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 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

Typical energies

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.

MCWE or Rotational Spectroscopy Classification of molecules Based on moments of inertia, I=mr2 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

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

Rotating diatomic molecule Degeneracy of Jth level is (2J+1) 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.

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

Populations of rotational levels

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 = 15 ie J=14 ® J=15 & 96.477 ¸ 6.028 = 16 ie J=15 ® J=16

Moment of inertia, I Þ (J s)/(s-1) = J s2 = kg m2 s-2 s2 = kg m2 DE = 2 h2/8p2I = hv = h(6.028´109 s-1) I = 2 h/(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 Þ r = Ö(I/m) = 218.6´10-12 m = 218.6 pm

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)

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

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 > 1 are allowed fundamental 0®1 overtone 0®2 hot band 1®2

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

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.

Vibrational + rotational changes in the IR

Hi-resolution spectrum of HCl Above the “gap”; DJ = +1 Below the “gap”: DJ = –1 Intensities mirror populations of starting levels

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

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

Raman Light source? Laser He-Ne 633 nm or Argon ion 488, 515 nm 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.

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

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

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

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

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

Rotational Raman

Raman vs IR CHCl3 Which? Very similar Diffs.?