Spectroscopy

Microwave (Rotational)

Infrared
(Vibrational)

Raman
(Rotational & Vibrational)

Texts
– “Physical Chemistry”, 6th edition, Atkins
– “Fundamentals of Molecular Spectroscopy”, 4th edition,
Banwell & McCash
1
Introduction-General Principles
Spectra - transitions between energy states
Molecule, Ef - Ei = hu photon
 Transition probability
– selection rules
 Populations (Boltzmann distribution)
– number of molecules in level j at equilibrium

n j  g j exp  j / kT
2

Typical energies
Region Frequency/Hz
nf / ni
7
4 mJ/mol
0.999998
10
11
40 J/mol
0.984
10
13
4 kJ/mol
0.202
UV-VIS
10
15
400 kJ/mol
X-RAY
1018
400 MJ/mol
RF
MCWE
IR
3
NA hu
10
-70
3x10
<10-99
Fate of molecule?



Non-radiative transition: M* + M M + M + heat
Spontaneous emission: M* M + hn (very fast for large DE)
Stimulated emission (opposite to stimulated absorption)
These factors contribute to linewidth & to lifetime of excited state.
4
MCWE or Rotational Spectroscopy
Classification of molecules

Based on moments of inertia, I=mr2
– IA IB IC
very complex eg H2O
– IA = IB = IC
no MCWE spectrum eg
CH4
– IA IB = IC
complicated eg NH3
– IA = 0, IB = IC linear molecules eg NaCl
J  J  1  2
EJ 
with J  0,1,2, also M J  0,1, J
2I
5
Microwave spectrometer
VACUUM
MICA WINDOW
BRASS TUBING
MCWE
SOURCE
DETECTOR
100 kHz
OSCILLATOR
FREQUENCY
SWEEP




6
AMPLIFIER
DISPLAY
MCWE 3 to 60 GHz
X-band at 8 to 12 GHz; 25-35 mm
Path-length 2 m; pressure 10-5 bar; Ts up to 800K; vapour-phase
Very high-resolution eg 12C16O absorption at 115,271.204 MHz
Stark electric field: each line splits into (J+1) components
Rotating diatomic molecule
Degeneracy of Jth level is (2J+1)
 Selection rules for absorption are:

 DJ = +1
 The molecule must have a non-zero dipole
moment, p 0. So N2 etc do not absorb
microwave radiation.

Compounds must be in the vapour-phase
– But it is easy to work at temperatures up to 800K since cell
is made of brass with mica windows. Even solid NaCl has
sufficient vapour pressure to give a good spectrum.
7
Rotational energy levels
For DJ=1

DE = 2 ( J+1)
h2/8p2I
J=4, M 4=9
01 DE = 2 h2/8p2I
12 DE = 4 h2/8p2I
E
23 DE = 6 h2/8p2I
J=2, M 2=5
etc., etc., etc.
Constant difference of:

8
DE = 2 h2/8p2I
J=3, M 3=7
0
J=1, M 1=3
J=0, M 0=1
Populations of rotational levels
nJ  g J exp  J / kT 
J
0
1
2
3
4
5
6
7
8
9
9
2J+1
1
3
5
7
9
11
13
15
17
19
exp ( - / kT )
1.000
0.981
0.945
0.893
0.828
0.754
0.673
0.590
0.507
0.428
nJ / n 0
1.00
2.94
4.73
6.25
7.45
8.29
8.75
8.85
8.62
8.13
Example
Pure MCWE absorptions at 84.421 , 90.449 and 96.477 GHz
on flowing dibromine gas over hot copper metal at 1100K.
What transitions do these frequencies represent?
Note:
96.477 - 90.449 = 6.028 and
also
90.449 - 84.421 = 6.028
 So, constant diff. of 6.028 GHz or 6.028109 s-1.
DE = 2 h2/8p2I = h (6.028109 s-1)
 So 84.421 6.028 = 14.00
ie J=13 J=14
 & 90.449 6.028 = 15
ie J=14 J=15
 & 96.477 6.028 = 16
ie J=15 J=16

10
Moment of inertia, I
DE = 2 h2/8p2I = hv = h(6.028109 s-1)
I = 2 h/(8p2 6.028109 )

I = 2 (6.62610-34)/(8p2 6.028109 )
I = 2.78410-45
Units?
 (J s)/(s-1) = J s2 = kg m2 s-2 s2 = kg m2
But I = mr2

m= (0.0630.079)/(0.063+0.079)NA = 5.8210-26 kg

11
 r = (I/m) = 218.610-12 m = 218.6 pm
Emission spectroscopy?

Radio-telescopes pick up radiation from interstellar
space. High resolution means that species can be
identified unambiguously.
Owens Valley Radio Observatory 10.4 m telescope

Orion A molecular cloud 300K, 10-7 cm-3
517 lines from 25 species
CN, SiO, SO2, H2CO, OCS, CH3OH, etc
 13CO
12
(220,399 MHz) and 12CO (230,538 MHz)
IR / Vibrational spectroscopy
Ev = (v + 1/2) (h/2p) (k/m)1/2
 v = 0, 1, 2, 3, …
Selection rules:

 Dv = 1 & p must change during vibration
Let we = wavenumber of transition then “energy”:
 v = (v + 1/2) we

Untrue for real molecules since parabolic potential does
not allow for bond breaking.
 v = (v + 1/2) we - (v + 1/2)2 we xe
– where xe is the anharmonicity constant
13
Differences?



Energy levels unequally
spaced, converging at
high energy. The
amount of distortion
increases with
increasing energy.
All transitions are no
longer the same
Dv > 1 are allowed
– fundamental
– overtone
– hot band
14
01
02
12
E
0
6
6
5
5
4
4
3
3
2
2
1
1
0
0
Example
HCl has a fundamental band at 2,885.9 cm-1, and an
overtone at 5,668.1cm-1.
Calculate we and the anharmonicity constant xe.

v = (v + 1/2) we - (v + 1/2)2 we xe
2 = (2 + 1/2) we - (2 + 1/2)2 we xe
1 = (1 + 1/2) we - (1 + 1/2)2 we xe
0 = (0 + 1/2) we - (0 + 1/2)2 we xe
2 - 0 = 2we - 6we xe= 5,668.1
1 - 0 = we - 2we xe= 2,885.9
\ we = 2,989.6 cm-1 we xe = 51.9 cm-1
15
xe = 0.0174
High resolution infrared
Ev = (v + 1/2) (h/2p) (k/m)1/2
v = (v + 1/2) w
EJ = J(J+1) (h2/8I)
J = J(J + 1) Bv
Vibrational + rotational energy changes
 v,J = (v + 1/2) wJ(J + 1) Bv
 Selection rule: Dv=+1, DJ=1
– Rotational energy change must accompany a vibrational
energy change.
16
Vibrational + rotational changes in the IR
J'=3
'
J =2
J'=1
VIBRATIONAL
EXCITED STATE
v=1, J'=0
J=3
J=2
J=1
17
VIB RATIONAL
GROUND STATE
v=0, J=0
Hi-resolution spectrum of HCl



18
Above the “gap”; DJ = +1
Below the “gap”: DJ = –1
Intensities mirror populations of starting levels
Example: HBr
Lines at … 2590.95, 2575.19, 2542.25, 2525.09, ... cm-1
 Difference is roughly 15 except between 2nd & 3rd
where it is double this. Hence, missing transition lies
around 2560 cm-1.
So 2575 is (v=0,J=0)  (v=1,J=1) & 2590 is (v=0,J=1)  (v=1,J=2)
So 2542 is (v=0,J=1)  (v=1,J=0) & 2525 is (v=0,J=2)  (v=1,J=1)
(2575.19 - 2525.25) = 6B0 B0=8.35 cm-1
(2590.95 - 2542.25) = 6B1 B1=8.12 cm-1
 Missing transition at 2542.25 + 2B0 = 2558.95 cm-1
19
Raman spectroscopy
MONOCHROMATIC
99.99%
RADIATION
TRANSPARENT DUST-FREE
SOLID, LIQUID or GAS


Different principles. Based on scattering of (usually) visible
monochromatic light by molecules of a gas, liquid or solid
Two kinds of scattering encountered:
– Rayleigh (1 in every 10,000)
– Raman (1 in every 10,000,000)
20
same frequency
different frequencies
Raman

Light source? Laser
– Monochromatic, Highly directional, Intense
He-Ne
633 nm or Argon ion
488, 515 nm
Cells? Glass or quartz; so aqueous solutions OK
 Form of emission spectroscopy
 Spectrum highly symmetrical eg for liquid CCl4 there
are peaks at 218, 314 and 459 cm-1 shifted from
the original incident radiation at 633 nm (15,800 cm-1).
» The lower wavenumber side or Stokes radiation tends
to be more intense (and therefore more useful) than
the higher wavenumber or anti-Stokes radiation.
21
22
Why?


23
Rayleigh scattering: no change in wavenumber of light
Raman scattering: either greater than original or less than
original by a constant amount determined by molecular energy
levels & independent of incident light frequency
Raman selection rules

Vibrational energy levels
– Dv =  1
– Polarisability must change during particular vibration

Rotational energy levels
– DJ =  2
– Non-isotropic polarisability (ie molecule must not be
spherically symmetric like CH4, SF6, etc.)

24
Combined
Vibrational Raman


Symmetric stretching vibration of CO2
Polarisability changes
– therefore Raman band at 1,340 cm-1

Dipole moment does not
– no absorption at 1,340 cm-1 in IR
25
Vibrational Raman


Asymmetric stretching vibration of CO2
Polarisability does not change during vibration
– No Raman band near 2,350 cm-1

Dipole moment does change
– CO2 absorbs at 2,349 cm-1 in the IR
26
Example
In an experiment jets of argon gas and tin vapour impinged on a
metal block cooled to 12 K in vacuo. The Raman spectrum of the
frozen matrix showed a series of peaks beginning at 187 cm-1
and with diminishing intensity at 373, 558, 743, etc cm-1.
What species is responsible for the observed spectrum?
 Shifts of ca. 200 cm-1 indicate vibrational energies; diatomic tin?
Is 187 the fundamental? With the second peak at 373 (note 2 x
187 = 374), the third at 558 being 3 x 187 = 561, etc.
Use v = (v + 1/2) we - (v + 1/2)2 we xe
 Substitute in v=0, v=1, v=2, etc then compute:
 1-0 = 187 20= 373 30 = 558 … & calculate we, xe

27
Pure Rotational Raman
Y
X
Z

Polarisability is not isotropic
– CO2 rotation is Raman active
– some 20 absorption lines are visible on either side of the
Rayleigh scattering peak with a maximum intensity for
the J=7 to J=9 transition.
– The DJ = +2 and DJ = -2 are nearly equal in intensity

28
Very near high intensity peak of exciting radiation;
needs good quality spectrometers
Rotational Raman
29
Raman applications

Structure of Hg(I) in aqueous solution
– Is it Hg+ ? or (Hg2)2+ ?
– Aqueous solutions of HgNO3 show Raman band at
169 cm-1 (as well as NO3- bands), solid HgCl
shows a band at 167 cm-1
– Conclusion: Hg(I) exists as a diatomic cation (note that a
symmetrical diatomic would vibrate but would not absorb in
the IR; different selection rule)
– Very little sample preparation required; easy to get good
quality spectra of: solids, powders, fibers, crystals
– Drawbacks: coloured samples may overheat & burn up
30
Raman spectra of KNO3
N.B. strong symmetric stretch band at 1,050 cm-1
31
Raman spectrum of aspirin
tablet; no sample preparation
32
Raman vs IR
33

CHCl3

Which?

Very similar

Diffs.?