Transcript Slides - IISER Pune

Introduction to Basics of
Raman Spectroscopy
Chandrabhas Narayana
Chemistry and Physics of Materials
Jawaharlal Nehru Centre for Advanced Scientific
Research, Jakkur P.O., Bangalore 560064, India
[email protected]
http://www.jncasr.ac.in/cbhas
Lecture at MASTANI Summer School, IISER, Pune, June 30, 2014 to July 12, 2014
July 11, 2014
What happens when light falls on
a material?
Transmission
Reflection
Absorption
Luminescence
Elastic Scattering
Inelastic Scattering
Raman Spectroscopy
1 in 107 photons is scattered inelastically
Scattered
Excitation
virtual
state
Rotational Raman
Vibrational Raman
Electronic Raman
v” = 1
v” = 0
Infrared
Raman
(absorption)
(scattering)
Raman visible through unaided eye
Raman, Fluorescence and IR
Scattering
Absorption
and emission
Absorption
Concept of normal modes in a
molecule
• There are 3N possible movements in a molecule made of N
atoms, each of which moving in one of three directions, x, y and
z.
– There are three transitional movements: all atoms in the
molecule moving in x, y or z direction at the same time.
– There are three rotational movements around x, y or z-axis
• Linear molecules are exceptions because two axes that
are perpendicular to the molecular axis are identical.
– The rest of movements are vibrational movements
• For linear molecules, 3N – 5 movements
• For non-linear molecules, 3N – 6 movements
– All vibrational movements of the sample can be described as
linear combinations of vibrational normal modes.
Vibrations in Molecules
Sym. Stretching
HCl n = 2991 cm-1
8086 cm-1 = 1 eV
HF
Asym. Stretching
Sym. Bending
n3 = 3939 cm-1
n2 = 1648 cm-1
H2O n1 = 3835 cm-1
n = 4139 cm-1
Asym. Bending
NH3 n1 = 3505.7 cm-1
SF6
n1 = 774.55 cm-1
n3 = 3573.1 cm-1
n5 = 643.35 cm-1
n3 = 947.98 cm-1
n2 = 1022 cm-1
n2 = 615.02 cm-1
n4 = 1689.7 cm-1
n6 = 348.08 cm-1
n4 = 523.56 cm-1
Vibrational Spectroscopy
DAB
For a diatomic molecule (A-B), the bond between the two
atoms can be approximated by a spring that restores the
distance between A and B to its equilibrium value. The
bond can be assigned a force constant, k (in Nm-1; the
stronger the bond, the larger k) and the relationship
between the frequency of the vibration, , is given by the
relationship:

0
rAB
k

or, more typically
2cn 
k

where , c is the speed of light, n is the frequency in “wave
numbers” (cm-1) and  is the reduced mass (in amu) of A
and B given by the equation:
m m

re
A
B
mA  mB
re = equilibrium distance between A and B
DAB = energy required to dissociate into A and B atoms
Vibrational Spectroscopy
2cn 
k

can be rearranged to solve for k (in N/m): k  5.89  10 5 n2
Molecule
n (cm-1)
k (N/m)
 (amu)
HF
3962
878
19/20
HCl
2886
477
35/36 or 37/38
HBr
2558
390
79/80 or 81/82
HI
2230
290
127/128
Cl2
557
320
17.5
Br2
321
246
39.5
CO
2143
1855
6.9
NO
1876
1548
7.5
2331
2240
7
N2
For a vibration to be active (observable) in an infrared (IR) spectrum, the
vibration must change the dipole moment of the molecule. (the vibrations for
Cl2, Br2, and N2 will not be observed in an IR experiment)
For a vibration to be active in a Raman spectrum, the vibration must change
the polarizability of the molecule.
Classical Picture of Raman
Induced Polarization
Polarizability
Stokes Raman
Anti-Stokes Raman
Mutually exclusive principle
Raman Scattering
Selection rule: v = ±1
Overtones: v = ±2, ±3, …
 z (t )   zzequil Emax cos 2n 0t 
1 d zz
rmax Emax cos 2 (n 0 n vib )t 
2 dr
1 d zz
rmax Emax cos 2 (n 0 n vib )t
2 dr
Must also have a change in polarizability
Classical Description does not suggest any difference
between Stokes and Anti-Stokes intensities

N1
e
N0
www.andor.com
hn vib
kT
Are you getting the concept?
Calculate the ratio of Anti-Stokes to Stokes scattering
intensity when T = 300 K and the vibrational frequency is
1440 cm-1.
h = 6.63 x 10-34 Js
k = 1.38 x 10-23 J/K
N1
e
N0
hn vib

kT
~ 0.5
Energy diagram and
Quantum picture
Virtual states
Raman cross section
S
photon
<eg,p2|Her|p2,eb> <eb,p2|Hep|p1,ea> <ea,p1|Her|p1,eg>
|Es-Eb|x|Ei-Ea|
a,b
ex
g
Electronic states
Vibrational states
If Ei = Ea or Es = Eb
We have Resonance Raman effect
•
•
•
•
•
Intensity of Normal Raman Peaks
The intensity or power of a normal Raman peak
depends in a complex way upon
the polarizability of the molecule,
the intensity of the source, and
the concentration of the active group.
The power of Raman emission increases with
the fourth power of the frequency of the
source; - photodecomposition is a problem.
Raman intensities are usually directly
proportional to the concentration of the active
species.
Raman Depolarization Ratios
Polarization is a property of a beam of
radiation and describes the plane in
which the radiation vibrates. Raman
spectra are excited by plane-polarized
radiation. The scattered radiation is found
to be polarized to various degrees
depending upon the type of vibration
responsible for the scattering.
Raman Depolarization Ratios
The depolarization ratio p is defined as
I
p
I
Experimentally, the depolarization ratio may be
obtained by inserting a polarizer between the
sample and the monochromator. The
depolarization ratio is dependent upon the
symmetry of the vibrations responsible for
scattering.
Raman Depolarization Ratios
Polarized band: p = < 0.76 for totally
symmetric modes (A1g)
Depolarized band: p = 0.76 for B1g and B2g
nonsymmetrical vibrational modes
Anomalously polarized band: p = > 0.76
for A2g vibrational modes
Raman spectra of CCl4
Isotope effect
Cl has two isotopes 35Cl and 37Cl
Relative abundance is 3:1
CCl4 Spectra
• 461.5 cm-1 is due to 35Cl4C
• 458.4 cm-1 is due to 35Cl337ClC
• 455.1 cm-1 is due to 35Cl237Cl2C
• What are the two question
marks?
• Why are these bands weak?
Raman Spectra of Methanol and Ethanol
CH
stretching
20000
Raman Intensity (arbitrary unit)
CCO
stretching
CH3 and CH2
deformation
15000
OH
stretching
10000
CO
stretching
CH3
deformation
5000
0
500
1000
1500
2000
2500
3000
3500
Raman Shift (cm-1)
CASR
Significant identification of alcohols which differ just in one CH2-g
Peak position – Chemical identity –
Similar Structures
CASR
3,4-Methylenedioxymethamphetamine (MDMA)
Methamphetamine
ecstasy
Raman Intensity (arbitrary unit)
500
1000
1500
2000
Raman Shift (cm-1)
2500
3000
3500
The Mass Effect on Raman Spectra
Significant identification of salts (SO42-) which
differ just in the metal ion employed
1200
1000
Intensity (counts/s)
800
600
400
1000
200
0
Wavenumber (cm-1)
Intensity (counts/s)
800
600
Mg - SO4
Na2 - SO4
400
200
Wavenumber (cm-1)
CASR
CASR
Peak positions – Chemical identity
Diasteromers
Ephedrine
Pseudoephedrine
Raman Intensity (arbitrary unit)
500
1000
1500
2000
Raman Shift (cm-1)
2500
3000
3500
Peak Position – Crystal Phases –
Polymorphs
Both Anatase and Rutile are TiO2 but of different
polymorphic forms - identical chemical composition,
different crystalline structures.
2 400
2 200
2 000
Rutile
1 800
Intensity (cnt)
1 600
1 400
1 200
1 000
800
600
400
200
200
400
600
800
1 000
1 200
Raman Shift (cm-1)
Anatase
1 400
CASR
Peak Shift – Stress and Strain
Larnite (b – Ca2SiO4) inclusion in Diamond
Nasdala, L., Harris, J.W. & Hofmeister, W. (2007): Micro-spectroscopy of diamond. Asia Oceania
Geosciences Society, 4th Annual Meeting, Bangkok, Thailand, August, 2007.
Nasdala, L., Raman barometry of mineral inclusions in diamond crystals. Mitt. Österr. Miner. Ges. 149
(2004)
CASR
Bandwidth – Crystallinity –
Structural order/disorder
Raman spectra of zircon, showing typical amorphous (blue)
and crystalline (red) spectra.
Intensity – Concentration
CASR
1341.0
4-Nitrophenol dissolved in CH2Cl2
3 500
Intensity (cnt/sec)
3 000
4-Nitrophenol in CH2Cl2_0,1 M
4-Nitrophenol in CH2Cl2_0,01 M
4-Nitrophenol in CH2Cl2_0,001 M
2 500
2 000
1 500
1 000
500
0
1 200
1 400
-1 )
Raman Shif t (cm
1 600
CASR
Raman technique – what
requirements are needed?
Requirements for Raman technique to determine peak position, peak shift,
bandwidth and intensity
- Laser Excitation
- Reduction of stray light
- Collecting Optics
- Spectral resolution and spectral coverage
- Spatial resolution and confocality
- Sensitivity: subject to detector
- Light flux: subject to dispersion
CASR
What do we need to make a
Raman measurement ?
Detector
•Monochromatic
typically a laser
Light
source
(A way of focusing the laser onto
the sample and then collecting the
Raman scatter.)
•Sampling optics
Grating
•Rejection filter
(A way of removing the
scattered light that is not
shifted( changed in colour).
Filter
Laser
Sample
•Spectrometer and detector
(often a single grating spectrometer
and CCD detector.)
Demonstration of the very high
CASR spectral resolution obtained in the
triple additive mode
Rotation-Vibration Spectrum of O2
Intensity (a.u.)
8000
Triple additive mode
Slit widths= 30 m
6000
4000
Triple subtractive mode . Slit=30 m
2000
1520
1540
1560
Wavenumber (cm-1)
1580
CASR
Laser excitation – Laser Selection to
avoid fluorescence
Laser wavelength, 3
Laser wavelength, 3
Laser
wavelength, 1

Fluorescence
Raman shift, 1-1+
Laser wavelength: 3 < 2 < 1
Raman shift, 3-1+
Laser excitation – Laser selection to
avoid fluorescence
CASR
60 000
Green spectrum: 532 nm laser
Red spectrum: 633 nm laser
Dark red spectrum: 785 nm laser
50 000
Intensity (cnt)
40 000
30 000
20 000
10 000
60 000
0
50 000
800
Wavelength (nm)
1 000
40 000
Intensity (cnt)
600
30 000
Fluorescence is wavelength dependent
Ordinary Raman is wavelength independent
20 000
10 000
0
1 000
2 000
Raman Shift (cm-1)
3 000
Laser
excitation
–
Laser
selection
to
CASR
avoid fluorescence
Commercial Hand Cream
x10 3
45
40
35
Intensity (cnt)
30
25
20
15
10
5
500
1 000
1 500
785 nm – 633 nm – 473 nm
2 000
Raman Shift (cm-1 )
2 500
3 000
3 500
Reduction of Fluorescence
CASR
Laser excitation – laser radiation power
Laser wavelength: 473 nm
Laser power at sample: 25.5 mW
Laser wavelength: 633 nm
Laser power at sample: 12.6 mW
Objective
N.A.
Objective
N.A.
Laser
spot
size
(µm)
Radiatio
n power
(kW/cm2)
100×
0.90
0.64
~7900
100×
0.90
0.85
~2200
50×
0.75
0.77
~5400
50×
0.75
1.03
~1500
10×
0.25
2.31
~600
10×
0.25
3.09
~200
Laser
Radiation
spot size
power
(µm)
(kW/cm2)
CASR
Laser excitation – laser radiation
power
• Keep in mind: the usage of high numerical
objective lenses causes a very small spot
size of the laser which results in a high
power density
• To avoid sample burning radiation power
has to be adapted INDIVIDUALY to the
sample
Collection solid angle
Large for high N.A. lens
Small for low N.A. lens
Low N.A. lens
Working distance
θ
High N.A. lens
Sampling volume
Small for high N.A. lens
Large for low N.A. lens
Laser spot size
Small for high N.A. lens
Large for low N.A. lens
NA = n · sin (Q)
n: refraction index
Q: aperture angle
θ
Working distance
Collecting Optics
CASR
CASR
Collecting Optics – Overview on
common objectives
Objective
N.A.
Working distance
[mm]
x100
0.90
0.21
x50
0.75
0.38
x10
0.25
10.6
x100 LWD
0.80
3.4
x50 LWD
0.50
10.6
Collecting Optics – what objective
should be used?
CASR
A distinction between opaque and transparent samples has to be made
For opaque samples, high N.A. lens works better because there is almost no
penetration of the laser into the sample. High N.A. lens enables
- High laser power density (mW/m3)  increases sensitivity
- Wide collection solid angle  increases sensitivity
Example for an opaque sample:
Silicon wafer
100 %
30 000
Silicon
x100 – NA = 0.9 – 31.350 C/s
x50 – NA = 0.75 - 21.995 C/s
x10 – NA = 0.25 - 1.462 C/s
25 000
70 %
20 000
Intensity (cnt/sec)
Y (µm)
-20
15 000
0
10 000
10 µm
20
5 000
5%
0
0
X (µm)
460
480
500
520
540
Raman Shif t (cm-1)
560
580
600
620
Collecting Optics – what objective
should be used?
CASR
A distinction between opaque and transparent samples has to be made
For transparent samples, low N.A. lens works better because of penetration of
the laser into the sample. Low N.A. lens enables
- Large sampling volume  increases sensitivity
x10 3
14
12
Sample: Cyclohexane
Instrument: ARAMIS
Red: x100LWD, 7,000 cts/s
Blue: Macro lens, 14,500 cts/s
cyclo_100xLWD
cyclo_macro
100 %
Intensity (cnt)
10
8
48 %
6
4
2
0
740
760
780
800
820
Raman Shift (cm-1 )
840
860
880
Spectral resolution and spectral
coverage
CASR
Schematic diagram of a Czerny-Turner spectrograph
Slit
Collimating
mirror
Grating
Detector
Focusing mirror
Focal Length
CASR
Spectral resolution and spectral
coverage
•
Spectral resolution is a function of 1. dispersion, 2. widths of entrance slit
and 3. pixel size of the CCD
•
Dispersion is the relation between refraction of light according to the
wavelength of light
•
Dispersion is a function of the 1. focal length of spectrograph the 2.
groove density of the grating and 3. the excitation wavelength
•
In general, long focal length and high groove density grating offer
high spectral resolution.
CCD Detector
Short focal length
Same grating
Same excitation wavelength
Long focal length
CCD Detector
CASR
Dispersion as a function of the focal
length
CASR
Dispersion as a function of the focal
length vis-a vis wavelength
Dispersion in cm-1 / pixel
1800 gr/mm Grating
LabRAM (F = 300 mm)
LabRAM HR (F = 800 mm)
Dispersion as a function of excitation
wavelength
CASR
Horizontal lines indicate a relative Raman Shift of 3800 cm
-1
244 - 269 nm (25 nm)
325 - 371 nm (46 nm)
457 - 553 nm (96 nm)
488 - 599 nm (111 nm)
514 - 639 nm (125 nm)
532 - 667 nm (135 nm)
633 - 833 nm (200 nm)
785 - 1119 nm (334 nm)
830 - 1210 nm (380 nm)
1064 - 1768 nm (704 nm)
200
400
600
800
1000 1200 1400 1600 1800
Wavelength [nm]
Short wavelength
CCD Detector
CCD Detector
Same focal length
Same grating
Long wavelength
CASR
Spectral coverage - dependence from
excitation wavelength
Length of CCD Chip
x10
3
473 nm – 633 nm – 785 nm
22
20
18
16
Intensity (cnt/sec)
Same focal length
Same grating
Length of CCD Chip
14
12
10
Length of CCD Chip
Relative Raman shift of 3100 cm-1
8 corresponds to 81 nm
6
Relative Raman shift of 3100 cm-1
corresponds to 154 nm
4
2
Relative Raman shift of 3100 cm-1
corresponds to 252 nm
0
500
600
700
800
Wavelength (nm)
900
1 000
Low density groove grating
High density groove grating
Same focal length
Same excitation wavelength
CCD Detector
Dispersion as a function of groove density
CCD Detector
CASR
CASR
Spectral resolution as a function slit
width
Slit
Slit
Slit
One of parameters that determines the spectral resolution is the entrance
slit width. The narrower the slit, the narrower the FWHM (full width at half
maximum), and higher the spectral resolution.
When recording a line whose natural width is smaller than the
monochromator’s resolution, the measured width will reflect the
spectrograph’s resolution.
Spectral resolution as a function of pixel
CASR
size
Detector
Intensity
Detector
600
650
Raman Shift (cm -1)
700
• Because a CCD detector
is made of very small
pixels, each pixel serves
as an exit slit (pixel size =
exit slit width)
• For the same size CCDs,
the CCD with a larger
number of smaller pixels
produces a larger number
of spectral points closer to
each other increasing the
limiting spectral resolution
and the sampling
frequency
• 26 m pixel vs. 52 m
pixel (simulation)
Choice of Laser for Raman
The choice of laser will influence
different parameters:
• Signal Intensity: (1/)4 rule applies to
Raman intensity.
• Probing volume: spot size
material penetration depth.
and
• Fluorescence: may overflow Raman
signal.
100
Silicium
Penetration depth (µm)
CASR
10
1
0,1
0,01
• Enhancement: some bounds only
react strongly at a certain wavelength.
• Coverage range and Resolution:
grating dispersion varies along the
spectral range.
0,001
200
300
400 500 600 700
Wavelength (nm)
800
900
CASR
Spatial resolution: penetration depth
Penetration depth in Silicon
244 nm
325 nm
0
2
4
6
Wavelength [nm]
244 nm
8
325 nm
457 nm
488 nm
514 nm
633 nm
0
10
500
Depth penetration [nm]
12
1000
1500
2000
2500
3000
 General: The larger the excitation wavelength,
the deeper the penetration.
 The exact values depend on material.
Spatial resolution: penetration depth
CASR
325 nm
Intensity [a.u.]
Strained Si
of top layer
325 nm
Si of SiGe
layer
~nm
~nm
488 nm
785 nm
Strained silicon layer
Uniform layer of SiGe
~µm
488 nm
Gradient SiGe layer
Si of silicon
substrate
400
450
500
550
785 nm
Pure Si substrate
Raman shift (cm-1)
 The higher the excitation wavelength, the deeper the penetration.
Spatial resolution: penetration depth
CASR
EXAMPLE
325nm laser results
The strain is not homogenous.
A characteristic, cross-hatch pattern is observed.
CASR
Spatial resolution and spot size
Laser spot size D is defined by the Rayleigh criterion:
excitation wavelength ()
D = 1.22  / NA
objective numerical aperture (NA)
With NA=n sinα
Spatial resolution is half of the laser spot diameter
Confocality
and
Spatial
Resolution
CASR
Nearly closed
confocal apertur
Sample
P P P
P'
P'
P'
Image P ' of laser focus P matches
exactly the confocal hole.
Length of
Laser Focus
Axial resolution of a Confocal Raman Microprobe
CASR
Confocality and spatial Resolution
Wide Hole
Laser focus
waist diameter
Depth of
laser focus
(d.o.f)
Sampling
Volume
Narrow Hole
Confocal z-scan against silicon
with different hole apertures
exc = 633 nm
CASR
Narrow Hole:
Collecting Raman radiation that
originates only from within a
diffraction limited laser focal volume
with a dimension of:
Focus waist diameter ~ 1.22  / NA
Depth of laser focus ~ 4  / (NA)2
Example of application using the confocality
principle : depth profile on a multilayer polymer
sample
50 00
Polyethylene
nylon
Polyethylene
75 m
Intensity (a.u.)
x
40 00
30 00
20 00
10 00
z
0
10 00
12 00
14 00
16 00
W avenumber (cm-1 )
Intensity (a.u.)
3000
2500
2000
1500
1000
500
0
CASR
1000
1200
1400
Wavenum ber (cm -1)
1600
Thank you for your attention!
Symmetry – Identity (E)
Identity operation (E)
This is a very important symmetry operation which is
where the molecule is rotated by 360º. In otherwords a
full rotation or doing nothing at all.
This appears in all molecules!!!
Symmetry – Rotation (Cn)
Rotations (Cn)
These are rotations about the axes of symmetry. n denotes
360º divided by the number for the rotation.
C2 = 180º
C3 = 120º
C4 = 90º
C5 = 72º
C6 = 60º
Symmetry – Reflections (s)
Reflections (sh, sd and sv) These are reflections in
a symmetry planes (x, y and z).
sh - Horizontal
Plane (y)
perpendicular
to the highest
rotation axis
sv - Vertical
Plane (z)
parallel
to the highest
rotation axis
sd - Diagonal
(dihedral) Plane (x)
the Diagonal Plane
that bisects two
axes
Symmetry – Inversion (i)
Inversion centre (i)
These are where the molecule can be inverted through the
centre of the molecule (atom or space).
Symmetry – Improper Rotation (Sn)
Improper rotations (Sn)
These are rotations about the axes of symmetry
followed by reflections.
Vibrational Spectroscopy
For polyatomic molecules, the situation is more complicated because there are more
possible types of motion. Each set of possible atomic motions is known as a mode.
There are a total of 3N possible motions for a molecule containing N atoms because
each atom can move in one of the three orthogonal directions (i.e. in the x, y, or z
direction).
A mode in which all the atoms are moving in the same
direction is called a translational mode because it is
equivalent to moving the molecule - there are three
translational modes for any molecule.
A mode in which the atoms move to rotate (change the
orientation) the molecule called a rotational mode - there
are three rotational modes for any non-linear molecule
and only two for linear molecules.
Translation
al modes
Rotational
modes
The other 3N-6 modes (or 3N-5 modes for a linear molecule) for a molecule
correspond to vibrations that we might be able to observe experimentally. We must
use symmetry to figure out what how many signals we expect to see and what atomic
motions contribute to the particular vibrational modes.
Vibrational Spectroscopy and Symmetry
We must use character tables to determine how many signals we will see
in a vibrational spectrum (IR or Raman) of a molecule. This process is
done a few easy steps that are similar to those used to determine the
bonding in molecules.
1. Determine the point group of the molecule.
2. Determine the Reducible Representation, tot, for all possible motions of
the atoms in the molecule.
3. Identify the Irreducible Representation that provides the Reducible
Representation.
4. Identify the representations corresponding to translation (3) and rotation
(2 if linear, 3 otherwise) of the molecule. Those that are left correspond
to the vibrational modes of the molecule.
5. Determine which of the vibrational modes will be visible in an IR or
Raman experiment.
Finding the Point Group
Vibrational Spectroscopy and Symmetry
Example, the vibrational modes in water.
The point group is C2v so we must use the appropriate character table
for the reducible representation of all possible atomic motions, tot. To
determine tot we have to determine how each symmetry operation
affects the displacement of each atom the molecule – this is done by
placing vectors parallel to the x, y and z axes on each atom and
applying the symmetry operations. As with the bonds in the previous
examples, if an atom changes position, each of its vectors is given a
value of 0; if an atom stays in the same place, we have to determine
the effect of the symmetry operation of the signs of all three vectors.
The sum for the vectors on all atoms is placed into the reducible
representation.
O
H
H
H
O
top view
H
Make a drawing of the molecule and add in vectors
on each of the atoms. Make the vectors point in
the same direction as the x (shown in blue), the y
(shown in black) and the z (shown in red) axes.
We will treat all vectors at the same time when we
are analyzing for molecular motions.
Vibrational Spectroscopy and Symmetry
Example, the vibrational modes in water.
z
O
The E operation leaves everything where it is so all nine
vectors stay in the same place and the character is 9.
H
H
The C2 operation moves both H atoms so we can ignore the
vectors on those atoms, but we have to look at the vectors on
the oxygen atom, because it is still in the same place. The
vector in the z direction does not change (+1) but the vectors
in the x, and y directions are reversed (-1 and -1) so the
character for C2 is -1.
H
C2V
E
C2
sv (xz)
s’v (yz)
tot
9
-1
3
1
O
H
x
C2
H
The sv (xz) operation leaves each atom where it was so we
H
have to look at the vectors on each atom. The vectors in the
z and x directions do not move (+3 and +3) but the vectors in
the y direction are reversed (-3) so the character is 3.
The s’v (yz) operation moves both H atoms so we can ignore
the vectors on those atoms, but we have to look at the vectors
on the oxygen atom, because it is still in the same place. The
vectors in the z and y directions do not move (+1 and +1) but
the vectors in the x direction is reversed (-1) so the character
is 1.
y
O
O
H
H
sv (xz)
H
H
O
O
H
H
s’v (yz)
H
O
H
Vibrational Spectroscopy and Symmetry
C2V
E
C2
sv (xz)
s’v (yz)
tot
9
-1
3
1
C2V
E
C2
sv (xz)
s’v (yz)
A1
1
1
1
1
z
x2,y2,z2
A2
1
1
-1
-1
Rz
xy
B1
1
-1
1
-1
x, Ry
xz
B2
1
-1
-1
1
y, Rx
yz
From the tot and the character table, we can figure out the number and types of modes
using the same equation that we used for bonding:
nX 


1
 # of operations in class)  (character of RR)  character of X)
order
This gives:






nA1 
1
1)9)1)  1)1)1)  1)3)1)  1)1)1)

4
nA 2 
1
1
1)9)1)  1)1)1)  1)3)1)  1)1)1) nB2 
1)9)1)  1)1)1)  1)3)1)  1)1)1)

4
4

nB1 

1
1)9)1)  1)1)1)  1)3)1)  1)1)1)
4
Which gives: 3 A1’s, 1 A2, 3 B1’s and 2 B2’s or a total of 9 modes, which is what we
needed to find because water has three atoms so 3N = 3(3) =9.
Vibrational Spectroscopy and Symmetry
Now that we have found that the irreducible representation for tot is (3A1 + A2 +
3B1+ 2B2), the next step is to identify the translational and rotational modes - this
can be done by reading them off the character table! The three translational modes
have the symmetry of the functions x, y, and z (B1, B2, A1) and the three rotational
modes have the symmetry of the functions Rx, Ry and Rz (B2, B1, A2).
Translation Rotational
al modes
modes
C2V
E
C2
sv (xz)
s’v (yz)
A1
1
1
1
1
z
x2,y2,z2
A2
1
1
-1
-1
Rz
xy
B1
1
-1
1
-1
x, Ry
xz
B2
1
-1
-1
1
y, Rx
yz
The other three modes (3(3)-6 = 3) that are left over for
water (2A1 + B1) are the vibrational modes that we might be
able to observe experimentally. Next we have to figure out if
we should expect to see these modes in an IR or Raman
vibrational spectrum.
Vibrational Spectroscopy and Symmetry
Remember that for a vibration to be observable in an IR
spectrum, the vibration must change the dipole moment of the
molecule. In the character table, representations that change
the dipole of the molecule are those that have the same
symmetry as translations. Since the irreducible representation
of the vibrational modes is (2A1 + B1) all three vibrations for
water will be IR active (in red) and we expect to see three
signals in the spectrum.
The three vibrational
modes for water. Each
mode is listed with a n
(Greek letter ‘nu’) and a
subscript and the energy
of the vibration is given
in parentheses. n1 is
called the “symmetric
stretch”, n3 is called the
“anti-symmetric stretch”
and n2 is called the
“symmetric bend”.
For a vibration to be active in a Raman spectrum, the vibration
must change the polarizability of the molecule. In the character
table, representations that change the polarizability of the
molecule are those that have the same symmetry as the second
order functions (with squared and multiplied variables). Thus all
three modes will also be Raman active (in blue) and we will see
three signals in the Raman spectrum.
C2V
E
C2
sv (xz)
s’v (yz)
A1
1
1
1
1
z
x2,y2,z2
A2
1
1
-1
-1
Rz
xy
B1
1
-1
1
-1
x, Ry
xz
B2
1
-1
-1
1
y, Rx
yz
The Geometry of the Sulfur Dioxide
Molecule
Cs structure: 3 normal modes, all
having A' symmetry
The Cs structure should have 3 IR active fundamental
transitions. These three fundamental transitions also should
be Raman active. We would expect to observe three strong
peaks in the IR and three strong peaks in the Raman at the
same frequency as in the IR. All of the Raman lines would be
polarized because they are totally symmetric (A' symmetry).
C2v structure: 3 normal modes, two
with A1 symmetry, one with B2
The C2v structure should have 3 IR active fundamental
transitions. These three fundamental transitions also should
be Raman active.We would expect to observe three strong
peaks in the IR and three strong peaks in the Raman at the
same frequency as in the IR.
Two of the Raman lines are totally symmetric (A1 symmetry)
and would be polarized. One Raman line would be
depolarized.
The Dooh structure should have two IR active
fundamental transitions. It will have one Raman
active fundamental transition at a different
frequency than either of the IR peaks.. The
Raman line will be polarized.
Experimental Observation
The experimental infrared and Raman bands of liquid
and gaseous sulfur dioxide have been reported in a
book by Herzberg 7 . Only the strong bands
corresponding to fundamental transitions are shown
below. The polarized Raman bands are in red.
Fundamental 2
1
3
IR (cm-1)
519
1151
1336
Raman (cm-1) 524
1151
1336
Conclusion
The existence of three experimental bands in the IR
and Raman corresponding to fundamental transitions
weighs strongly against the symmetrical linear (Dooh)
structure. We usually do not expect more strong bands
to exist than are predicted by symmetry.
Group theory predicts that both bent structures would
have three fundamental transitions that are active in
both the IR and Raman. However all three of the
Raman lines would be polarized if the structure were
unsymmetrical (Cs symmetry). The fact that one
Raman line is depolarized indicates that the structure
must be bent and symmetrical (C2v symmetry).
The sulfur dioxide molecule has C2v symmetry.
Problems with Raman:
a)Very Weak – for every 106 photons only 1
photon Raman
a)Resonant Raman not feasible with every sample.
b)Absorption a better process than scattering
International and National Patent (2007), G.V. Pavan Kumar et al Current Science (2007) 93, 778.
Micro–Raman
setup
Raman
Spectrometers
Optical fiber
CCD
Monochromator
Focusing lens
Edge filter
Camera
Computer
Dichroic
Mirror
LASER
Objective lens
Stage