Spectroscopy - PowerPoint PPT Presentation

About This Presentation
Title:

Spectroscopy

Description:

Microwave spectrometer. MCWE 3 to 60 GHz X-band at 8 to 12 GHz; 25-35 mm ... near high intensity peak of exciting radiation; needs good quality spectrometers ... – PowerPoint PPT presentation

Number of Views:502
Avg rating:3.0/5.0
Slides: 34
Provided by: johns167
Category:

less

Transcript and Presenter's Notes

Title: Spectroscopy


1
Spectroscopy
  • Microwave (Rotational)
  • Infrared (Vibrational)
  • Raman (Rotational Vibrational)
  • Texts
  • Physical Chemistry, 6th edition, Atkins
  • Fundamentals of Molecular Spectroscopy, 4th
    edition, Banwell McCash

2
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

3
Typical energies
4
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.

5
MCWE or Rotational SpectroscopyClassification of
molecules
  • Based on moments of inertia, Imr2
  • IA ¹ IB ¹ IC very complex eg H2O
  • IA IB IC no MCWE spectrum eg CH4
  • IA ¹ IB IC complicated eg NH3
  • IA 0, IB IC linear molecules eg NaCl

6
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 (J1)
    components

7
Rotating diatomic molecule
  • Degeneracy of Jth level is (2J1)
  • Selection rules for absorption are
  • DJ 1
  • The molecule must have a non-zero dipole
    moment, p ¹ 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.

8
Rotational energy levels
  • For DJ1
  • DE 2 ( J1) h2/8p2I
  • 01 DE 2 h2/8p2I
  • 12 DE 4 h2/8p2I
  • 23 DE 6 h2/8p2I
  • etc., etc., etc.
  • Constant difference of
  • DE 2 h2/8p2I

9
Populations of rotational levels
10
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 6.028109 s-1.
  • DE 2 h2/8p2I h (6.028109 s-1)
  • So 84.421 6.028 14.00 ie J13 J14
  • 90.449 6.028 15 ie J14 J15
  • 96.477 6.028 16 ie J15 J16

11
Moment of inertia, I
  • DE 2 h2/8p2I hv h(6.028109 s-1)
  • I 2 h/(8p2 6.028109 )
  • I 2 (6.62610-34)/(8p2 6.028109 )
  • I 2.78410-45
  • Units?
  • Þ (J s)/(s-1) J s2 kg m2 s-2 s2 kg m2
  • But I mr2
  • m (0.0630.079)/(0.0630.079)NA 5.8210-26 kg
  • Þ r Ö(I/m) 218.610-12 m 218.6 pm

12
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)

13
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
  • ev (v 1/2) we
  • Untrue for real molecules since parabolic
    potential does not allow for bond breaking.
  • ev (v 1/2) we - (v 1/2)2 we xe
  • where xe is the anharmonicity constant

14
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 gt 1 are allowed
  • fundamental 01
  • overtone 02
  • hot band 12

15
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.
  • ev (v 1/2) we - (v 1/2)2 we xe
  • e2 (2 1/2) we - (2 1/2)2 we xe
  • e1 (1 1/2) we - (1 1/2)2 we xe
  • e0 (0 1/2) we - (0 1/2)2 we xe
  • e2 - e0 2we - 6we xe 5,668.1
  • e1 - e0 we - 2we xe 2,885.9
  • \ we 2,989.6 cm-1 we xe 51.9 cm-1 xe
    0.0174

16
High resolution infrared
  • Ev (v 1/2) (h/2p) (k/m)1/2
  • ev (v 1/2) we
  • EJ J(J1) (h2/8I)
  • eJ J(J 1) Bv
  • Vibrational rotational energy changes
  • e(v,J) (v 1/2) we J(J 1) Bv
  • Selection rule Dv1, DJ1
  • Rotational energy change must accompany a
    vibrational energy change.

17
Vibrational rotational changes in the IR
18
Hi-resolution spectrum of HCl
  • Above the gap DJ 1
  • Below the gap DJ 1
  • Intensities mirror populations of starting levels

19
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 (v0,J0) (v1,J1) 2590 is
    (v0,J1) (v1,J2)
  • So 2542 is (v0,J1) (v1,J0) 2525 is
    (v0,J2) (v1,J1)
  • (2575.19 - 2525.25) 6B0 B08.35 cm-1
  • (2590.95 - 2542.25) 6B1 B18.12 cm-1
  • Missing transition at 2542.25 2B0 2558.95 cm-1

20
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

21
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.

22
(No Transcript)
23
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

24
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.)
  • Combined

25
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

26
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

27
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 ev (v 1/2) we - (v 1/2)2 we xe
  • Substitute in v0, v1, v2, etc then compute
  • e1-e0 187 e2-e0 373 e3-e0 558
    calculate we, xe

28
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 J7 to J9 transition.
  • The DJ 2 and DJ -2 are nearly equal in
    intensity
  • Very near high intensity peak of exciting
    radiation needs good quality spectrometers

29
Rotational Raman
30
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

31
Raman spectra of KNO3N.B. strong symmetric
stretch band at 1,050 cm-1
32
Raman spectrum of aspirin tablet no sample
preparation
33
Raman vs IR
  • CHCl3
  • Which?
  • Very similar
  • Diffs.?
Write a Comment
User Comments (0)
About PowerShow.com