Post on 10-Apr-2023
transcript
Jyväskylä 2008
V q fq( ) = 12
2α
ν= f
h
− +�
��
��
� =hm
ddq
V q q E q2
2
2
28π( ) ( ) ( )Ψ Ψ
E h vv e= +ν ( )12
Ψv v vqq N H q e( ) ( ) /= −1 2 2 2α
νπ µe
f= 12
Harmonic oscillatorSchrödinger’s equationV(r)
rre
Define q = r - re
Solution:Quantum number v = 0, 1, 2, ...
Jyväskylä 2008
E h e012= ν
∆E v v h e( )+ ← =1 ν
Harmonic oscillator
Zero-point energy
Same distance between energy levels
Isotope effect: E.g.�
e(H2) = 4395 cm-1
�
e(D2) = 3118 cm-1
Jyväskylä 2008
Isotope effectExample: Water
4000 3500 3000 2500 2000 1500 1000 5000
100
T %
Wavenumber (1/cm)
4000 3500 3000 2500 2000 1500 1000 5000
100
T %
Wavenumber (1/cm)
H2O
D2O
Jyväskylä 2008
( )V r D eer( ) = − −1 2α
(
� �
)T V E+ =Ψ Ψ
E h v h x vv e e e= + − +ν ν( ) ( )12
12
2
Anharmonic oscillator
V(r)
r
Realistic form of potential energy curve.Many possible functions: Most popularare polynomial and the Morse potential.
If V(r) is Morse potential the Schrödingerequation can be solved analytically.
Solutions: quantum number v = 0,1,2,...
Jyväskylä 2008
∆E v v h h x ve e e( ) ( )+ ← = − +1 2 1ν ν
Anharmonic oscillator
Distance between energy levels varies
Almost the same transition energy for1 � 0but large discrepances for higher levels
Harmonic approximation is quite good forthe lowest transition and much simpler soharmonic approximaton is used.
Jyväskylä 2008
Degrees of freedom
Translations of the wholemolecule require 3degrees of freedom.
Totally 3N degrees offreedom.
Rotations require further3 degrees of freedom
Internal motions, 3N-6degrees of freedom, 3N- 5 in linear molecules
Jyväskylä 2008
Independent of each otherCommon starting level, all unexcitedSymmetry adapted
Normal vibrations
Jyväskylä 2008
Independent of each otherSymmetry adaptedInvolve the whole molecule
Normal vibrations
Jyväskylä 2008
For quantum chemists: the zero-pointenergy may be quite large
E.g., CO2: normal frequencies 1340, 667 and2349 cm-1. Each contributes to the zero-pointenergy by ½ v~e. Thus we obtain
Zero-point energy
670 + 333 + 1174 = 2177 cm -1
Spectroscopy measures distances between energy levels,not absolute energies, so the zero-point energy isirrelevant.
Jyväskylä 2008
Given the stiffnesses (force constants) ofindividual bonds (valence coordinates,really) and possible interrelations,calculatehow the atoms move and what are theharmonic energy levels.
Normal coordinate analysis
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Use molecular mechanics calculations
Interactions
� Bond distances
� Bond angles
� Torsion angles
� 1-4 interactions
� Cross-terms
Out come harmonic (even anharmonic)energy levels and shapes of the motions
Normal coordinate analysisModern method
Jyväskylä 2008
� Use good basis sets (TZP, 6-311G(d,p) orsuch)
� There is a systematic error, apply acorrection factor to vibrational energies
� Only harmonic frequencies are available(anharmonic are very expensive tocalculate)
Normal coordinatesQuantum chemical methods
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Wilson’s GF matrix method
Requires a good force constant matrix
Requires carefully defined valencecoordinates
Normal coordinatesThe traditional method
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Force constant matrixExample: Kim Palmö, HU, 1987
�
CC
�
CH1
�
CH2
�
CH3�
CH4
�
H1CH2
�
H3CH4
�
CH1,2
�
CH3,4�
CC 9.3881 0.0683 0.0683 0.0683 0.0683 -0.2697 -0.2697 0 0�
CH1 0.0683 5.0844 0.0273 0 0 0 0 0 0�
CH2 0.0683 0.0273 5.0844 0 0 0 0 0 0�
CH3 0.0683 0 0 5.0844 0.0273 0 0 0 0�
CH4 0.0683 0 0 0.0273 5.0844 0 0 0 0�
H1CH2 -0.2697 0 0 0 0 0.6786 0.0177 0 0�
H3CH4 -0.2697 0 0 0 0 0.0177 0.6786 0 0�
CH1,2 0 0 0 0 0 0 0 1.1113 0.1930�
CH3,4 0 0 0 0 0 0 0 0.1930 1.1113
C CH2
H1
H3
H4
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� Out come
� Exact motions
� Correctsymmetries
� Vibrationenergies
� (The selectionrules wereanalyzedseparately)
Normal coordinatesWilson’s method
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Ag Ag Ag Au
R 1623 R 3019 R 1342
B1g B1g B1u B2g
R 3272 R 1050 IR 949 R 943
B2u B2u B3u B3u
IR 3105 IR 995 IR 1443 IR 2989
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� Basically all vibrations involve all atoms inthe molecule
� However, some vibrations are mostlyconcentrated in a certain functional groupof the molecule => group frequencies
Group vibrations
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� Copy from character tables the characters of the xyztranslations, �
xyz.
� For each class of symmetry operations, count thestationary atoms, Nstat.
� Multiply the lines �
xyz and Nstat => �
tot.
� Subtract the translations of the whole molecule, �
xyz.
� Subtract the rotations of the whole molecule, �
rot.
� Reduce the resulting representation with characters �
vib.
Symmetry of the vibrations
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Symmetry of the vibrationsExample: equilaterial triangle
D3h E 2C3
�
h 2S3 3C2 3 �
vA1' 1 1 1 1 1 1A2' 1 1 1 1 -1 -1 RzE’ 2 -1 2 -1 0 0 (x,y)A1" 1 1 -1 -1 1 -1A2" 1 1 -1 -1 -1 1 zE” 2 -1 -2 1 0 0 (Rx, Ry)�
rot 3 0 -1 2 -1 -1
�
xyz 3 0 1 -2 -1 1Nstat 3 0 3 0 1 1�
tot 9 0 3 0 -1 1
- �
xyz -3 0 -1 2 1 -1- �
rot -3 0 1 -2 1 1
�
vib 3 0 3 0 1 1
aA1' = 1/12(3+0+3+0+3+3)=1aA2' = 1/12(3+0+3+0-3-3)=0aE' = 1/12(6+0+6+0+0+0)=1aA1" = 1/12(3+0-3+0+3-3)=0aA2" = 1/12(3+0-3+0-3+3)=0aE” = 1/12(6+0-6+0+0+0)=0
�
vib = A1' + E’
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Γ Γ Γ Γstatev v
MvM= ( ) ( ) ( )1 2
1 2 �
� No vibration is excited: totally symmetrical
� One vibration excited: the vibrational statehas the symmetry of the excited state
� Generally: Let the vibrations have thesymmetries �
1, �
2, ...
�
M. Let vibration nbe excited to state vn. The the totalvibrational state has the symmetry
Vibrational states
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� Vibrational motions belong to A1' and E’
� Ground state, none excited: �
state = A1'
� Fundamental excitation, A1' excited 1 � 0: �
state = A1'
� Fundamental excitation, E' excited 1 � 0: �
state = E'
� Overtone, E' excited 2 � 0: �state = E' �E’ = A1'+[A2']+E’ =
A1'+E’ (N.B. Special rules for powers of E!)
� Combination band, both A1' and E' excited 1 � 0: �
state =A1' �E' = E’
Vibrational statesExample: Equilateral triangle
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ν1 ν2 ν3
Possible transitionsCarbon dioxide
v=0
IR667 cm-1R1340 cm-1 IR2349 cm-1
Fundamental: 2349 cm-1
Overtone: 2001 cm-1
Combination band: 2007 cm-1
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Vibrational transition take place betweenvibrational states.
! Usually fundamental transition: initial state is thetotally symmetrical ground state and final statehas the same symmetry as the vibration
Two measuring techniques
! Infrared spectroscopy– Absorption; use the dipole operator
! Raman spectroscopy– Use the polarizability operator
Selection rules
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Ψ Ψ Γ Γ Γf i f O iO|"
| #⊗ ⊗
Selection rulesExample: the equilateral triangle
D3h E 2C3
$
h 2S3 3C2 3 $
vA1' 1 1 1 1 1 1 X2+y2, z2
A2' 1 1 1 1 -1 -1 RzE’ 2 -1 2 -1 0 0 (x,y) (X2-y2, xy)A1" 1 1 -1 -1 1 -1A2" 1 1 -1 -1 -1 1 zE” 2 -1 -2 1 0 0 (Rx, Ry) (xz, yz)
D3h A1 A2 EA1 A1 A2 EA2 A1 EE A1+[A2]+E
Infrared Vibration A1': Ground state A1', excited state A1'%
x: A1' &E’ &A1'=E’; not allowed%
y: the same%
z: A1' &A2" &A1'=E’; not allowedVibration E': Ground state A1', excited state E’%
x: A1' &E’ &E’=A1'+A2'+E’; ALLOWED!%
y: the same%
z: A1' &A2" &E'=E’; not allowed
Evaluate the symmetry of integral
Operator O is % for infraredspectroscopy and ' for Ramanspectroscopy.
If there is a totally symmetricalcomponent in the product, thetransition is allowed.
Jyväskylä 2008
Selection rulesExample: the equilateral triangle
Raman Vibration A1': Ground state A1', excited state A1'(
x2+y2: A1' )A1' )A1'=A1'; ALLOWED!(
z2: the same(
x2-y2: A1' )E’ )A1'=E’; not allowed(
xy: the same(
xz: A1' )E” )A1'=E”; not allowed(
yz: the sameVibration E': Ground state A1', excited state E’(
x2+y2: E’ )A1' )A1'=E’'; not allowed(
z2: the same(
x2-y2: E’ )E’ )A1'=A1'+A2'+E’; ALLOWED!(
xy: the same(
xz: E’ )E” )A1'=A1"+A2"+E”; not allowed(
yz: the same
SUMMARY:Excitation of vibration A1'- not allowed in infrared spectroscopy- allowed in Raman spectroscopyExcitation of vibration E’- allowed in infrared spectroscopy- allowed in Raman spectroscopy
Jyväskylä 2008
ν1 ν2 ν3
Fermi resonanceExample: Carbon dioxide
v=0
IR667 cm-1R1340 cm-1 IR2349 cm-1
First overtone 1334 cm-1 hastwo components,*
u
+ *
u=
,
g++ -
g. The firstinteracts with the .
1 linebecause the symmetry is thesame.