chemistry 2 lecture 11 electronic spec of polyatomic molecules: chromophores and fluorescence
TRANSCRIPT
Chemistry 2
Lecture 11 •Electronic spec of polyatomic molecules: chromophores and fluorescence
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
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.
Assumed knowledge
Lecture 7
Some random images of last lecture…
re’re”
De”
De’
e”
e’
Franck-CondonPrinciple
Lecture 7
Franck-Condon Principle (reprise)
6543210
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 A diatomic (or other small) molecule
A large molecule
1s
2s
2p
3s
Lecture 7
Electronic states(thick lines)
Vibrational levels(thin lines)
Jablonski Diagrams…
Etot = Eelec + Evib
Again, this is the Born-Oppenheimer approximation.
Lecture 7
Correlation between diatomic PES and Jablonski diagram
Jablonski
Lecture 7
Nomenclature and spin states
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
Name symbol
0 singlet S
½ doublet D
1 triplet T
In polyatomic molecules, the total electron spin, S, is one of the few good quantum numbers.
If the total electron spin is zero: S = 0, then there is only one way to arrange the spins, and we have a singlet state, denoted, S(c.f. L=0 for atoms)
If there is one unpaired electron: the total spin is S = ½ and there are 2 ways the spin can be aligned (up and down), and we have a doublet state, denoted, D
Total spin is also called “multiplicity”.
Lecture 7
Jablonski Diagrams…
Only the ground state gets the symbol “0”Other states are labelled in order, 1, 2, … according to their multiplicity
S0
S1
T1
“0” = ground state(which is a singlet in this case)
First excited singlet state
First excited triplet state
Lecture 7
The x-axis doesn’t mean anything in a Jablonski diagram.Position the states to best illustrate the case at hand.
S0
S1
T1
Correlation between diatomic PES and Jablonski (again)
Lecture 7
Chromophores
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.
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.
Lecture 7
Types of valence electrons
-electrons are localised between two atoms and tightly bound. Transitions from -orbitals are therefore quite high in energy (typically vacuum-UV, 100-200 nm).
-electrons are more delocalised (even in ethylene) than their counterparts. They are bound less tightly and transitions from orbitals occur at lower energy (typically far UV, 150-250nm, for a single, unconjugated -orbital).
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 -orbitals with E =
Lecture 7
Transitions involving valence electrons
Vacuum (or far) uv
Nearuv
Visible Near IR
n
n
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(H2O)62+
Ni(H2NCH2CH2CH3)32+ Mn(H2O)6
2+
Lecture 7
Chromophores in the near UV and visible
2. Delocalised -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, delocalised orbitals, with lower energy. Transitions involving larger chromophores occur at lower energy.
-carotene (all trans)
Lecture 7
400 450 500 550 600 650 700 750 800
Wavelength (nm)
Chromophore
Effect of chromophore size
Lecture 7
Chromophores in the near UV and visible
Aromatic chromophores:
Benzene Tetracene
What is this structure?
Lecture 7
Chromophores at work
O
N
C
O
N
C
Dibenzooxazolyl-ethylenes(whiteners for clothes)
O
CH3
CH3
CH3
CH3 CH3
trans-retinal(light absorber in eye)
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 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 2&3 computer labs)
Lecture 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 2&3 computer lab)
Lecture 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 3 computer lab)
Lecture 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 3 computer lab)
Lecture 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 3 computer lab)
Lecture 7Slide taken from “Invitrogen” tutorial (http://probes.invitrogen.com/resources/education/, or Level 3 computer lab)
Lecture 7
S0
S1
~10-12 s
~10-8 s
1 = Absorption2 = Non-radiative decay3 = Fluorescence
Summary
Fluorescence is ALWAYS red-shifted (lower energy) compared to absorption
Lecture 7
Franck-Condon Principle (in reverse)
Lecture 7
Franck-Condon Principle (in reverse)
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
Stokes shift
Absorption
The shift between max(abs.) and max(fluor) is called the STOKES SHIFTA bigger Stokes shift will produce more dissipation of heat
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The Origin of the Stokes Shift and 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 04 is the most intense in absorption, 04 is also most intense in emission.The Stokes shift here is G’(4) + G”(4)
Mirror symmetry
v”=0
v’=0
v’=4
v”=4
sum = Stokesshift
Stokes shift
Lecture 7
Different Stokes
Note mirror symmetry in most, but not all dyes.
Which dye dissipates most heat when excited?
A. B. C.
D. E. F.
Lecture 7
Fluorescence spectrum f(exc)
… animation …
NRD
Lecture 7
500 550 600 650 700
Wavelength (nm)
Absorption Fluorescence
Real data…
- Fluorescence is always to longer wavelength- Stokes shift = (abs. max.) – (fluor. max.) [= 50 nm here]
- Mirror symmetry
Revision: The Electromagnetic Spectrum
Revision: Light as a EM field
wavefunctions and classical vibrationA 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 vibrationA molecule which is in a superposition of v = 0 and v =1 will be in a non-stationary state:
0
1
The animation shows the time-dependence of an admixture of 20% v=1 into the v=0 wavefunction. Mixing 1 into the 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”.
+ 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.
Hydrogen
Which electronic transitions are allowed?
The allowed transitions are associated with electronic vibration giving rise to an oscillating dipole
Electronic 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, e, exe, De, D0, etc
Depicting other electronic states
Ground Electronic State
Excited Electronic States1. Unbound
2. Bound
There is an infinite number of excited states, so we only draw the ones relevant to the problem at hand.
Notice the different shape potential energy curves including different bond lengths…
Ladders upon ladders…Each electronic state has its own set of vibrational states.
Note that each electronic state has its own set of vibrational parameters, including:- bond length, re
- dissociation energy, De
- vibrational frequency, we
re’re”
De”
De’
e”
e’
Notice: single prime (’) = upper statedouble prime (”) = lower state
The Born-Oppenheimer ApproximationThe total wavefunction for a molecule is a function of both nuclear and electronic coordinates:
(r1…rn, R1…Rn)
where the electron coordinates are denoted, ri , and the nuclear coordinates, Ri.
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:
(r1…rn, R1…Rn) = elec (r1…rn; Ri) x vib(R1…Rn)
Electronic wavefunction at ×each geometry
Total wavefunction = Nuclear wavefunction
The Born-Oppenheimer Approximation
(r1…rn, R1…Rn) = elecr1…rn; Ri)xvibR1…Rn)
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!
Electronic wavefunction at ×each geometry
Total wavefunction = Nuclear wavefunction
Spectroscopic implications of the B-O approx.
1. The total energy of the molecule is the sum of electronic and vibrational energies:
Etot = Eelec + Evib
Eelec
Etot
Evib
Spectroscopic implications of the B-O approx.
• In the IR spectroscopy lectures we introduced the concept of a transition dipole moment:
|1
|2upper state
wavefunctionlower state
wavefunctiontransition
dipole moment
dipole moment operator
integrate over all coords.
0),( ˆ),( 1*221 drdRRrRr jjii μμ
dRRRR
drdRRrrR
jvib
ivib
jvib
jelec
ielec
ivib
)( )()(
)()( ˆ)()(
1*2
11*2
*221
μ
μμusing the B-O approximation:
2. The transition moment is a smooth function of the nuclear coordinates.
Spectroscopic implications of the B-O approx.
|1
|2
dRRR
dRRRR
drdRRrrR
jvib
ivib
jvib
ivib
jvib
jelec
ielec
ivib
)( )(
)( )()(
)()( ˆ)()(
1*20
1*2
11*2
*221
μ
μ
μμ
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 v selection rule).
Electronic AbsorptionThere are no vibrational selection rules, so any v is possible.
But, there is a distinct favouritism for certain 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, re.”
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.
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).
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.
Quantum (mathematical) description of FC principle
approximately constant with
geometry
Franck-Condon (FC) factor
μ21 = constant × FC factor
FC factors are not as restrictive as IR selection rules (v=1). As a result there are many more vibrational transitions in electronic spectroscopy.
FC factors, however, do determine the intensity.
dRRR jvib
ivib )( )( 1
*2021 μμ
Franck-Condon Principle (quantum idea)
In the ground state, what is the most likely position to find the nuclei?
2Prob
Max. probability at Re
0
1 2 3 4
Wave number
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.
v’ = 0
1
2
v” = 0
Look at this more closely…
• Excellent overlap everywhere
Negative overlap to left, postive overlap to right
overall zero overlap
Negative overlap in middlePositive overlap at edgesoverall very small overlap
Franck-Condon Factors
0
1
2
34
Wave number
Franck-Condon Factors
v=10
Note: analogy with classical picture of FC principle! v. poor v=0 overlap
Electronic Absorption
There are no vibrational selection rules, so any v is possible.
μ21 = constant × FC factor
Relative vibrational intensitiescome from the FC factor
Absorption spectrum of binaphthyl•Example of real spectra showing FC profile
30100 30200 30300 30400 30500 30600 30700 30800 30900
8
9
10
11
12
13
14
15
1617 18
19
2021
22
23
24
25 26 27
28
(3) (4) (5) 67
Wave number (cm-1)
Absorption spectrum of CFCl
If the excited state is dissociative, e.g. a * state, then there are no vibrational states and the absorption spectrum is broad and diffuse.
Unbound states (1)
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.
Unbound states (2)
Some real examples…A purely dissociative state
leads to a diffuse spectrum.HI
The dissociation limit observed in the spectrum!I2
Some real examples…
Analyzing the spectrum…All transitions are (in principle) possible. There is no v selection rule
Vibrational structure
cm-1
18327.8
18405.4
18480.9
18555.6
18626.8
18706.3
18780.0
18846.6
18911.5
18973.9
19037.5
v’25
26
27
28
29
30
31
32
33
34
35
eee xG 22
12
1 vv)v( v”0
0
0
0
0
0
0
0
0
0
0
How would you solve this?(you have too much data!)
1. Take various combinations of v’ and solve for e and exe simultaneously. Average the answers.2. Fit the equation to your data (using XL or some other program).
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
v’25
26
27
28
29
30
31
32
33
34
35
eeeelecelectot xEGEE 22
12
1 vv)v(
v”0
0
0
0
0
0
0
0
0
0
0
Eelec = 15,667 cm-1
e = 129.30 cm-1
exe = 0.976 cm-1
Dissociation energy = 19950 cm-1
vibelectot EEE
Analyzing the spectrum…
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 3rd floor computer lab and with the
online simulations:http://assign3.chem.usyd.edu.au/spectroscopy/index.php
Practice Questions1. Which of the following molecular parameters are likely to change when a molecule is
electronically excited?(a) ωe (b) ωexe (c) μ (d) De (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) Which depicts a transition in a molecule that has a larger bond length in the excited state?(c) Which would show the largest intensity in the 0-0 transition?(d) Which represents molecules that can dissociate after electronic excitation?
(e) Which represents the states of a molecule for which the v”=0 → v’=3 transition is strongest?