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1
Spectroscopy
Microwave (Rotational)
Infrared (Vibrational)
Raman (Rotational & Vibrational)
Texts– “Physical Chemistry”, 6th edition, Atkins
– “Fundamentals of Molecular Spectroscopy”, 4th
edition, Banwell & McCash
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Introduction-General Principles
Spectra - transitions between energy statesMolecule, Ef - Ei = h photon
Transition probability– selection rules
Populations (Boltzmann distribution)– number of molecules in level j at
equilibrium n g kTj j j exp /
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Typical energies
Region Frequency/Hz NA h nf / ni
RF 10 7 4 mJ/mol 0.999998
MCWE 1011 40 J/mol 0.984
IR 1013 4 kJ/mol 0.202
UV-VIS 1015 400 kJ/mol 3x10-70
X-RAY 1018 400 MJ/mol <10-99
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Fate of molecule?
Non-radiative transition: M* + M M + M + heat Spontaneous emission: M* M + h (very fast for large E) Stimulated emission (opposite to stimulated absorption)
These factors contribute to linewidth & to lifetime of excited state.
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MCWE or Rotational SpectroscopyClassification of molecules
Based on moments of inertia, I=mr2
– IAIBIC very complex eg H2O
– IA = IB = IC no MCWE spectrum eg CH4
– IAIB = IC complicated eg NH3
– IA = 0, IB = IC linear molecules eg NaCl
EJ J
IJ M JJ J
1
20 1 2 0 1
2 with also, , , , ,
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Microwave spectrometer
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
M C W ES O U R C E
D ETEC TO R
FR EQ UENC YS W EEP
A M PL I FI ER1 0 0 k H z
O S C I L L A TO R
D I S PL A Y
B R A S S TUB I NG
V A C UUM
M I C A W I ND O W
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Rotating diatomic molecule Degeneracy of Jth level is (2J+1) Selection rules for absorption are:
J = +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.
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Rotational energy levelsFor J=1
E = 2 ( J+1) h2/82I
01 E = 2 h2/82I
12 E = 4 h2/82I
23 E = 6 h2/82I
etc., etc., etc.
Constant difference of:
E = 2 h2/82I
E
J = 4 , M 4= 9
J = 3 , M 3= 7
J = 2 , M 2= 5
J = 1 , M 1= 3J = 0 , M 0= 10
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Populations of rotational levels n g kTJ J J exp /
J 2J+1 exp ( - / kT ) nJ / n0
0 1 1.000 1.001 3 0.981 2.942 5 0.945 4.733 7 0.893 6.254 9 0.828 7.455 11 0.754 8.296 13 0.673 8.757 15 0.590 8.858 17 0.507 8.629 19 0.428 8.13
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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 andalso 90.449 - 84.421 = 6.028 So, constant diff. of 6.028 GHz or 6.028109 s-1.
E = 2 h2/82I = h (6.028109 s-1) So 84.421 6.028 = 14.00ie J=13 J=14 & 90.449 6.028 = 15 ie J=14 J=15 & 96.477 6.028 = 16 ie J=15 J=16
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Moment of inertia, I E = 2 h2/82I = hv = h(6.028109 s-1)I = 2 h/(82 6.028109 )
I = 2 (6.62610-34)/(82 6.028109 )
I = 2.78410-45
Units? (J s)/(s-1) = J s2 = kg m2 s-2 s2 = kg m2
But I = r2
= (0.0630.079)/(0.063+0.079)NA = 5.8210-26 kg
r = (I/) = 218.610-12m= 218.6pm
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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 (220,399 MHz) and 12CO (230,538 MHz)
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IR / Vibrational spectroscopy Ev = (v + 1/2) (h/2) (k/)1/2
v = 0, 1, 2, 3, …Selection rules:
v = 1 & p must change during vibration
Let e = wavenumber of transition then “energy”:v = (v + 1/2) e
Untrue for real molecules since parabolic potential does not allow for bond breaking.v = (v + 1/2) e - (v + 1/2)2 e xe
– where xe is the anharmonicity constant
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Differences? Energy levels unequally
spaced, converging at high energy. The amount of distortion increases with increasing energy.
All transitions are no longer the same
v > 1 are allowed– fundamental 01– overtone
02– hot band
12
E
0
6
5
4
3
2
1
0
6
5
4
3
2
1
0
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Example HCl has a fundamental band at 2,885.9 cm-1, and an
overtone at 5,668.1cm-1.Calculate e and the anharmonicity constant xe.
v = (v + 1/2) e - (v + 1/2)2 e xe
2 = (2 + 1/2) e - (2 + 1/2)2 e xe
1 = (1 + 1/2) e - (1 + 1/2)2 e xe
0 = (0 + 1/2) e - (0 + 1/2)2 e xe
2 - 0 = 2e - 6e xe= 5,668.1 1 - 0 = e - 2e xe= 2,885.9e = 2,989.6 cm-1 e xe = 51.9 cm-1 xe = 0.0174
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High resolution infraredEv = (v + 1/2) (h/2) (k/m)1/2
v = (v + 1/2)
EJ = J(J+1) (h2/8I)
J = J(J + 1) v
Vibrational + rotational energy changes v,J = (v + 1/2) J(J + 1) v
Selection rule: v=+1, J=1– Rotational energy change must accompany a vibrational
energy change.
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Vibrational + rotational changes in the IR
v= 1 , J '= 0J '= 1
J '= 2
J '= 3
J = 1
J = 2
J = 3
v= 0 , J= 0
V I B R A T I O N A LG R O U N D S T A T E
V I B R A TI O NA LEX C I TED S TA TE
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Hi-resolution spectrum of HCl
Above the “gap”; J = +1 Below the “gap”: J = –1 Intensities mirror populations of starting levels
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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
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Raman spectroscopy
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) same frequency– Raman (1 in every 10,000,000) different frequencies
9 9 .9 9 %M O NO C H R O M A TI C
R A D I A TI O N
TR A NS PA R ENT D US T-FR EE S O L I D , L I Q UI D o r G A S
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Raman Light source? Laser
– Monochromatic, Highly directional, Intense
He-Ne 633 nm or Argon ion 488, 515 nmCells? 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.
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Why?
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
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Raman selection rules Vibrational energy levels
– v = 1– Polarisability must change during particular vibration
Rotational energy levels– J = 2– Non-isotropic polarisability (ie molecule must not be
spherically symmetric like CH4, SF6, etc.)
Combined
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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
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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
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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) e - (v + 1/2)2 e xe
Substitute in v=0, v=1, v=2, etc then compute: - = 187 = 373 = 558 … & calculate e, xe
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Pure Rotational Raman
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 J = +2 and J = -2 are nearly equal in intensity Very near high intensity peak of exciting radiation;
needs good quality spectrometers
X
Y
Z
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Rotational Raman
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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
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Raman spectra of KNO3N.B. strong symmetric stretch band at 1,050 cm-1
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Raman spectrum of aspirin tablet; no sample preparation
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Raman vs IR
CHCl3
Which?
Very similar
Diffs.?