Photochemistry

Question 1

Describe light as a reagent.

Answer 1

Absorbing a photon excites a molecule from the electronic ground state to the first electronically excited state.

Absorbing a green photon could provide the energy required for a molecule to react. The concentration of molecules with > E_a depends on light intensity and the rate of loss of the excited species.

Question 2

Describe:

Electronic excitation

Stimulated absorption

Stimulated emission

Spontaneous emission

Answer 2

_Electronic excitatio_n - visible light absorption changes the population of electronic states of molecules.

Stimulated absorption - photon excites a molecule from a low-energy to a high-energy state

For a two-state system:

Stimulated emission - photon induces a molecule to transfer from high-energy to low-energy state

Spontaneous emission - transition from high-energy to low-energy state, independent of radiation

Question 3

Describe electronic excitation in a simple system.

Answer 3

• each electronic state has a potential well and an associated set of vibrational levels
• when we excite an electronic transition, we move the molecule from the well in ground state to well in first excited state
• Born-Oppenheimer approximation states that electrons move infinitely fast compared to nuclei, so the nucleus are assumed static upon transition = ‘vertical’ transition
• Frank-Condon principle describes transition intensities and most likely transitions

Question 4

What can happen once we’ve created an electronically excited state?

Answer 4

• radiative decay - photon emitted as molecule relaxes in a single step
• non-radiativ e decay - normally multiple step process involving many energy levels but no photons are emitted

Question 5

Describe quantum yield.

Answer 5

The number of reactant molecules consumed for each photon of light absorbed.

• measure of ‘efficiency’ of photochemical reactions

Question 6

Distinguish between primary and secondary photochemistry via quantum yields.

Answer 6

The Stark Einstein Law states that only one molecule can be decomposed in the primary step, so:

φ > 1 suggests secondary reaction

φ > 2 suggests chain reaction

Question 7

Define primary (φ) and overall (Φ) quantum yields.

Answer 7

A primary QY is stated for a specific primary process.

Unless otherwise defined, the overall QY relates to the removal of reactant.

Question 8

Describe the solvent cage.

Answer 8

Solvent effects are often the cause of very low QY values or high ‘geminate’ recombination, though high radiative or non-radiative relaxation of the initial excited state may also cause this.

The sum of QYs for all primary processes must add up to one.

Question 9

Describe crystal field theory.

Answer 9

Treats ligands as point charges, leading to an inability to predict the spectrochemical series accurately. A basic description of what happens to the d-orbitals when in a complex.

Question 10

Describe ligand field theory.

Answer 10

Applies a MO approach to the structure/bonding of complexes. It allows much better comparison with spectroscopic data.

1. Begin with TM d-orbitals
2. Form symmetry adapted linear combinations of d-orbitals with ligand orbitals

Question 11

Describe the MO description of bonding in an octahedral complex given by ligand field theory for sigma orbital overlap.

Answer 11

• some orbitals are of mainly ‘ligand’ character, e.g. the bonding orbitals at the bottom of the diagram
• some orbitals are of mainly ‘metal’ character, such as e_g and t_2g orbitals
• retains the ligand field splitting parameter, Δ_o

Question 12

What happens to ligand field theory MO description of an octahedral complex when π-bonding from the ligands is introduced?

Answer 12

• the π-orbital combinations have t_2g symmetry so they overlap with the previously non-bonding metal t_2g, changing Δ_o
• π-donor ligands decrease Δ_o
• π-acceptor ligands increase Δ_o
• therefore, π-bonding has an impact on the spectrochemical series

Question 13

Describe prompt photochemical reactions.

Answer 13

Very short amount of time between absorption and the following photochemistry.

E.g. dissociation of CO ligand from group VI hexacarbonyls, where the ligand field transition weakens both pi- and sigma-bonding which encourages dissociation.

Question 15

Describe d-d reactions.

Answer 15

Excitation of a d-d transition leads to redistribution of electron density within a d-shell and often to increased occupation of anti-bonding e_g orbitals.

Question 16

Describe delayed reactions.

Answer 16

If an excited state has a long lifetime (time taken to drop from excited state to ground state), then it can be considered to ‘equilibrate’ (relaxes to vibrational ground state in electronic excited state) and remains long enough to undergo chemistry.

e.g. intersystem crossing from ¹MLCT to ³MLCT leads to a long lifetime state.

Question 17

Describe charge transfer excitation.

Answer 17

In general, charge transfer excitation (MLCT or LMCT) causes a radial re-distribution of electrons between the metal and ligands.

MLCT excitation corresponds to metal oxidation.

LMCT excitation corresponds to metal reduction.

This can lead to photoredox reactions.

Question 18

Compare d-d transitions and charge transfer transitions.

Answer 18

Generally, d-d transition lead to photosubstitution and CT transitions lead to photoredox behaviour, but not always.

Question 19

Compare a molecule in the ground state vs. in an excited electronic state.

Answer 19

Ground state:

IE is high, Electron affinity (EA) is low.

Excited electronic state:

IE is reduced, EA is increased.

This means that excited state molecules are both better electron acceptors and better electron donors. This means that excitation with light creates molecules that are capable of acting as both reductant and oxidant.

Question 20

Give the equation describing the oxidation of molecule A and describe it.

Answer 20

E⁰ (A⁺/*A) = E⁰ (A⁺/A) - E₀₀ (*A/A)

Says that the energy to oxidise *A is lower than the energy needed to oxidise A by the energy needed to excite A to A*.

Question 21

Why is it sometimes preferable to undergo reactions using excited state molecules rather than ground state?

Answer 21

Reactions with ground state molecules can sometimes be non-spontaneous (ΔG > 0). However, by exciting the molecule into an electronically excited state, the reaction may then become spontaneous (ΔG < 0).

e.g. disproportionation of [Ru(bpy)₃]²⁺ in the ground state is non-spontaneous, but it’s spontaneous for *[Ru(bpy)₃]²⁺

Question 22

What reactions can the *[Ru(bpy)₃]²⁺ excited state undergo with other species?

Answer 22

Oxidative quenching:

*[Ru(bpy)₃]²⁺ + Q –> [Ru(bpy)₃]³⁺ + Q^-

Reductive quenching:

*[Ru(bpy)₃]²⁺ + Q –> [Ru(bpy)₃]⁺ + Q⁺

In principle these two processes are in competition with one another, but usually only one is thermodynamically allowed. If both are thermodynamically possible, then kinetic factors must be considered.

Question 23

Describe photosensitisation.

Answer 23

The use of molecules which absorb light to promote light-driven reactions of molecules which don’t absorb light. The photosensitiser (PS) directs light energy to where it is needed in the chemical process.

Two types of photosensitisation exist:

LAS - light absorption sensitisers - act to absorb light and direct its use in chemistry.

LES - light emission sensitisers - act to emit light, exploiting chemical energy produced in a reaction by molecules which can’t emit light.

Question 24

How can photosensitisation help with harvesting solar energy?

Answer 24

Solar energy can be harvested by semiconductors such as TiO₂. However, TiO₂ has a large bandgap (about 450 nm), which limits the amount of the solar spectrum that can be absorbed as this is the point where solar radiation drops off.

If a LAS could be used to harvest more energy, then the process could be improved. An Ru LAS is able to absorb at around 500 nm, which is close to the peak of solar radiation. This maximises the amount of solar energy that we can use.

Question 25

Describe the process in a Gratzel cell.

Answer 25

1. Light absorbed by LAS (PS)
2. Electron transferred from *PS state to TiO₂ conduction band, then is injected into the anode (excitation to reactive state then oxidative quenching) (ps)
3. PS⁺ is reduced by a redox mediator (R⁺/R) in solution (ns)
4. R is reduced at the cathode

Question 26

Describe dye sensitised solar cells.

Answer 26

DSSC is a more general type for the Graztel type cell.

Advantages:

• Absorption spectrum is separated from the bandgap of the semiconductor.
• Cell sensitivity can be tuned to match the solar spectrum using molecular manipultion of the LAS (PS)
• Low cost construction

Question 27

Describe tandem DSSCs.

Answer 27

Similar to normal DSSCs, but uses a tandem arrangement to exploit light to assist at both electrodes of the cell (not just the anode).

Question 28

Describe the light harvesting complex (LHC) in photosynthetic systems.

Answer 28

• combination of proteins and chromophores
• antenna pigments transfer energy to one Chl-a molecule, called the reactionc entre
• energy transfer occurs via resonance energy transfer
• in PSII, chl-a molecules absorb, excite an electron and transfer it to pheophytin
• chl-a then replaces the electrons via oxidation of water

Question 29

Describe the Z-scheme.

Answer 29

• Involves electron transfer to form NADPH using PSII and PSI.
• One electron can be transferred by the action of one photon
• the full reaction requires a minimum of 4 e^- i.e. needs 4 individual photons
• QY measurements suggests that it’s nearer to 8 photons, suggesting the presence of intermediates/multistep processes

Question 30

Describe water splitting in the Z-scheme.

Answer 30

Involves the excitation of PSII:

*PSII + e^- –> PSII^-

So *PSII is a strong reducing agent. The overall effect leaves PSII as electron deficient PSII⁺, leading to water oxidation to regain the electrons:

2H₂O –> 4H⁺ + 4e^- + O₂

Question 31

Describe producing NADPH in the Z-scheme.

Answer 31

Involves the excitation of PSI:

*PSI + e^- –> PSI^-

So *PSI is a strong reducing agent. Electrons are transferred onwards to NADP⁺ via ferredoxin:

2FD_red + H⁺ + NADP⁺ –> 2FD_ox + NADPH

So electron obtained from PSII are used to create NADPH via another photon.

Question 32

Describe CO₂ reduction for making fuel.

Answer 32

e.g. via light absorption in semi conductors.

• Photon leads to excitation from conduction band to valence band
• Relative sizes of band gap (must lie in the visible) and energies of HOMO/LUMO dictate whether electron or hole transfer will occur
• Rates must compete with electron-hole combination
• CO₂ reduction is a multistep process involving up to 8 electrons to cleave CO bonds and create CH bonds - possible for multiple products

Question 33

What problem arises using CO₂ reduction for making fuel and how can this be overcome?

Answer 33

• CO₂ is stable and inert, therefore it has a very low electron affinity
• addition of an electron leads to a bent configuration and lone pair repulsion (O), so the LUMO of CO₂ is high in energy - very few semiconductors have the band gap required to achieve this
• surfaces offer a way to overcome the obstacle of attaching the first electron by forming a CO₂^δ- species which is bent, so the LUMO energy is lowered encouraging first electron attachment

Question 34

Describe water splitting for making fuel.

Answer 34

CO₂ reduction can be achieved via reaction with molecular H₂, leading to focus on the process of water splitting which requires 4 photons.

Water splitting is described by two processes, both of which are multi-electron steps involving more than one water molecule. A single photocatalyst must have the ability to overcome the energy gap of 1.23 V, e.g. TiO₂.

Question 35

How has the method for artificial water splitting improved?

Answer 35

Biological water splitting uses two catalysts. Alternative artificial strategies now use one catalyst each for H₂ and O₂ generation.

This means that narrower band gaps can be used, moving absorption into the visible. Spatial separation of the two processes also avoids back reactions. This method, however, requires very precise arrangements of components and consideration of relative rates.