lec16 Normal Modes of vibration.pdf

7/27/2019 lec16 Normal Modes of vibration.pdf

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Normal Modes of Vibration

A molecule having N atoms will have 3N degrees of freedom

For a nonlinear molecule: 3 degrees correspond to x, y, z translations3 degrees correspond to R x, R y, R z rotations

The remainder (3N-6) are vibrational motions (normal modes)

These normal modes form basis functions for a 1-D representation in themolecular symmetry group. (An irreducible representation)

In general, all normal modes can be resolved into either one or acombination of these normal modes.

For Linear Molecules, all vibrations can be resolved into 3N-5 normal modes

Consider SO 2: 3(3)-6 = 3 vibrational (normal) modes

S

O O

S

O O

S

O O

S

O O

S

O O

S

O O

1 2 3

SymmetricStretch

AsymmetricStretch

Bend


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Experimentally

(cm -1) assignment519 2

difference band

6061-2

1151 11361 3

combination band

1871 2+3

2305 21combination

band2499 1+3

Convention for Designating Vibrations

- symmetric/antisymmetric stretch - Bends (in plane) -Bends (out of plane)

Subscriptsas asymmetrics symmetricd - degenerate

For CO 2: normal modes3N-5 = 4 Exptl/cm-1

O C O

O C O

O C O

O C O

1

2

3

(degenerate)

1

2

3 - 2349-1340

-667

Note: For linear molecules, 3N-5 degrees of freedom


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By convention1 sym. Stretch2 sym. Bend

3 asym. Stretch4 asym. Bend

Consider 1 (1340 cm -1) and 2 2 (667 2 = 1344 cm -1) are nearly degenerate.

The wavefunctions describing these states will interact causing a resonance.

A

B A & B are now admixtures of these vibrational modes1388 cm

-1

~ 1340 cm -1

1286 cm -12 2

1

E

cannot call either " " or "2 2"

called: Fermi Resonance

How does one assign vibrations? Optional Material

Normal coordinate analyses solves the problem of the vibrating moleculewavefunctions in terms of internal coordinates

O

H O

L11 L22

L

Example:

H2OL11 , L22 are internal coordinatesfor OH bonds

F11 & F 22 are corresponding four constants

L , F - bending motion

The wavefunctions & force constants can be related by a series of secular equations/determinant.

GF-E = 0 Where: G matrix contains information about the kineticenergy of the molecule


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G-matrix can be calculated from bond distances/anglesF-matrix elements describe the potential energies of the vibrations

F-matrix is usually incomplete because force constants are not wellestablished. While =422 (from experiment)

If meaningful force constants can be obtained, then the vibrationalwavefunctions are determined by multiplying by the basis sets (internaldisplacements).

GF-E L = 0 G = 3 3, F = 3 3, L = 3 1 (L 11 , L 22, L)E = unit matrix

This yields the wavefunctions for the normal modes: For H 2O1 = NL + N (L11 + L 22) A1 symmetric stretch2 = N(L 11 + L 22) + N L A1 symmetric bend3 =

2

1 (L11 L 22) B1 asymmetric stretch

For N >> N

For a diatomic molecule: all matrices are 1 1 thereforeF11G11 - = 0 where F 11 is only unknown

This becomes: 11F = 422 = 2114

F


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Raman Spectroscopy visible photons to obtain normal modes

Energy Level Diagram molecule scatters a photon off at energies h O

12

vibrational state

21

0

1. Stokes Shift: h - o ( o virtual state)2. AntiStokes Shift: h + o ( 1 virtual state)

2 produces much weaker transitions because there

are fewer molecules in 1 than o

antistokes

Rayleigh line

stokes

Raman Shift/ cm -1

h + o

h - o

h

** Stokes lines are always atenergies = 1 vibrational quantumlower than Rayleigh line

Rayleigh line is the elastic scattering (h = h) and very intense .

Raman lines (stokes and antistokes) do not result from absorption and re-emission scattering phenomenon (T 10 -15 s)

Keypoint : h - o = vibrational energy for given normal mode(h + o) too weak to use as probe

*very useful for large molecules that absorb broadly in IR region, but only 1in 10 6 Raman photons produced inefficient


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Selection Rules for Raman Intensity f( ) molecule polarability

Intensity of scattered radiation: I = CI os4 ij ij

where: C- constantIo- intensity of incident radiations- frequency of scattered radiation (I is greater for blue photons)

For a Raman active vibration, there must be a change in the polarizabilityof the molecule during the vibration

Polarizability is a measure of deformability: Consider CO 2 (1)

C OO O C O O C O

extended equilibrium compressed

Now: Consider 3 (asymmetric stretch)

O C O OCO- ot changing polarizability(but dipole moment changes: IR active)

Summary for CO 2 1) Symmetric Stretch: Polarizability changes ( ) Raman allowed;

Dipole constant ( ), IR forbidden2) Asymmetric Stretch: - constant, Raman forbidden;

- changing, IR-active

Note: 1. If molecule has an inversion center, no vibrational mode can be both IR & Raman active

2. Raman bands that originate from the highest symmetry vibrationsof a pt. group (A 1, A 1g, A 1, etc) are polarized (Input of polarizedexcitation gives polarized scattering).


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Quantum Mechanically: The polarizability of a molecule by radiation alongthe direction ij (recall xx, yy, zz polarizability along x, y, z)

ij = h1

e oeenme j )Mi()M(

[ + seenme )Mj()Mi(

+ ]

Where: m & n are initial & final states; e is an intermediate excited state(virtual) Mi & Mj are transition moments (electron) along i & j; e is energyof transition to e; s & o represent scattered & incident radiation

Therefore: dependent upon o, M ij In order for a vibration to be Raman active:

rer

0 (must be a change in ) where: r is displacement along

normal mode

Can plot vs. r obtainr

o

r re

A B

o

re

If r near r e is small, Raman inactive mode (B)

If r near r e is larger, Raman active (A)

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lec16 Normal Modes of vibration.pdf