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Spectroscopy

Spectroscopy. Microwave (Rotational) Infrared (Vibrational) Raman (Rotational & Vibrational) Texts “Physical Chemistry”, 6th edition, Atkins “Fundamentals of Molecular Spectroscopy”, 4th edition, Banwell & McCash. Introduction-General Principles.

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Spectroscopy

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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

  2. Introduction-General Principles Spectra - transitions between energy states Molecule,Ef - Ei = huphoton • 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, I=mr2 • IA ¹ IB ¹ ICvery complex eg H2O • IA = IB = ICno MCWE spectrum eg CH4 • IA ¹ IB = ICcomplicated eg NH3 • IA = 0, IB = IClinear 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 (J+1) components

  7. Rotating diatomic molecule • Degeneracy ofJth level is(2J+1) • Selection rules for absorption are: • DJ= +1 • The molecule must have a non-zero dipole moment,p¹ 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 800Ksince 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 DJ=1 • DE = 2 ( J+1) h2/8p2I 0®1 DE = 2 h2/8p2I 1®2 DE = 4 h2/8p2I 2®3 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 6.028´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 = 15ie J=14® J=15 • & 96.477 ¸ 6.028 = 16ie J=15® J=16

  11. Moment of inertia, I • DE = 2 h2/8p2I = hv = h(6.028´109 s-1) I = 2h/(8p2 6.028´109 ) I = 2(6.626´10-34)/(8p2 6.028´109 ) I = 2.784´10-45 Units? • Þ (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 • Þr = Ö(I/m) = 218.6´10-12m= 218.6pm

  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 & pmust 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)2wexe • 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> 1are allowed • fundamental 0®1 • overtone 0®2 • hot band 1®2

  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)2wexe • e2= (2 + 1/2)we - (2 + 1/2)2wexe • e1= (1 + 1/2)we - (1 + 1/2)2wexe • e0= (0 + 1/2)we - (0 + 1/2)2wexe • e2-e0= 2we - 6wexe= 5,668.1 • e1-e0= we - 2wexe= 2,885.9 • \ we= 2,989.6cm-1wexe= 51.9cm-1xe= 0.0174

  16. High resolution infrared Ev = (v + 1/2) (h/2p) (k/m)1/2 ev = (v + 1/2)we EJ = J(J+1) (h2/8I) eJ = J(J + 1)Bv Vibrational + rotational energy changes • e(v,J) = (v + 1/2)we + J(J + 1)Bv • Selection rule: Dv=+1, DJ=±1 • 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 (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

  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. 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

  23. 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

  24. 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 ev= (v + 1/2)we - (v + 1/2)2wexe • Substitute in v=0, v=1, v=2, etc then compute: • e1-e0 = 187 e2-e0 = 373 e3-e0 = 558 …& calculate we, xe

  27. 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 DJ = +2 and DJ = -2 are nearly equal in intensity • Very near high intensity peak of exciting radiation; needs good quality spectrometers

  28. 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 KNO3N.B. strong symmetric stretch band at 1,050 cm-1

  31. Raman spectrum of aspirin tablet; no sample preparation

  32. Raman vs IR • CHCl3 • Which? • Very similar • Diffs.?

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