PESMO

Question 1

Describe the fundamental principles behind MO theory.

Answer 1

Each electron is describe by a one-electron spatial wavefunction, φ_i, know as the MO. The electron itself is indistingushable and only represents a share of the electron density.

The MOs are polycentric and stretch across many atoms. Each MO is defined by a set of quantum numbers which also define their shape and energy, ε_i.

Each electron has a spin of m_s = ±1.

Question 2

Describe how the spacial wavefunction is affected by the individual atoms with an equation and what are the requirements for the atomic orbitals to combine as such?

Answer 2

φ = c₁χ_A + c₂χ_B where χ are the atomic one-electron orbitals.

They may combine effectively if they have comparable energies, they have resonable overlap and they have the same symmetry.

Question 3

Define a basis set and describe it for a 2nd row diatomic.

Answer 3

The basis set is a set of equations which describe the bonding orbitals of the electrons in a molecule. They generally only describe the valence orbitals that can engage in bonding so this is typically the 2p and 2s orbitals.

Question 4

Describe the effects and draw diagrams showing the change in MO shape from weak to strong sp mixing of a 2nd row diatomic.

Answer 4

This is summerised by combinding each of the wavefunctions,

φ_σ = c₁2s_A + c₂2s_B + c₃2p_{z A} + c₄2p_{z B}

Question 5

Outline, for the 2nd row diatomics, how sp mixing changes.

Answer 5

Starting from Li₂ the orbitals are close in energy so there is very strong sp mixing. This decreases down to N₂ which is still mixed but only closely. The next diatomic, O₂, is weakly mixed so the MO energies change but the 1π is above the 3σ.

Question 6

Briefly describe the photoelectric effect and the associated graph and equation.

Answer 6

Electrons are emitted from a metal when light is shone onto it. Photolelectrons are only released after a threshold frequency is reached and the kinetic energy of the photoelectrons increases linearly with photon frequency. This is described by hv = ½m_ev² + Φ.

Question 7

Describe the way that photoelectron spectroscopy was developed from the observation of the photoelectron effect.

Answer 7

Much higher energy light sources were developed using x-rays to ionise core electrons and deep UV rays to ionise valence electrons. The UV source was a He(I) lamp with a very specific source wavelength of 21.22 eV.

Question 8

Describe generally how PES occurs and Koopmans theorem for how it can be analysed.

Answer 8

The electrons are ionised from the valence bands and the kinetic energies are measured. The rest of the photons energy is assumed to have been used in the ionisation of the electron from its band.

Koopmans theorem states that ‘For a closed shell molecule, the ionisation energy for an electron in a particular orbital is approximately equal to the negative of the orbtial energy.’ However this relies on the theory of one electron orbitals and neglets eletron reorganisation such as the repulsion been electrons. It also cannot be used for open shell molecules.

Question 9

Why must the monochromatic light source be carefully selected.

Answer 9

It must have an appropriate amount of energy and have a wavelength that will propagate in air. It must also only produce light from one transition.

Question 10

How is the kinetic energy of the emitted photonelectron analysed and how can it be adapted?

Answer 10

Using a magnetic or electric field, the photoelectron can be curved around a path between hemispherical plates where the number of photoelectrons that have the correct kinetic energy can be detected. A small range of energies being detected at once is desireable.

The resolution, the range of energies, can be determined by ES/2R where E is the kinetic energy, R is the radius of the path and S is the slit width at the start and end of the path.

Question 11

Describe the origin and meaning of the vibrational structure produced when an ionisation takes place.

Answer 11

Assuming the Born-Oppenheimer approximation, the electron movement will be much faster than nuclei movement. Hence the molecular structure will electronically and structurally reorganise after the transition.

If the electron is removed from a non-bonding orbital, the bond length will be the same in both species and a sharp peak will be seen for the transition to the v = 0 level in the product as this will have the greatest overlap with the ground species.

If the electron is removed from a bonding orbital, the bond is weakened, and will vibrate outwards since the best overlap is with upper vibration states of the cation produced. Hence a vibrational progression is seen in the spectra.

If the electron is removed from an anti-bonding orbital, the bond is strengthened and the progession will be shorter as the potential energy curve for the cation will be narrower.

Question 12

For F₂, weak sp mixing, describe the change in bond length and the progressions for each of the MO ionisations.

Answer 12

• π* orbital - bond is shortened as the bonding energy increases. Short progression with middle peak highest as PE curve narrows.
• π orbital - bond is lengthened as the bonding energy decreases. Long progression.
• σ orbital - bond is even more lengthened as the bonding energy is more significantly reduced. Long progression.

The intensity of the bands decreases the deeper they are.

Question 13

What does the Λ symbol represent and how can it cause complications to the spectra?

Answer 13

Λ represents the quantum number l (angular momentum) associated with the symmetry of the species produced. Π labelled species have a value of 1 and Σ labelled species have a value of 0.

This can the couple to the electron spin, called spin orbit coupling and they add together to give the total angular momentum of the species. Ω = |Λ| + M_s. For Π this gives states of 3/2 and 1/2 where the higher value is lower energy.

This causes 2 peaks to form at each vibrational level and the effect increases with z⁴. For I₂, there is 2 distinct, seperate bands.

Question 14

Describe the PES spectrum and the change in bond length associated with N₂.

Answer 14

The orbital ordering of N₂ is sp mixed and the occupancy is up to 3σ.

• 3σ (sp mixed to weakly bonding) - small increase in bond length. Short progression.
• 1π bonding orbital - strongly bonding so large increase in bond length with long progression.
• 2σ* slightly anti-bonding - small decrease in bond length, short vibrational structure.

Question 15

Describe the complexities in the PES spectrum of O₂.

Answer 15

The state where both unpaired electrons are spin up is favoured. For the first band, one is ionised, one band is formed.

For the other bands, where an additional unpaired electron is produced, two resultant states can form, where the unpaired electron can either go with or against the spin of the other unpaired electrons. The state where the unpaired electrons are the same spin is lower in energy and has a multiplicity of 4. The states are a multiplicity of either doublet or quartet, hence a band is formed for each.

The bond lengths will change but the amount that it does is quite complex.

Question 16

How can sp mixing be determined from a PES spectrum?

Answer 16

The bonding character of the orbitals can be determined by the interatomic bond wavenumber (measured from the gap between peaks). This can be determined to be strongly bonding for weak sp mixing or weakly/non-bonding for strong sp mixing.

Question 17

Describe how you would construct a MO diagram for an AH₂ molecule, including which MOs are formed and how the relative energies of each are determined.

Answer 17

1. Define the basis set of orbitals that are going to overlap, typically 2s, 2p of A and 1s of H.
2. Find the point group given the shape.
3. Classify each of the orbitals on the centre atom with a symmetry, s will be totally symmetric, p will be determined from the z, x and y column.
4. With the ligand atoms, find the symmetries of the SALCs - the difference combinations of phase.
5. Coordinate the symmetries of the central atom with the ones formed by the SALCs, this determines which orbtials will form MOs
6. Draw the MOs and order them based on: Fewer nodes = Lower energy, More s character = Lower energy

Question 18

Describe the role and use of a Walsh diagram. What are the most important changes for AH₂?

Answer 18

The Walsh diagram shows the different possible orbital structures for two different shapes and how the MO energies will change. They are used to predict the shape that will be taken by a molecule, given the number of valence electrons and can predict a shape change on ionisation.

1. 1π_u - 3a₁: the linear form has the non-bonding orbital but as the shape changes a bonding orbital forms
2. 1σ_u - 1b₂: when the out of phase 1s orbitals are brought together they are destabilised
3. 2σ_g - 2a₁: in phase 1s orbitals are brought together hence more stabilisation

Question 19

How are walsh diagrams dependant on the central atom?

Answer 19

The sp mixing influences the MO energies. Additionally the difference in energy between the states changes so overlap with the H 1s orbitals will change as well. BeH₂ has much more H 1s and 2p_z character in 1σ orbital than H₂O whose 2s orbitals are more seperated from the other orbitals.

Question 20

For H₂O, what can be described for each of the bands on the PES spectrum?

Answer 20

The orbital character of the O-H and H-H can be described for the molecule.

The vibrational frequency as a stretch and a bend can be described, hence the bond length and angle can be compared to the original species.

A reduction in the wavenumber of the stretch vibration is associated with a longer bond. When the bend wavenumber is decreased, the stability has decreased which may either be in-phase 1s orbitals moving away from each other or out-of-phase 1s orbitals moving closer together.

Question 21

Describe how the SALCs of a planar AH₃ system is determined.

Answer 21

Find the reducible and reduce by inspection to A₁’ and E’. Find 2 of the SALCs by applying the A₁’ point group and E’ point group and finding the resulting combinations.

Now rotate the E’ combination to find the varients, and take one from the other. This forms the 3rd SALC also with E’ symmetry.

Question 22

How does the MO change when bringing together two bent AH₂ molecules to form H₂AAH₂? How does this apply to ethene?

Answer 22

Each orbital will pair up with is equivalent symmetry orbital. The non-bonding p_x orbital can form a π bond with the equivalent and as it it singly occupied for carbon, the H₂AAH₂ form is more stable than the indivdual states, the ethene structure is favourable.

Question 23

Describe the basic process of finding the basic MO.

Answer 23

Question 24

Give the Huckel approximations to solve a secular determinant.

Answer 24

Question 25

Describe how to find the coefficients c₁ and c₂ from the calculated energy levels and the secular equations.

Answer 25

Sub in the epsilon values into the secular equations. The relation between the orbitals can then be found. Normalisation is required to find the exact value.

Question 26

How do you work out the delocalisation energy of a molecule?

Answer 26

Multiply the energy of a state by its occupancy, do this for all involved species. For butadiene to dissociate to two ethenes, take the π energy of the two ethenes from the butadiene.

Question 27

Give the shortcut method to finding the substituted secular determinant.

Answer 27

Draw the molecule and label the unsaturated atoms with numbers. Fill a NxN matrix with the N as the number of unsaturated atoms with X on the leading diagonal and 1s where there is a bond between the atoms (row 1, column 2). Place a zero everywhere else.

Question 28

Give Frosts rule of aromaticity.

Answer 28

For a cyclic conjugated system of N atoms, with the shape on a point, within a circle (radius 2) with the distance from a midpoint marked. The point should be the radius of distance 2 away from the midpoint.

The distances are the roots of the equation.