Chemistry 2

Lecture 11

• Electronic spec of polyatomic molecules: chromophores and fluorescence

Assumed knowledge

Excitations in the visible and ultraviolet correspond to excitations of electrons between orbitals. There are an infinite number of different electronic states of atoms and molecules.

Learning outcomes

• Be able to draw the potential energy curves for excited electronic states in diatomics that are bound and unbound • Be able to explain the vibrational fine structure on the bands in electronic spectroscopy for bound excited states in terms of the classical Franck-Condon model • Be able to explain the appearance of the band in electronic spectroscopy for unbound excited states

Some random images of last lecture…

w e ’ w e ” r e ” r e ’ D e ” D e ’

Franck-Condon Principle

Lecture 7

Franck-Condon Principle (reprise) 0 0 1 1 2 2 R R 3 3 4 4 5 5 0 1 2 R 3 4 5 Lecture 7

Electronic spectra of larger molecules…

An atom A diatomic (or other small) molecule A large molecule A molecule has 3N-6 different vibrational modes. When you have no selection rules any more on vibrational transitions then the spectrum quickly becomes so complicated that the vibrational states cannot be readily resolved.

Lecture 7

Electronic spectra of larger molecules…

An atom 3s 2s 1s 2p A diatomic (or other small) molecule A large molecule Lecture 7

Jablonski Diagrams…

Electronic states

(thick lines) E tot = E elec + E vib Again, this is the Born-Oppenheimer approximation.

Vibrational levels

(thin lines) Lecture 7

Correlation between diatomic PES and Jablonski diagram Jablonski Lecture 7

Nomenclature and spin states

In polyatomic molecules, the total electron spin, S, is one of the few good quantum numbers.

If the total electron spin is zero: way to arrange the spins, and we have a singlet state, denoted, S

(c.f. L=0 for atoms)

S = 0, then there is only one If there is one unpaired electron: there are 2 ways the spin can be aligned (up and down), and we have a doublet state, denoted, D the total spin is S = ½ and If there are two unpaired spins: then there are 3 ways the spins can be aligned (c.f. 3 x p- orbitals for L=1). This is a triplet state that we denote, T Total Spin 0 ½ Name singlet doublet symbol S D 1 triplet T Total spin is also called “multiplicity”.

Lecture 7

Jablonski Diagrams…

First excited singlet state

S

1

T

1

First excited triplet state

S

0

“0” = ground state (which is a singlet in this case) Only the ground state gets the symbol “0” Other states are labelled in order, 1, 2, … according to their multiplicity Lecture 7

Correlation between diatomic PES and Jablonski (again)

S

1

S

0

T

1

The x-axis doesn’t mean anything in a Jablonski diagram.

Position the states to best illustrate the case at hand.

Lecture 7

Chromophores

Any electron in the molecule can be excited to an unoccupied level. We can separate electrons in to various types, that have characteristic spectral properties. A chromophore is simply that part of the molecule that is responsible for the absorption.

Core electrons: These electrons lie so low in energy that it requires, typically, an X-ray photon to excite them. These energies are characteristic of the atom from which they come.

Valence electrons: These electrons are shared in one or more bonds, and are the highest lying occupied states (HOMO, etc). Transitions to low lying unoccupied levels (LUMO, etc) occur in the UV and visible and are characteristic of the bonds from which they come.

Lecture 7

Types of valence electrons

s -electrons are localised between two atoms and tightly bound. Transitions from s -orbitals are therefore quite high in energy (typically vacuum-UV, 100-200 nm).

p -electrons are more delocalised (even in ethylene) than their s counterparts. They are bound less tightly and transitions from single, unconjugated p -orbital).

p orbitals occur at lower energy (typically far UV, 150-250nm, for a n-electrons are not involved in chemical bonding. The energy of a non bonding orbital lies typically between that for bonding and antibonding orbitals. Transitions are therefore lower energy. n-orbitals are commonly O, N lone pairs, or Hückel p -orbitals with E = a Lecture 7

Transitions involving valence electrons

Vacuum (or far) uv pp* Near uv ss*

n

p*

n

s* 100 200 Visible 300 400 500 Wavelength (nm) 600 Near IR 700 800 Lecture 7

Chromophores in the near UV and visible There are two main ways that electronic spectra are shifted into the near-UV and visible regions of the spectrum: 1. Having enough electrons that higher-lying levels are filled. Remember that the electronic energy level spacing decreases with increasing quantum numbers (e.g. H-atom). Atoms/molecules with d and f electrons often have spectra in the visible and even near-IR. Large atoms (e.g. Br, I) have electrons with large principle quantum number.

Ni(H 2 O) 6 2+ Ni(H 2 NCH 2 CH 2 CH 3 ) 3 2+ Mn(H 2 O) 6 2+ Lecture 7

Chromophores in the near UV and visible 2. Delocalised delocalised p -electrons. From your knowledge of Hückel theory and “particle-in-a-box”, you should understand that larger Hückel chromophores have a larger number of more extended, p orbitals, with lower energy. Transitions involving larger chromophores occur at lower energy.

b -carotene (all trans) Lecture 7

Effect of chromophore size

400 450 500 550 600 650

Wavelength (nm)

700 750 800 Chromophore Lecture 7

Chromophores in the near UV and visible Aromatic chromophores: What is this structure?

Benzene Tetracene Lecture 7

Chromophores at work

CH 3 CH 3 CH 3 CH 3 CH 3 O

trans

-retinal (light absorber in eye) N N C C O O Dibenzooxazolyl-ethylenes (whiteners for clothes) Lecture 7

After absorption, then what?

After molecules absorb light they must eventually lose the energy in some process. We can separate these energy loss processes into two classes: • radiative transitions (fluorescence and phosphorescence) • non-radiative transitions (internal conversion, intersystem crossing, non-radiative decay) Let’s use a Jablonski diagram (ie large molecule picture) to explore these processes… Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 2&3 computer labs) Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 2&3 computer lab) Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 3 computer lab) Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 3 computer lab) Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 3 computer lab) Lecture 7

Slide taken from “Invitrogen” tutorial ( http://probes.invitrogen.com/resources/education/ , or Level 3 computer lab) Lecture 7

Summary

~10 -12 s

S 1

~10 -8 s

S 0 Fluorescence is ALWAYS red-shifted (lower energy) compared to absorption 1 = Absorption 2 = Non-radiative decay 3 = Fluorescence

Lecture 7

Franck-Condon Principle (in reverse) 0 0 1 1 2 2 R R 3 3 4 4 5 5 0 1 2 R 3 4 5 Lecture 7

Franck-Condon Principle (in reverse) 0 0 1 1 2 2 R R 3 3 4 4 5 5 Note: If vibrational structure in the ground and exited state are similar, then the spectra look the same, but reversed -> the so-called “mirror symmetry” Lecture 7 0 1 2 R 3 4 5

Stokes shift

Absorption

The shift between l max (abs.) and l max (fluor) is called the STOKES SHIFT A bigger Stokes shift will produce more dissipation of heat Lecture 7

The Origin of the Stokes Shift and mirror symmetry

v’=4 v’=0 v”=4 v”=0 Stokes shift

sum = Stokes shift Mirror symmetry If the vibrational level structure in the ground and excited electronic states is similar, then the absorption and fluorescence spectra look similar, but reversed. Notice that if 0  4 is the most intense in absorption, 0  4 is also most intense in emission.

The Stokes shift here is G’(4) + G”(4) Lecture 7

Different Stokes

A.

B.

Which dye dissipates most heat when excited?

C.

D.

E.

F.

Note mirror symmetry in most, but not all dyes.

Lecture 7

Fluorescence spectrum



f(

l exc

)

NRD … animation … Lecture 7

Real data…

Absorption Fluorescence

500 550 600 650 Wavelength (nm)

- Fluorescence is always to longer wavelength - Stokes shift = (abs. max.) – (fluor. max.) [= 50 nm here] - Mirror symmetry

700

Lecture 7

Revision: The Electromagnetic Spectrum

Revision: Light as a EM field

wavefunctions and classical vibration

A molecule in a particular solution to the vibrational Schrödinger equation has a stationary probability distribution: So why do we call this vibration?

wavefunctions and classical vibration

A molecule which is in a superposition of v = 0 and v =1 will be in a non-stationary state: Y 0 Y 1 |Y| 2 The animation shows the time-dependence of an admixture of 20% v=1 into the v=0 wavefunction. Mixing Y 1 into the Y 0 wavefunction shifts the probability distribution to the right as drawn (red). If the molecular dipole changes along the coordinate then the vibration brings about an oscillating dipole.

wavefunctions and electronic vibration

+ 0.2

× = If mixing some excited state character into the ground state wavefunction changes the dipole, then electric fields can do this. The transition is said to be “allowed”.

Hydrogen

+ 0.2

× = + 0.2

× = Is to 2p transition is allowed. Electrical dipole is brought about by mixing 1s and 2p.

Is to 22 transition is forbidden. No electrical dipole is brought about by mixing 1s and 2s.

Which electronic transitions are allowed?

     The allowed transitions are associated with electronic vibration giving rise to an oscillating dipole

E lectronic spectroscopy of diatomics

• For the same reason that we started our examination of IR spectroscopy with diatomic molecules (for simplicity), so too will we start electronic spectroscopy with diatomics.

• Some revision: – there are an infinite number of different electronic states of atoms and molecules – changing the electron distribution will change the forces on the atoms, and therefore change the potential energy, including

k

, w e , w e

x

e ,

D

e ,

D

0 , etc

Depicting other electronic states

Notice the different shape potential energy curves including different bond lengths… Excited Electronic States

1. Unbound 2. Bound

Ground Electronic State

There is an infinite number of excited states, so we only draw the ones relevant to the problem at hand.

Ladders upon ladders…

Each electronic state has its own set of vibrational states.

w e ’ Note that each electronic state has its own set of vibrational parameters, including: - bond length, r e - dissociation energy, D e - vibrational frequency, w e D e ” w e ”

Notice: single prime (

’)

= upper state double prime (”) = lower state

r e ” r e ’ D e ’

The Born-Oppenheimer Approximation

The total wavefunction for a molecule is a function of both nuclear and electronic coordinates: Y (r 1 …r n

, R

1 …R n ) where the electron coordinates are denoted, r

i

nuclear coordinates, R

i .

, and the The Born-Oppenheimer approximation uses the fact the nuclei, being much heavier than the electrons, move ~1000x more slowly than the electrons. This suggests that we can separate the wavefunction into two components: Y (r

1 …r n , R 1 …R n ) =

 elec (r 1

…r

n

; R

i ) x  vib (R 1 …R n ) Total wavefunction = Electronic wavefunction at

×

each geometry Nuclear wavefunction

The Born-Oppenheimer Approximation

Y (

r 1 …r n , R 1 …R n ) =

 elec (

r

1

…r

n

; R

i ) x  vib (

R

1 …

R

n ) Total wavefunction = Electronic wavefunction at

×

each geometry Nuclear wavefunction The B-O Approximation allows us to think about (and calculate) the motion of the electrons and nuclei separately. The total wavefunction is constructed by holding the nuclei at a fixed distance, then calculating the electronic wavefunction at that distance. Then we choose a new distance, recalculate the electronic part, and so on, until the whole potential energy surface is calculated.

While the B-O approximation does break down, particularly for some excited electronic states, the implications for the way that we interpret electronic spectroscopy are enormous!

Spectroscopic implications of the B-O approx.

1. The total energy of the molecule is the sum of electronic and vibrational energies:

E tot

E

tot

= E

elec

+ E

vib

E elec E vib

Spectroscopic implications of the B-O approx.

• In the IR spectroscopy lectures we introduced the concept of

μ

21 a transition dipole moment:   Y 2 * (

r i

,

R i

)

μ

Y 1 (

r j

,

R j

)

drdR

 0 |2  transition dipole moment upper state wavefunction dipole moment operator lower state wavefunction integrate over all coords.

|1  using the B-O approximation:

μ

21      2 2 *

vib

*

vib

( (

R i R i

)  2 *

elec

(

r i

)

μ

(

R

)  1

vib

) ˆ  1

elec

(

r j

(

R j

)

dR

)  1

vib

(

R j

)

drdR

2. The transition moment is a smooth function of the nuclear coordinates.

Spectroscopic implications of the B-O approx.

μ

21      2 *

vib

(

R i

)  2 *

elec

(

r i

) ˆ  1

elec

(

r j

)  1

vib

(

R j

)

drdR

  2 *

vib

(

R i

)

μ

(

R

)  1

vib

(

R j

)

dR

μ

0   2 *

vib

(

R i

)  1

vib

(

R j

)

dR

|2  |1 

2. The transition moment is a smooth function of the nuclear coordinates. If it is constant then we may take it outside the integral and we are left with a vibrational overlap integral. This is known as the Franck-Condon approximation.

3. The transition moment is derived only from the electronic term. A consequence of this is that the vibrational quantum numbers, v, do not constrain the transition (no

D

v selection rule).

Electronic Absorption

There are no vibrational selection rules, so any D v is possible.

But, there is a distinct favouritism for certain D v. Why is this?

Franck-Condon Principle (classical idea)

Classical interpretation:

“Most probable bond length for a molecule in the ground electronic state is at the equilibrium bond length,

r

e .” 0 1 2 R 3 4 5

Franck-Condon Principle (classical idea)

The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

In the ground state, the molecule is most likely in v=0.

0 1 2 R 3 4 5

Franck-Condon Principle (classical idea)

•The Franck-Condon Principle

states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The electron excitation is effectively instantaneous; the nuclei do not have a chance to move. The transition is represented by a VERTICAL ARROW on the diagram (R does not change).

0 1 2 R 3 4 5

Franck-Condon Principle (classical idea)

•The Franck-Condon Principle

states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The most likely place to find an oscillating object is at its turning point (where it slows down and reverses). So the most likely transition is to a turning point on the excited state.

0 1 2 R 3 4 5

Quantum (mathematical) description of FC

μ

21 

μ

0 

principle

 *

vib

2 (

R i

)  1

vib

(

R j

)

dR

approximately constant with geometry Franck-Condon (FC) factor

μ

21

= constant × FC factor

FC factors are not as restrictive as IR selection rules ( D v=1). As a result there are many more vibrational transitions in electronic spectroscopy.

FC factors, however, do determine the intensity.

Franck-Condon Principle (quantum idea)

In the ground state, what is the most likely position to find the nuclei?

Prob   2 Max. probability at

R e

3 2 1 0 0 1 2 R v=0 3  (0) 4

Franck-Condon Factors

If electronic excitation is much faster than nuclei move, then wavefunction cannot change. The most likely transition is the one that has most overlap with the excited state wavefunction.

0 1 2 Wave number 3 4 1 v’ = 0 2 v” = 0

Look at this more closely…

Negative overlap in middle Positive overlap at edges overall very small overlap Negative overlap to left, postive overlap to right overall zero overlap • Excellent overlap everywhere

Franck-Condon Factors

1 0 2 Wave number 3 4

Franck-Condon Factors

Note: analogy with classical picture of FC principle!

v=10 v. poor v=0 overlap

Electronic Absorption

There are no vibrational selection rules, so any D v is possible.

Relative vibrational intensities come from the FC factor

μ

21

= constant × FC factor

Absorption spectrum of binaphthyl

•Example of real spectra showing FC profile 16 17 18 19 15 20 21 22 14 23 10 11 12 13 24 25 26 27 28 (3) (4) (5) 6 7 8 9 30100 30200 30300 30400 30500 30600 Wave number (cm -1 ) 30700 30800 30900

Absorption spectrum of CFCl

= CCl 2 peaks 5 3 17000 5 3 4 2 18000 (0,n,0) (0,n,1) (0,n,2) 10 } (0,0,1) hot bands 19000 20000 Wave number (cm -1 ) 21000 (0,n,0) (0,n,1) (0,n,2) (1,n,0)

Unbound states (1)

If the excited state is dissociative, e.g. a p * state, then there are no vibrational states and the absorption spectrum is broad and diffuse.

Unbound states (2)

Even if the excited state is bound, it is possible to access a range of vibrations, right into the dissociative continuum. Then the spectrum is structured for low energy and diffuse at higher energy.

Some real examples…

HI A purely dissociative state leads to a diffuse spectrum.

Some real examples…

I 2 The dissociation limit observed in the spectrum!

0.25

0.20

0.15

0.10

0.05

0.00

16000 18000 20000 Wave number (cm -1 ) 22000 I 2

Analyzing the spectrum…

All transitions are (in principle) possible. There is no D v selection rule 0.15

0.10

0.05

0.00

0.25

Vibrational structure

0.20

16000 18000 20000 Wave number (cm -1 ) 22000 I 2

v’ 25 26 27 28 29 30 31 32 33 34 35 0 0 0 v” 0 0 0 0 0 0 0 0

Analysing the spectrum…

cm -1 18327.8

18405.4

18480.9

18555.6

18626.8

18706.3

18780.0

18846.6

18911.5

18973.9

19037.5

G

( v  )  ( v   1 2  w

e

 ( v   1 2  2 w

e x e

How would you solve this?

(you have too much data!) 1. Take various combinations of v’ and solve for w e and w e x e simultaneously. Average the answers.

2. Fit the equation to your data (using XL or some other program).

0 0 0 0 0 0 0 0 0 0 v” 0

Analyzing the spectrum…

E tot E tot

 

E elec E elec

 

E vib G

( v  ) 

E elec

 ( v   1 2  w

e

 ( v   1 2  2 w

e x e

v’ 25 cm -1 18327.8

20000

26 18405.4

19500

27 18480.9

19000

28 18555.6

29 18626.8

18500

30 31 32 18706.3

18780.0

18846.6

18000 17500 E elec w e w e x e = 15,667 cm -1 = 129.30 cm -1 = 0.976 cm -1

33 18911.5

34 35 18973.9

19037.5

17000 10 20 30 40 50 v' Dissociation energy = 19950 cm -1 60 70

Summary

• The potential energy curve and the equilibrium geometry in an electronic excited state will be different to the ground state • An excited state may have no equilibrium geometry: unbound • For bound excited states, transitions to the individual vibrational levels of the excited state are observed • The energies of these transitions depend on the vibrational levels of the excited state • The intensities of the lines depend on the Franck-Condon factors with ‘vertical’ transition being the strongest • For unbound excited states, the electronic spectrum is broad and diffuse

The take home message from this lecture is to understand the (classical) Franck-Condon Principle

Next lecture

• The vibrational spectroscopy of polyatomic molecules.

Week 12 homework

• Vibrational spectroscopy worksheet in tutorials • Practice problems at the end of lecture notes • Play with the “IR Tutor” in the 3 rd online simulations: floor computer lab and with the http://assign3.chem.usyd.edu.au/spectroscopy/index.php

Practice Questions

1. Which of the following molecular parameters are likely to change when a molecule is electronically excited?

(a) ω e (b) ω e x e (c) μ (d)

D e (e) k

2.

Consider the four sketches below, each depicting an electronic transition in a diatomic molecule. Note that more than one answer may be possible (a) Which depicts a transition to a dissociative state?

(b) (c) (d) (e) Which depicts a transition in a molecule that has a larger bond length in the excited state?

Which would show the largest intensity in the 0-0 transition?

Which represents molecules that can dissociate after electronic excitation?

Which represents the states of a molecule for which the v”=0 → v’=3 transition is strongest?