Quantum Chemistry. NC State University. Lecture 5. The electronic structure of molecules Absorption spectroscopy Fluorescence spectroscopy

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1 Quantum Chemistry Lecture 5 The electronic structure of molecules Absorption spectroscopy Fluorescence spectroscopy NC State University

2 3.5 Selective absorption and emission by atmospheric gases (source: P&O fig 4.2) Absorption by gases in the atmosphere 5500 o K 266 o K Electronic Vibrational Rotational

3 The electronic absorption of oxygen LUMO p* HOMO p

4 Potential energy surfaces oxygen dimer In addition to transitions Between states, absorption Of light can also lead to O O bond breaking. This is important for Production of oxygen atoms Needed to make ozone. Schumann-Runge bands nm Proc. R. Soc. Lond. A, 226, , (1954)

5 The electronic absorption of ozone LUMO p * HOMO non

6 The importance of the ozone continuum The fact that ozone can photodissociate leads to an intense Absorption of UV radiation over a wide spectral range. The ozone filter is very effective and protects the surface of Earth from receiving UV radiation below about 270 nm.

7 Molecular Orbital Theory In MO theory electrons are treated as including the entire molecule. Each MO is built up from a linear combination of atomic orbitals (LCAO). MO = I i = 1 c i i where i are atomic orbitals The coefficients are optimized by the selfconsistent field (SCF) method. The variational principle justifies minimization of the energy by adjustment of the coefficients c i.

8 The molecular orbitals in a diatomic molecule are formed from linear combinations of atomic orbitals + = Bonding s - = s s Anti-Bonding s*

9 Constructive overlap between two atomic orbitals gives rise to a bonding state NO NODES + = s s => 1/ s + s )

10 Destructive overlap between two atomic orbitals gives rise to an anti-bonding state ONE NODE - = s s => 1/ s - s )

11 For diatomic hydrogen we consider the s and s* molecular orbitals in the following energy diagram s* s s s

12 MO treatment of H 2 anti-bonding E 1s E 1s bonding The two electrons must have opposite spins,a and b. The wave function must be anti-symmetric with respect to electron exchange. MO = b (1) b (2) Spatial part 1 a(1)b(2) a(2)b(1) 2 Spin part (anti-symmetric)

13 The wavefunction is composed of electronic and nuclear parts = electronic nuclear Total Electronic Nuclear The wavefunction represents the probability amplitude of electrons and nuclei.

14 The wave equation can be separated into electronic and nuclear parts Hamiltonian Energy Operator Eigenvalue Energy value H elec = E elec H nucl = E nucl Wavefunctions

15 The electronic part

16 For diatomic hydrogen we consider the s and s* molecular orbitals in the following energy diagram s* s s s

17 Absorption of visible or ultraviolet radiation leads to electronic transitions s* Transition moment s s The change in nodal structure also implies a change in orbital angular momentum. s

18 The highest occupied molecular orbital of ethylene p

19 The lowest unoccupied molecular orbital of ethylene p* 1 p node

20 Separation of electronic and nuclear parts of the transition moment The transition moment, -e< 1 q 2 > can be separated into the electronic wavefunction that depends on q and the nuclear wavefunction that does not. e e q = q 1v 2v v These enter the rate expression as the square e 2 q 2 v 1v 2v 2 = M 12 2 FC

21 The electronic transition moment The electronic transition moment is M 12 = -e< 1 q 2 > Light will be absorbed when the electric vector is aligned with the transition moment. Light will not be absorbed when the electric vector is perpendicular to the transition moment.

22 Transition moment of lowest p-p* transition of ethylene p p* The transition moment is perpendicular to the change in nodal structure. Electromagnetic polarized along this direction will give the maximum transition probability.

23 Calculate the transition moment for a p-p* transition using a simple model The p and p * states are: p = and p * = The transition moment is: M p p* = e p x p* d

24 The magnitude of the transition moment for C 2 H 4 can be calculated from a simple model M p p* = e x 1 2 d = e 2 1 x 1 d 2 x 2 d = e 2 x 1 x 2 where x 1 - x 2 is the C=C bond length of 1.35 Å. One charge displaced through 1 Å has a dipole moment of 4.8 D. M p-p* = 3.24 D for C 2 H 4.

25 The nuclear part

26 The Franck-Condon factor is due to the overlap of ground state v=0 with excited state v =0, 1, etc. Excited state 0-0 Ground state

27 The Franck-Condon factor is due to the overlap of ground state v=0 with excited state v =0, 1, etc. Excited state 0-1 Ground state

28 The Franck-Condon factor is due to the overlap of ground state v=0 with excited state v =0, 1, etc. Excited state 0-2 Ground state

29 The Franck-Condon factor is due to the overlap of ground state v=0 with excited state v =0, 1, etc. Excited state 0-3 Ground state

30 The Franck-Condon factor is due to the overlap of ground state v=0 with excited state v =0, 1, etc. Excited state 0-4 Ground state

31 Based on the FC factors we can construct a stick spectrum Calculated assuming E(0-0 ) = 8000 cm -1 and vibrational mode of 1000 cm ev = cm -1.

32 The Franck-Condon factor determines the envelop of the absorption lineshape D D D S = D 2 /2 S is electron-phonon coupling D is nuclear displacement

33 Analytical expression for the FC factor The Franck-Condon factor is a vibrational overlap term. It depends on nuclear displacement, which is treated using the parameter S, the electron-vibration (or electron-phonon) coupling. The big the displacement the bigger is S. There is a progression of lines with relative intensities given by: The delta function gives a stick spectrum spaced by energy equal to the vibrational frequency. This function is also known as a Poisson distribution. For large S this function approaches a Gaussian shape.

34 The absorption cross section, s A The absorption cross section has units of area (cm 2 ). It gives a probability for absorption. We have discussed the probability in terms of the transition dipole moment. M 12 and shape in terms of the Franck-Condon factor, FC. The absorption cross section is proportional to the well known extinction coefficient. The extinction coefficient has units of M -1 cm -1.

35 I = I 0 10 A A = Cd Beer-Lambert Law A is the absorbance. ( ) is the extinction coefficient. The unit of ( ) is M -1 cm -1. C is the concentration (M). d is the pathlength (cm). The exponential attenuation of the intensity is shown in the Figure. I 0 x dx I I +di d I

36 Fluorescence usually occurs after vibrational relaxation 1. Absorption 2. Vibrational relaxation 3.Fluorescence

37 The Franck-Condon principle: Transitions are vertical in both absorption and emission 1. Absorption 2. Vibrational relaxation 3.Fluorescence

38 The Franck-Condon factor is the same for absorbance and fluorescence 1. Absorption 2. Vibrational relaxation 3.Fluorescence

39 This leads to a mirror image relationship between absorption and fluorescence bands Absorption Fluorescence Energy Wavelength

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