spectroscopy_3

8/8/2019 spectroscopy_3

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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 states

Molecule, Ef - Ei = hY photon

Transition probability

selection rules

Populations (Boltzmann distribution)

number of molecules in leveljat equilibrium

n g kT j j jw exp /I


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Typical energies

Region Frequency/Hz NA hY nf / ni

RF10

7

4 mJ/mol 0.999998

MCWE 1011 40 J/mol 0.984

IR 10 4 kJ/mol 0.202

UV-VIS 1015 400 kJ/mol 3x10

-70

X-RAY 1018 400 MJ/mol


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Fate of molecule?

Non-radiative transition: M* + M pM + M + at

Spontaneous emission:M* pM + R (very fast for large

(E) Stimulated emission (opposite to stimulated absorption)

These factors contribute to linewidth & to lifetime of excited


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MCWEor 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 egCH4

IA {IB= IC complicated eg NH3

IA= 0, IB = IC linear molecules eg NaCl E

J J

I J M J

J J!

! ! s s

1

2012 0 1

2J

- -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 10-5bar; Ts up to 800K; vapour-phase

Very high-resolution eg 12C16O absorption at 115,271.204MHz

Stark electric field: each line s lits into J+1 com onents

MCWE

SOURCEDETECTOR

FREQUENCY

SWEEP

AMPLIFIER

100 kHz

OSCILLATOR

DISPLAY

BRASS TUBING

VACUUM

MICA WINDOW


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Rotating diatomic molecule

Degeneracy of Jt level is (2J+1)

Selection rules for absorption are:

(J = +1

The molecule must have a non-zero dipolemoment, p {0. So N2 etc do not absorbmicrowave 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 hassufficient vapour pressure to give a good spectrum.


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Rotational energy levelsFor (J=1

(E = 2(J+1)

h2/8T2I

0p1 (E = 2 h2/8T2I1p2 (E = 4 h2/8T2I

2p3 (E = 6 h2/8T2I

etc., etc., etc.Constant difference of:

(E = 2 h2/8T2I

E

J=4, 4=9

J=3, 3=7

J=2, 2=5

J=1, 1=3

J=0, 0=10


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Populations of rotational levels

n g k J J J exp /I

J 2J+1 exp ( -I / kT ) n J / n00 1 1.000 1.00

1 3 0.981 2.942 5 0.945 4.73

3 7 0.893 6.25

4 9 0.828 7.45

5 11 0.754 8.29

6 13 0.673 8.75

7 15 0.590 8.85

8 17 0.507 8.62

9 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 and

also 90.449 - 84.421 = 6.028

So, constant diff. of6.028 GHz or6.028v109 s-1.(E = 2 h2/8T2I = h (6.028v109 s-1)

So 84.421 z6.028 = 14.00 ie J=13 pJ=14 & 90.449 z6.028 = 15 ie J=14 pJ=15 & 96.477z6.028 = 16 ie J=15 pJ=16


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Moment of inertia, I

(E = 2 h2/8T2I = hv = h(6.028v109 s-1)I = 2 h/(8T2 6.028v109)

I = 2 (6.626v10-34)/(8T2 6.028v109)

I = 2.784v10-45Units?

(J s)/(s-1) = J s2 = kg m2 s-2 s2 = kg m2

But I = Qr2

Q= (0.063v0.079)/(0.063+0.079)NA= 5.82v10-26 kg

r = (I/Q) = 218.6v10-12 m = 218.6 pm


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Emission spectroscopy?

Radio-telescopespick up radiation from interstellar

space. High resolution means that species can be

identified unambiguously.

Owens ValleyRadio Observatory 10.4 m telescope

Orion A molecular cloud}300K, }10-7cm-3

517 lines from 25 species

CN, SiO, SO2, H2CO, OCS,CH3OH, etc

13CO (220,399 MHz) and12CO (230,538MHz)


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IR / Vibrational spectroscopy Ev = (v + 1/2) (h/2T) (k/Q)1/2

v = 0, 1, 2, 3,

Selection rules:

(v = 1 &p must change during vibrationLet[e = wavenumber of transition then energy:

Iv= (v+1/2) [e Untrue for real molecules since parabolicpotential

does not allow for bond breaking.

Iv = (v+1/2) [e - (v+1/2)2 [e xe where xe isthe anharmonicity constant


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Differences? Energy levels

unequally spaced,converging at highenergy. The amount of

distortion increaseswith increasingenergy.

All transitions are nolonger the same

(v > 1 are allowed fundamental 0 p1

overtone 0 p2

hot band 1p2

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 anovertone at 5,668.1cm-1.

Calculate [e and the anharmonicity constant xe.Iv = (v+1/2) [

e

- (v+1/2)2 [ex

ejI2 = (2+1/2) [e - (2+1/2)2 [e xejI1 = (1+1/2) [e - (1+1/2)2 [e xejI

0 = (0 +1/2) [e - (0 +1/2)2 [e xe

jI2 - I0= 2

[e -6

[ex

e=5,66

8.1jI1 - I0 = [e - 2[e xe= 2,885.9

@ [e = 2,989.6 cm-1 [e xe = 51.9 cm-1 xe = 0.0174


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High resolution infrared

Ev = (v+1/2)(h/2T)(k/m)1/2

Iv= (v+1/2) [IEJ = J(J+1)(h

2/8I)

IJ = J(J+1) &vVibrati al+r tati al energy changes

Iv,J = (v+1/2) [IJ(J+1) &v Selecti n r le: (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'=0

J'=1

J'=2

J '=3

J=1

J=2

J=3

v=0, J=0

VIBRATIONAL

GROUND STATE

VIBRATIONAL

EXCITED STATE


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Hi-resolution spectrum ofHCl

Ab ve the gap;(J =+1 Bel wthe gap: (J =1

Intensitiesmirr rp p lati ns ofstarting 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) p (v=1,J=1) & 2590 is (v=0,J=1) p (v=1,J=2)

So 2542 is (v=0,J=1) p (v=1,J=0) & 2525 is (v=0,J=2) p (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 at2542.25 + 2B0 = 2558.95 cm-1


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Raman spectroscopy

Differentprinciples. 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

99.99%MONOCHROMATIC

RADIATION

TRANSPARENT DUST-FREE

SOLID, LIQUID or GAS


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Raman Light source? Laser

Monochromatic, Highly directional,Intense

He-Ne 633 nm orArgon ion 488, 515 nm

Cells? Glass or quartz; so aqueous solutions OK

Form of emission spectroscopy

Spectrum highly symmetrical eg for liquidCCl4there arepeaks ats218, s314 ands459 cm-1 shifted

from the original incident radiation at 633 nm(15,800 cm-1).

The lowerwavenumberside orStokesradiation tends

to be more intense (and therefore more useful)than

the higherwavenumberoranti-Stokesradiation.


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Raman selection rules

Vibrational energy levels

(v = s 1

Polarisability must change duringparticular vibration

Rotational energy levels (J = s 2

Non-isotropicpolarisability(ie molecule must not be

spherically symmetric like CH4, SF6, etc.)

Combined


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Vibrational Raman

Symmetric stretching vibration ofCO2

Polarisability changes

therefore Raman band at1,3 0 cm-1

Dipole moment doesnot

no absorption at1,3 0 cm-1 in IR


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Vibrational Raman

Asymmetric stretching vibration ofCO2

Polarisability doesnotchange during vibration

No Raman band near2,350 cm-1

Dipole moment does change

CO2 absorbsat2,3 9 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 Ramanspectrum of the frozen matrix showed a series ofpeaksbeginning 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 secondpeak at 373

(note 2 x 187 = 374), the third at 558 being 3 x 187 = 561, etc.

Use Iv = (v+1/2) [e - (v+1/2)2 [e xe Substitute in v=0, v=1, v=2, etc then compute:

I-I = 187 II= 373 II = 558 & calculate[e, xe


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Pure Rotational Raman

Polarisability is notisotropic CO2 rotation isRaman active

some 20absorption linesare visible on eitherside oftheRayleighscattering peak withamaximumintensity forthe J=7to J=9transition.

The (J =+2and (J =-2are nearly equalin intensity

Very nearhighintensity peak of exciting radiation;needs good quality spectrometers

X

Y

Z


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Rotational Raman


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Raman applications

Structure ofHg(I) in aqueoussolution IsitHg+ ? or(Hg2)

2+ ?

Aqueoussolutions ofHgNO3 showRaman band at169 cm-1 (aswellasNO3

- bands),solid HgClshowsaband at167 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 samplepreparation required; easy to get goodquality spectra of: solids,powders, fibers, crystals

Drawbacks: coloured samples may overheat & burn up


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Raman spectra ofKNO3N.B. strong symmetric stretchband at1,050 cm-1


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Raman spectrum ofaspirin

tablet;no sample preparation


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Raman vs IR

CHCl3

Which?

Very similar

Diffs.?

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