLE 6: Combination Transitions of acetylene (C2H2)



required. Two fundamental are combined. They (101 and 30

1)correspond to symmetry

representations A1g and A2u respectively. Thus the combination corresponds to symmetry

A1g x A2u = A2u. Thus, the combination has a symmetry representation that contains one of the

translational representations, thus the band is IR active.



required. One fundamental and one overtone are combined. The fundamental (201 and 30

1)

correspond to symmetry representations A1g and A2u respectively. Thus the combination

corresponds to symmetry A1g x (A2u x A2u) = A1g . Thus, the combination has a symmetry

representation that DOES NOT contain one of the translational representations, thus the band is IR inactive.


Polyatomic Molecules: Selection Rules and Symmetry for Raman

Given that the Raman effect is associated to an induced dipole proportional to the polarizability change of the molecule as it vibrates, then the symmetry of the molecule also determines the probability of a Raman vibrational transition. In this case, the selection rule is given by:

Г(v’) x Г(i,j) x Г(v”) ⊇ A

In this case, the symmetry representation of the upper state must include the symmetry components of the polarizability tensor, which are also represented in the right hand side of the character tables.

Example: water has polarizability representations with A1, A2, B1 and B2 symmetry, meaning that

all fundamental transitions (2A1 and B2) are Raman active.

Example: the molecule trans-diazene (HN=NH) is C2h and has Гvib = 3Ag + Au + 2Bu. Three

fundamental modes are IR active (Au + 2Bu) and 3 IR inactive. Interestingly, there are 3

fundamental modes (3Ag) that are Raman active, and three (Au + 2Bu) that are Raman inactive.

Molecules with inversion centers (centrosymmetric) behave this way.


Polyatomic Molecules: Selection Rules and Symmetry for Raman

Given that the Raman effect is associated to an induced dipole proportional to the polarizability change of the molecule as it vibrates, then the symmetry of the molecule also determines the probability of a Raman vibrational transition. In this case, the selection rule is given by:

Г(v’) x Г(i,j) x Г(v”) ⊇ A

In this case, the symmetry representation of the upper state must include the symmetry components of the polarizability tensor, which are also represented in the right hand side of the character tables.

Example: water has polarizability representations with A1, A2, B1 and B2 symmetry, meaning that

all fundamental transitions (2A1 and B2) are Raman active.Example: the molecule trans-diazene (HN=NH) is C2h and has Гvib = 3Ag + Au + 2Bu. Three

fundamental modes are IR active (Au + 2Bu) and 3 IR inactive. Interestingly, there are 3

fundamental modes (3Ag) that are Raman active, and three (Au + 2Bu) that are Raman inactive.

Molecules with inversion centers usually behave this way.


Polyatomic Molecules: Vibrational Rotational Spectra

Linear MoleculesBased on symmetry considerations, the allowed (i.e. IR active) transitions for linear molecules can be classified as:

A1g (g+) → A2u (u

+) and A1g (g+) → E1u (u) (symmetric molecules: D∞h)

A1 (+) → A2 (+) and A1 (+) → E1 () (asymmetric molecules: C∞v)


+ +- -v5, E1u

v3, A2u


Linear MoleculesSigma Type Bands: A1g (g

+) → A2u (u+) or A1 (+) → A2 (+)


v3, A2u

•Presence of P and R bands, absence of Q band: J = ± 1•For symmetric molecules: alternation in the intensity due to spin statistics•Data is treated as done for a linear diatomic molecule (i.e. Bv and DJ can be obtained)

A1g (g+)

A2u (u+)

J” = 0

J’ = 0

J” = 3

J’ = 3


Linear MoleculesPi Type Bands: A1g (g

+) → E1u (u) or A1 (+) → E1 ()


•Presence of P, Q and R bands: J = 0, ± 1•Parity (related to symmetry of nuclear exchange upon rotation) must be taken into account: + ↔ – is allowed.•There is no J’ = 0, because degenerate (E orin this case) vibrational states do not allow for total zero angular momentum•Rotational levels in degenerate states are split due to Coriolis forces. The effect of Coriolis is included in the energy equation:

EJ = BvJ(J + 1) – Bv ± ζi/2 J(J + 1)

•For symmetric molecules: alternation in the intensity due to spin statistics

+ +- -v5, E1u

J’ = 1

A1g (g+)

E1u (u)

J” = 0

J” = 3

J’ = 4

+–+–

–++––+

+–


Symmetric RotorsAllowed transitions depend on the symmetry of the molecule. Many symmetric rotors are C3v, so

let us use this as an example. According to the character table allowed vibrational transitions are:

A1 → A1 and A1 → E

Bands associated to the A1 symmetry are called parallel bands, while those associated to the E

symmetry are called perpendicular bands



Symmetric Rotors (C3v case)Parallel Bands: A1→ A1

•Presence of P, Q and R bands: J = 0, ± 1 and K = 0 allowed for K ≠ 0 J = ± 1 and K = 0 allowed for K = 0

•Band is similar to that of a A1→ E band of linear molecule, but splitting is due to different K values.



Symmetric Rotors (C3v case)Perpendicular Bands: A1→ E•Presence of P, Q and R bands: J = 0, ± 1 and K = ± 1 allowed for all K.•P, Q and R branches spread out more than in a parallel band. Spacing between Q branches is approximately equal to 2(A – B). The example below shows a low resolution spectrum of the 60

1

band of SiH3F



Spherical RotorsAllowed transitions depend on the symmetry of the molecule. Most spherical rotors are Td and Oh.

According to the character table allowed vibrational transitions are:

A1 → T2 (for Td)

A1g → T1u (for Oh)

For example, methane has Гvib = A1 + E + 2T2, thus two fundamentals are IR active. They are the

ones corresponding to v3 (asym. strecth) and v4 (asym. bending).



Spherical Rotors•Only one type of band: A1 → T2

•The upper state is triply degenerate, thus there are P, Q and R bands•The upper rotational levels are triply split due to Coriolis forces (+, 0, –). The parity rule is:

for J = 0, A1 → T20

for J = +1, A1 → T2+

for J = –1, A1 → T2–

• It is possible to obtain Bv and Dj in a similar fashion as with linear diatomic molecules, expect that a Coriolis term must be included in the rotational energy.



Asymmetric Rotors•They are quite complex due to the low symmetry of the molecule. The transition rule is J = 0, ±1, thus the bands have P, Q and R branches. Since K is not a good quantum number, the bands are classified according to the orientation of the vibrational motion with respect to the rotational axes a, b, and c. Thus the bands are notated as A-type, B-type and C-type. •Transition rules:

A-type band: J = ±1 for Ka = 0 and J =0, ±1 for Kc = ±1

B-type band: J = ±1 for Ka = ±1 and J =0, ±1 for Kc = ±1

C-type band: J = ±1 for Kc = 0 and J =0, ±1 for Ka = ±1



Asymmetric Rotors•For example ethylene is a prolate asymmetric rotor.


A-type band, strong PQR bands B-type band, very weak Q band

C-type band, very weak PR bands

Polyatomic Molecules: Anharmonicity and Potential Energy Surfaces

The description of potential energy function in a molecule requires as many dimensions as there are vibrations plus one. For instance in the diatomic molecule, the surface is two-dimensional

(V(x) vs. x). In triatomic molecule like H3, the potential energy is a function of 3 vibrational

motions (2 stretching and one degenerate bending), thus it is four-dimensional. Since our brains are wired to see three dimensions at once, we must represent these potential energy functions as three-dimensional surfaces, or a two-dimensional slice at a constant potential (called a contour).



In a triatomic molecule like H3 shown below, point X is a saddle point, this represents a transtion

state in the collision process: H + H2 → H···H···H → H2 + H. At the bottom of the surface lie H2

molecules. Thus H3 is not expected to be a stable molecule and will dissociate spontaneously to

make H2.


X

X


On the contrary, CO2 is a stable molecule which would lie at the bottom of the surface relative to

the collision of CO and O: O + CO → OCO → OC + O


Trajectory Q1: symmetric stretchingTrayectory Q3: asymmetric stretching

Symmetric stretching motion leads to dissociation into atoms

Asymmetric stretching motion does not lead to dissociation

Polyatomic Molecules: Local Modes

The description of vibrational motion using normal modes is usually satisfactory. There are some cases, usually for vibrationally excited stretching motion (i.e. overtones) of symmetrically equivalent terminal atoms, in which the normal mode description fails. In these cases, a local mode description is more useful. For example, consider the normal mode of the symmetric stretch of

methane (v1, A1 symmetry).

In a normal mode description, vibrational excitation would lead to simultaneous dissociation of all

hydrogen atoms: CH4 + light → C + 4H.

However, anharmonicity tends to localize the motion in one of the C-H bonds leading to

dissociation of one of the hydrogen atoms CH4 + light → CH3 + H : In such case the vibrational

behavior resembles that of a diatomic molecule (A-H, where A = CH3) and it is straightforward to

describe the anharmonicity using a Morse potential.


H

C

HHH

H

C

HHH

Normal ModeLocal Mode

Documents

Laser Molecular Spectroscopy CHE466 Fall 2009