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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

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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.?