Spectroscopy The Light Spectrum. Vibrational Spectroscopy D(t) r(t) Band structure The higher the BO: i) the deeper the Well, ii) the wider the spacing.

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Presentation transcript:

Spectroscopy The Light Spectrum

Vibrational Spectroscopy D(t) r(t) Band structure The higher the BO: i) the deeper the Well, ii) the wider the spacing between vibrational states, and iii) the higher the frequency.

Scanning Infrared Spectrometer Grating Orientation (  ) Absorption I 0 = Blank I = Sample   Determines wavelength selected

Vibrational Spectra of Molecules k related to Bond Order Reduced Mass  = frequency ~ 1/  (1/cm)  Wave numbers  = wave length (cm) C-H C=O O-H C-CC-H

5 Photoelectron Spectroscopy Another form of spectroscopy can be used to corroborate the orbital occupancies predicted by MO theory – photoelectron spectroscopy (PES). The principles behind photoelectron spectroscopy are the same as those behind the photoelectric effect:

6 As the energies of interest typically correspond to that of ultraviolet light PES is often referred to as UPES (ultraviolet photoelectron spectroscopy) spectrometer. Similar to a mass spectrometer. Photoelectron Spectroscopy UV to X-ray K.E. (electron) = h UV  - IE IE = Ionization Energy KE = Kinetic Energy= 1/2 mv 2

The PES spectrum Neon is very simple, showing only one line for each orbital. Photoelectron Spectroscopy of Neon Unbound State 2s 2p High KE Low KE High IE Low IE h O

Photoelectron Spectroscopy of Neon Unbound State 2s 2p High KE Low KE High IE Low IE h O If in this UV PES a source with wavelength of 20 nm is used what would be the kinetic energy of an electron removed from the 2s and 2porbitals. Would 30 nm still work? K.E. (electron) = h UV  – IEWe need uv and EI from 2s and 2p orbitals

9 Photoelectron Spectroscopy of Neon = c/  3.0*10 8 m/s)/(20*10 -9 m) = 1.5*10 16 s -1 From the PES Spectrum: EI (2s) = 4750 kJ/mol & EI (2p) = 2100 kJ/mol K.E.(2s) = h UV  – IE(2s) = (6.626* Js)(1.5*10 16 s -1 ) – (4,750,000 J/mol)/(6.02*10 23 mol -1 ) = 9.93* J – 7.89* J = 2.05* J 1 eV = 1.602* J = (2.05* J) / (1.602* J/eV) = 12.8 eV K.E.(2p) = h UV  – IE(2p) = 9.93* J – (2,100,000 J/mol)/(6.02*10 23 mol -1 ) = 9.93* J – 3.49* J = 6.44* J = (6.44* J)/ (1.602* J/eV) = 40.2 eV

11 22 33 11 22 33 11 22 Photoelectron Spectroscopy of Diatomic Molecules N2N2 O2O2 22 11 33

The PES spectrum of diatomics are more complex resulting from the are vibrational energy levels. Without vibrational energy levels the lines in the PES spectrum would not be split as seen with neon. When the UV photon is absorbed, and as the electron is emitted transitions take place from the ground state vibrational state of the molecule to an excited vibrational state of the cation. The change in vibrational state reduces the KE of the electron ejected as a consequence of the conservation of energy. Hence there is a distribution of KE’s of the electron from a particular MO is according to the vibrational transitions that are possible Ionization Energy (eV) Vibrational Fine Structure

12 Vibrational fine structure indirectly gives the vibrational spectrum for the cation produced. Comparison of the two vibrational spectra tells us how the bond was affected: i) Loss of a nonbonding electron: the shape of the potential energy diagram for the molecule changes little. ii) Loss of a bonding electron: the shape of the potential energy diagram changes dramatically – indicating weakening of the bond. iii) Loss of an antibonding electron is removed: the shape of the potential energy diagram again changes dramatically – this time, indicating strengthening of the bond. Photoelectron spectroscopy allows us to measure energies of orbitals – as well as confirming behaviour predicted by MO. Vibrational Fine Structure

Experimental Evidence for MO Predictions 11 2  11 22 33 44 N2N2 CO 11 2  11 1s1s 11 11 2s2s 3p3p 3p3p 11 22 11 PES shows N 2 and CO are isoelectronic. As the vibrational frequency of N 2 + is higher than CO +, the splitting is wider. Increase in BO of the cation leads to large spacing in the 2    signal. What is the vibrational frequency of N 2 + and CO + ?