Excited States and Photochemistry

Question 1

Describe what happens during an electronic transition. Why does this matter?

Answer 1

An electron moves from one MO to another. The electron density is redistributed and the bonding and structure change.

This means the chemistry occuring to the excited state is quite different to that occuring at the ground state.

Question 2

What are the equations relating frequency, wavenumber and wavelength?

Answer 2

E = hν = hc/λ = hcṽ

Question 3

How does conjugation of molecules affect the wavelength it absorbs light at?

Answer 3

It shifts the absorbance to a longer wavelength, this is because the HOMO and LUMO become closer together.

Question 4

What defines the total spin of the system, S, and the multiplicity of the system?

Answer 4

Spin is determined by the unpaired electrons in a system, however if 2 unpaired electrons are in opposite spin states they can will still produce a spin of 0. In a system with 2 unpaired electrons in the same spin state, S = 1.

The multiplicity = 2S + 1. This creates 2 main spin states, the singlet state, S and the triplet state, T.

Question 5

How are excitation states represented?

Answer 5

With state diagrams and labels.

Singlet = S, Triplet = T, Ground = 0, Excited = 1, 2, 3…

Each excited state has a S and T state where the T state is slightly lower energy (due to Hund’s rule, electrons in different orbitals spin allign to form the lowest energy state)

A state diagram is drawn with an energy y axis and labelled S₀, S₁, S₂, T₁, T₂… states

Question 6

How do atoms differ from molecules in terms of their electronic energy levels and UV-Vis spectra?

Answer 6

Atoms have no rotational or vibrational degrees of freedom so the energy levels are clearly defined and the UV-Vis spectra consists of sharp lines.

Molecules have rotational and vibrational degrees of freedom so there is many energy levels and the UV-Vis spectra has many lines and broad bands. The many energy levels and hence, possible excited states is due to the several MOs that transitions can occur via.

Question 7

What is the Born-Oppenheimer approximation? How do their energy magnitudes add up?

Answer 7

The total energy and wavefunction is the sum of the individual wavefunctions and energies of the electronic, vibrational and rotational properties which are independant of each other. The approximate wavenumbers of each property is electronic = 10⁴, vibrational = 10³ and rotational = 10¹.

Question 8

From which ground state does each transition occur over and what state does it transition to?

Answer 8

Electronic - the S₀ ground state to many possible S and T states but mostly S₁.

Vibrational - the most populated level is v’’ = 0 (v’’ represents the ground state, v’ represents the excited state) to several v’ levels. This expresses the vibrational fine structure.

Rotational - many ground state rotational levels, J’’ are occupied so many transition state rotational levels are occupied, J’. This can heavily congest the spectrum.

Question 9

When can rotational and vibrational fine structure be observed?

Answer 9

Rotational can be seen from gases, vibrational can be seen in gas, liquid and some solids. However for vibrational peaks may be broad depending on the solvent polarity.

Question 10

What determines the strength of the absorption coefficient for a transition?

Answer 10

The probability of the transition which is determined by the selection rules.

Question 11

What is the overall selection rule for an electronic transition? What is the equation for this?

Answer 11

There must be a displacement of charge, causing a transition dipole moment (TM).

TM = ∫ Ψ’* μ^ Ψ’’ dτ

Where Ψ’’ is the wavefunction of the inital state, Ψ’ is the wavefunction of the final state, μ^ (meant to be mu bar) is the dipole moment operator and ∫ dτ is the intergral over all space.

Question 12

How can the specific selection rules be determined from the overall selection rule?

TM = ∫ Ψ’* μ^ Ψ’’ dτ

Answer 12

The Born-Oppenheimer approximation is used to seperate the wavefunction into its electronic and vibrational wavefunctions (rotational not considered).

The electronic wavefunction can be split into the electron spin, S, and orbtial, φ, wavefunctions.

Noting that the dipole moment operator acts only on the orbital wavefunction:

TM = ∫ S’* S’’ dτ_S • ∫ φ’* μ^ φ’’ dτ_φ • ∫ Ψ_v’* Ψ_v’’ dτ_v

This gives the 3 spin selection rules, if any one is zero, the transition is disallowed by the selection rule.

Question 13

What is the spin selection rule and when is it non-zero?

Answer 13

Electrons must retain their spin (normalised) to be 1, if they change spin (orthogonal), they are 0. ΔS = 0

Therefore triplet excited states cannot be efficiently by absorption.

Question 14

What is the symmetry aspect of the orbital selection rule and when is it non-zero?

Answer 14

The orbital selection rule can be split into two parts, the symmetry and spatial parts.

Symmetry can be split into componants along the x, y and z axes. TM_orbital = TM_x + TM_y + TM_z. To be symmetry forbidden all three must be zero.

• Find the point group
• Deduce the symmetries of the inital and final states
• Deduce the symmetries of the x, y and z components (in the point group table)
• Deduce the three direct products: Γ_f x Γ_x/y/z x Γ_i
• If any direct product gives the totally symmetric representation (A₁ or A_g depending on the molecule) then the transition is non-zero.

Question 15

In working out the symmetry aspect of the orbtial selection rule, how do you work out the symmetries of the ground states and the excited states?

Answer 15

Orbitals with spin paired electrons (all ground states apart from paramagnetic species) are totally symmetric.

For the excited states, the cross product of the symmetries of the singly occupied orbitals determines the excited state symmetry.

They are then multiplied by the irreducible with the correct symmetry label (x, y or z).

This determines the direction that the molecule is polarised along and can be used in conjuction with the spin selection rule.

Question 16

How does the spacial aspect of the orbital selection rule determine the likelihood of transitions?

Answer 16

If the initial and final orbitals occupy a similar region of space, the overlap integral will be large and the transition is allowed. If they are different, the spacial overlap integral will be small and will be forbidden.

This determines the magnitude of the orbital integral.

Question 17

When is the vibrational overlap integral non-zero?

Answer 17

The shape of the vibrations are sine functions with the number of nodes in each is the vibrational level. The shape squared is the probabilities of the electrons at each point. Electrons transition to levels that have a good overlap with the ground state wavefunction, however there is also a change in equlibrium bond length in the excited molecule so this won’t be centered.

The electrons move faster than nuclei so after the transition the nuclei will move to favourable positions. Hence the electrons move vertically up.

Question 18

How do bond lengths change when electrons are excited to higher energy orbitals?

Answer 18

The bonding becomes weaker as the anti-bonding orbital is filled. Hence the bond will lengthan.

Question 19

How can the vibrational fine structure be used to determine the change in equlibrium bond length?

Answer 19

If the bond lengths are about the same, the best overlap will be with the v’ = 0 and the highest peak will be at the lowest energy.

If there is a larger change, the highest peak will be one of the inner ones.

Question 20

How is the probability of the transition represented? How is this related to the Beer-Lambert law?

Answer 20

By the oscillator strength, f. f ∝ TM² and ∫ ε_υ dυ which is related to experimental absorbance coefficients. Hence the selection rules relate to experimental ε values via f.

Question 21

How can the oscillator strength, f, be broken into components and what are the implications if the selection rules are broken?

Answer 21

The oscillator strength can be broken down into its spin and orbital components.

f ≈ f_spin • f_symmetry • f_spatial

If the transition is allowed by each of the rules, the value of f is 1. If disallowed they take a much smaller value to reflect the much lower chance of the transition occuring. f_spin = 10^-8 - 10^-5, f_symmetry = 10^-3 - 10^-1 and f_spatial = 10^-3 - 10^-1.

Hence a spin disallowed transition is much less likely to occur than one which is disallowed by the other rules.

Question 22

When and why do the selection rules “breakdown”?

Answer 22

In some cases the Born-Oppenheimer approximation is not suitable. Spin and orbital motions of electrons are not independant, this is know as spin-orbit coupling and its magnitude ∝ z⁴ where z is the atomic number. A spin-flip is strongly forbidden but in the presence of heavy atoms on the atom or in the solvent, (such as iodine) it is weakly allowed. All spin changing processes increase in rate.

Vibrational and electronic motions are not independant and have vibronic coupling. When asymmetric vibrations distort a molecule the symmetry is reduced which weakly allows some forbidden transitions.

Question 23

What are the two general and 5 specific photophysical processes of the decay of excited states? Give a brief description of each and its typical rate.

Answer 23

Relaxation can either be radiative - emission of light, or non-radiative.

Radiative processes:

• Fluorescence (f) - radiative transition between states of the same multiplicity, k_f = 10⁶ - 10⁹ s^-1.
• Phosphorescence (p) - radiative transition between states of different multiplicity (spin disallowed), k_p = 10^-3 - 10⁴ s^-1

Non-radiative processes:

• Vibrational relaxation (vr) - excess vibrational energy is given to surrounding molecules (solvent) by collisions, k_collision = 10¹³ s^-1
• Internal conversion (ic) - isoenergetic transition (ΔE = 0) between states of the same multiplicity typically from a low vibrational level in a more excited state to a high vibrational level of a less excited state,the rate is much faster between excited states; k_ic (excited⇒excited) = 10¹² - 10¹³ s^-1, k_ic (excited⇒ground) = 10⁵ - 10⁹ s^-1
• Intersystem crossing (isc) - isoenergetic transition between states of different multiplicity which may involve a change in MO (represented by isc’) (spin disallowed), this is also much faster between excited states for the same reasons as ic, k_isc (excited⇒excited) = 10⁶ - 10¹¹ s^-1, k‘_isc (excited⇒ground) = 10^-3 - 10³ s^-1

Question 24

What are the most common photophysical relaxation paths?

Answer 24

vr to the lowest energy state, then either f or ic, followed by vr back to the ground state.

Question 25

What can be interpreted from a vibrational fine structure emission and absorption spectrum?

Answer 25

The gaps between peaks are the difference in energy between the vibrational states, the absorption gaps are for the excited state and the emission gaps are for the ground state.

Question 26

How does the excited state ⇒ ground state transition occur at a longer wavelenth for the emission spectrum (where the middle peak of the structure doesn’t overlap)?

How does the solvent choice affect this?

What is the name of this effect?

Answer 26

The solvent rearranges around the molecule before the excitation and before the fluorescence. This means that after excitation and emission, the solvent then redistributes around the molecule which moves it to a lower energy. This means that the energy absorbed is greater than the actual energy difference and the energy emitted is less than the actual energy difference.

More polar solvents typically give larger shifts.

This is called the Stokes shift

Question 27

Why is internal conversion much faster between excited states than it is between excited and ground states?

Answer 27

For transitions between excited states, the energy gap is small and the overlap integral is large giving a very large rate (10¹² - 10¹³ s^-1). For a transtion to the ground state, the energy gap is large, hence the vibrational level that is transisitioned to is high meaning the overlap integral is small, giving a lower rate (10⁵ - 10⁹ s^-1).

Question 28

What is Kasha’s rule?

Answer 28

For excitation states larger that 1, ic and vr occur very quickly down to S₁.

Fluorescence and other processes exclusively occur from the S₁ state.

Question 29

What are the competing decay processes for the S₁ state?

Answer 29

Fluorescene and internal conversion to the S₀ state and intersystem crossing to the T₁ state.

Question 30

How would an overlay spectrum of a molecule absorbing and phosphorescing compare to one that fluoresces?

Answer 30

The gap between the peaks of absorbance and phosphorescence would be larger since the T₁ energy state is lower than the S₁ state.

Question 31

Draw a basic Jablonski diagram showing the 5 types of photophysical relaxation.

Answer 31

Question 32

What is chemiluminescence?

Answer 32

Where luminescence occurs from an excited state formed as the product of a chemical reaction. This requires a strongly exothermic reaction.

Question 33

Define the quantum yield, φ, for a process. How are photophysical quantum yields related? How can they be defined by rate constants?

Answer 33

The number of molecules undergoing the process/Number of photons absorbed

or Rate of process/Rate of absorption

The sum of the quantum yields for each process occurring will equal 1.

Quantum yield can be calculated from the rate of the process by k_process/k_total where k total is the sum of all the decay rates of a state.

Question 34

How are quantum yields calculated for T₁ decay processes?

Answer 34

The formation of the T₁ state as well as the decay competition by phosphorescence and isc’. The formation of the T₁ state is φ_T = φ_isc which is then multiplied by the quantum efficiency, θ, which is the rate of decay from the T₁ state.

Question 35

Define the rate law and lifetime for the decay of an excited state.

Answer 35

The rate law follows first order kinetics, [A*]_t = [A*]₀e^{-Σ<em>kt </em>}

Hence the lifetime, τ, is when t = 1/e

Question 36

Define the observed lifetime and the radiative lifetime. How do the lifetimes of S₁ and T₁ states compare?

Answer 36

The observed lifetime is 1/Σk = 1/k_obs

The radiative lifetime, τ⁰, is the decay by emission only and can be calculated for fluorescence and phosphorescence.

τ_f⁰ = 1/k_f , τ_p⁰ = 1/k_p

T₁ states generally are longer lived than S₁ states as their decay processes are spin disallowed. T₁ lifetime = 10^-6 - 10⁰ s, S₁ lifetime = 10^-12 - 10^-9.

Question 37

How does temperature affect the rate of decay by photophysical processes?

Answer 37

It doesn’t affect the rate of the photophysical processes much but has a large effect on other decay processes such as physical quenching, therefore changing the quantum yields of photophysical rates.

Question 38

Why do T₁ states decay faster than predicted by the rates of phosphorescence and isc’ I’m fluid solution? How can this be accounted for?

Answer 38

Non-reactive decay can be caused by quenchers that diffuse and collide with the excited states. Quenchers can outcompete typical T₁ state decay rates, as a result the rate constant for quenching, k_q[Q], is often included in observed lifetimes of T₁ states.

Question 39

How is phosphorescence studied? Why is this required for phosphorescence but not for fluorescence?

Answer 39

Studied in a rigid matrix, where quenchers cannot diffuse, at 77 K using a solvent that becomes a rigid glass (EtOH). In fluid solution, fluorescence is fast enough that the quenching is not significant whereas phosphorescence disappears almost completely when quenchers are present.

Question 40

Briefly describe the physical methods used to study rates. What can they measure?

Answer 40

Steady state techniques: UV-Vis absorption gives absorption coefficients and emission gives quantum yields and both give spectra.

Time-resolved techniques: Excite a sample with a light pulse and measure either the emission or absorbance intensity

Question 41

Define photochemistry. How is the reactivity of an excited state different to the ground state?

Answer 41

The reactive decay of excited states which compete with the decay via photophysical states.

The excited state is different by its electronic configuration (bonding, structure, redox ability), excess energy (increased ability to overcome activation barriers) and the lifetime of the reacting molecules is much shorter.

Question 42

List the 8 catagories of ways excited electronic states can decay.

Answer 42

Question 43

Describe how the different types of dissociative unimolecular photochemistry take place. How can they be distinguished? What can occur that will reduce the yield of these products?

Answer 43

Photodissociation: the energy absorbed is so large that the vibration is long enough to overcome the bonding energy holding the molecule together. The emitted fragments have a continuous range of kinetic energies, the absorbance spectrum will have vibrational fine structure up to the point where it becomes smooth at the dissociative point.

Photopredissociation: an excited state is initially created but then undergoes ic to a dissociative state. Part of the vibrational fine structure is lost at the vibrational level where ic occurs.

Question 44

Asides from dissociation, describe 2 other unimolecular processes that may occur.

Answer 44

Photoisomerisation: the high barrier to isomerisation may be overcome by the increased energy, or the change in bonding upon ionisation may favour a change in structure (such as in alkenes, the weakening of the double bond can favour a rotation). A pure sample of a single isomer can become racemic, also known as a photostationary state where the composition depends on the relative absorbance of the isomers (can be 100% one isomer if only one absorbs light)

Photochromism: a light induced, reversible change of colour which is often associated with a change in isomer or a ring opening/closing.

Question 45

What is an excimer? How can you tell they’re being formed?

Answer 45

A system that forms dimers but mostly in the excited state - an excited state dimer. Typically these are conjugated rings giving π stacked structures.

For self dimerising systems, at low concentrations of the molecule, the monomer fluorescence dominates, at high concentrations, the excimer fluorescence dominates at a shorter wavelength. They are also time dependant as relaxation occurs.

Question 46

What are exciplexes? How do you know when they’re formed?

Answer 46

Systems that form complexes in the excited state that don’t form in the ground state.

At low concentrations of the ligand, the monomer fluorescence with vibrational structure is seen. At high concentrations, exciplex fluorescence dominates at a longer wavelength with no vibrational structure.

Question 47

What are the similarities and differences of excimers and exciplexes?

Answer 47

Similarities:

Fluorescence and vibrational characteristics when changing concentrations

Potential energy curve (favouring dimerisation in the excited state)

Concentration and time dependence of fluorescence.

Differences:

Exciplexes are polar, charge-transfer complexes

There is significant electron transfer between exciplexes

Question 48

Describe the two ways bimolecular quenching can occur.

Answer 48

Reactively, where the excited state and quencher interact and change, or non-reactively where only energy is transferred.

Question 49

Give the Stern-Volmer equation in terms of rates, lifetimes and quantum yields, and describe its derivation.

How is the equation used?

Answer 49

Φ_f⁰/Φ_f^q = 1 + k_q[Q]/(k_f + k_nr) = 1 + τ_obs⁰k_q[Q] = τ_obs⁰/τ_obs^q

The derivation is from the quatum yield of fluorescence when no quenching is present divided by when the quencher is present.

The rate of quenching, k_q, is found by plotting [Q] against the ratio of quantum yields.

Question 50

How can the rate of quenching be estimated for non-viscous solvents?

Answer 50

The rate of quenching approaches the diffusion controlled limit since this is the rate determining factor. This can be estimated from the Debye equation:

k_diff = 8RT/3η

Where η is the solvent viscosity.

Question 51

Describe the process of electron transfer by quenching. How can you estimate the thermodynamics of the process? How fast is the reaction and when is it favoured?

Answer 51

A quencher exchanges electrons with the excited state either oxidatively or reductively with respect to the excited state. When Q is an electron acceptor, the excited electron transfers from the excited state into the quenchers LUMO. When Q is an electron donor, an electron from its HOMO moves into the half empty HOMO of the excited atom.

By taking the electronic excitation energy from the free energy change of ground state electron transfer. Therefore the reaction is more favourable in the excited state.

The rate approaches the diffusion rate and the ion products (rather than the exciplex) is favoured in more polar solvents.

Question 52

Briefly describe the three mechanisms of energy transfer. What are the requirements for efficient transfer and general characteristics of the mechanism?

Answer 52

Radiative: Donor emits a photon, acceptor absorbs it. The emission of D* must overlap with the absorption of A. It is the dominant mechanism at large separation of unlimited distance.

Long-range Coulombic: Energy transfer by a dipole-dipole interaction without photon emission or contact. The energy gaps between D* to D and A to A* must be well matched, and occurs over a distance of 2-10 nm. The rate is larger than the diffusion controlled limit as collision is not required and it is genially observed between singlet states.

Short-range electron exchange: An electron from each species is exchanged without photon emission at <2 nm. The energy gaps must match and the rate approaches the diffusion controlled limit.

Question 53

How can T₁ states be created when the quantum yield of isc is low?

Answer 53

By photosensitisation. A triplet state on a donor undergoes energy transfer with an acceptor. The triplet electron swaps for a ground state electron on the acceptor molecule. The donor must have a high quantum yield of isc and a T₁ state higher in energy than the acceptor T₁ state.