LCAO, MO diagrams, and BDE

Question 1

What is a wave function? How can it be used?

Answer 1

An equation, called a molecular orbital, which describes an electron

The square of the wave function gives the probability distribution of an electron at a point in space

Question 2

What does it mean to be polycentric? How is energy associated to an orbital?

Answer 2

Polycentric: electron density is centred around more than 1 nucleus

Each orbital has a specific energy, which is defined and quantised

Question 3

What are the two main approximations used in the linear combination of atomic orbitals?

Answer 3

Born-Oppenheimer approximation: the nuclear and electron positions can be treated independently
Nuclei can be assumed to be fixed but electrons can move (lighter)

Orbital Approximation: Wavefunction of N electrons can be written as the product of N one electron wavefunctions

i.e ψ (1,2,3,4,…)=ψ(1) x ψ (2) x ψ (3) …

Question 4

What is the equation for calculating LCAO?

Answer 4

ψ = ∑ᵢ cᵢϕᵢ

where c is a weighting factor of the contribution of the wavefunction for the bonding/antibonding orbital

Question 5

How can the equation for LCAO be applied to H₂ + ?

Answer 5

Let one of the orbitals be 1a and the other 1b
As below, weighting factor x wave function, and squared for born interpretation
(Xa1a + Xb1b)²

= (Xa1a)² + (Xb1b)² + 2XaXb1a1b

Close to A, resemble the first term, close to B the second, and the internuclear region is the final term

Question 6

How does LCAO give rise to bonding and antibonding MOs, using H₂+ as an example?

Answer 6

The contribution to the wavefunction must be equal as we are using two H wave functions
Using the born interpretation:

(Ca)² = (Cb)²
Ca= Cb or Ca=-Cb

Same sign=bonding, Opposite, out of phase= anti-bonding

Question 7

What does the electron density along the bonding H₂+ axis look like relative to H alone?

Answer 7

H alone: high electron density at the nucleus alone
H₂+: Electron density at the nuclei reduced but increased in the internuclear region, stabilising

Question 8

What does the electron density along the antibonding H₂+ axis look like relative to H alone?

Answer 8

H alone: high electron density at the nucleus alone
H₂+: Electron density at the nuclei increased and steep decline in internuclear region, falling to 0 as there exists a nodal plane

Question 9

What are the balance of forces which determine the bond length in a molecule?

Answer 9

Coulombic nuclear repulsion decreasing with increasing radius
Nuclear-electronic attraction decreasing with increasing radius

There is a minimum of the sum of these forces at the bond length, where the energy is minimised

There exists no minimum for the antibonding orbitals as increasing radius will stabilise both factors

where B= repulsion A= attraction C= sum

Question 10

What is the variation principle? How can it be applied to H₂+ ?

Answer 10

The variation principle states that an estimated wavefunction will always be at the same or higher energy than the true wavefunction
This means lower energy estimates are more reflective of the actual wavefunction

Contracted orbitals give lower energies so a closer representation of the wavefunction

Question 11

How do you name the MOs based on the orbitals which form?

Answer 11

S orbitals will always form σ bonds
P orbitals can form either σ bonds or π bonds
D orbitals can form σ, π or δ bonds

If a molecule consists of the same species i.e O2, then the MOs can be classified with u or g to distinguish bonding / antibonding

σ g= bonding σ u = antibonding
π u = bonding π g= antibonding

From g meaning the same on inversion

Question 12

How does the electron density change from the H atoms alone to the bonding MO of H₂ ?

Answer 12

Decreased electron density at the nuclei and increased in the internuclear region

Question 13

Why is the bond strength of H₂ not twice that of H₂ + ?

Answer 13

Whilst there is twice the amount of stabilising binding energy with the additional electron - nuclear attraction, there is also additional inter electron repulsion weakening the bond
The net effect is strengthening of the bond compared to with 1 electron, but not twice as strong

Question 14

What does the electron density along the nuclear axis of an antibonding molecule look like for H₂ ?

Answer 14

Higher electron density at the nucleus
Falls to 0 at the centre of the axis, and changes sign
Opposite phase peak at the other nucleus

Question 15

How do you calculate bond order?

Answer 15

Bond Order= 1/2 ( n of bonding electrons - n of antibonding electrons)

Question 16

Compare the bond strength for H₂ and He₂ and ions? Why is this the case?

Answer 16

H₂ > H₂ + > He₂ + > He₂

H₂ : 1σg 2 1
H₂ + : 1σg 1 1/2
He₂ +: 1σg 2 1σu 1 1/2
He ₂ : 1σg 2 1σu 2 0

Bond strength lies with the increasing bond order, as there is greater nuclear-electron attraction, stabilising the bond and increasing strength

When comparing the ions, they have the same bond order, but there is an electron in the antibonding for He but not in H
As antibonding orbitals are more destabilising than bonding orbitals are stabilising, He would be at a higher energy, and so weaker bond

Question 17

What does the MO diagram for H₂ look like?

Answer 17

Larger gap between atoms and antibonding than atoms and bonding

Question 18

What does the MO diagrams for Li₂ look like? How does the bond strength of Li₂ compare to H₂ ?

Answer 18

Still the same diagram with1σg 2 configuration
This is because the 1s orbitals are too contracted to be involved with bonding to a large extent, as there is a greater nuclear charge

However, Li₂ is a weaker bond, lower bond dissociation energy, and so the distance from bonding/antibonding is smaller
This is because electrons are removed from a larger atom, possessing larger bond lengths
There is a reduced nuclear attraction between the bonding electrons and nuclei, so a weaker bond

Question 18

How can the p orbitals overlap in bonding? How do the energies of these compare?

Answer 18

The Pz orbitals can overlap head on, forming a σ bond
The Px and Py can only overlap sideways, forming π bonds
These form bonding and antibonding MOs

σg< πu < πg < σu

Question 19

What characteristics favours MO formation?

Answer 19

Energies of the atomic orbitals are similar
Orbitals should overlap well
Atomic orbitals should have the same same symmetry relative to the molecular axis

Question 20

What does the MO diagram for O₂ look like?

Answer 20

Question 21

How do the bond disassociation energy for O₂ and its first ions compare? Why?

Answer 21

O₂ : 2σg2 2πu4 2πg2 2
O₂ + : 2σg2 2πu4 2πg1 5/2
O₂ 2+ : 2σg2 2πu4 3
O₂ - : 2σg2 2πu4 2πg3 3/2

As before, increasing bond order reflects a greater nuclear-electron attraction and so stronger bond, larger bond dissociation energy
This arises from removing antibonding electrons which would otherwise destabilise the bond
Adding electrons increases the number of antibonding electrons, destabilising the bond
If further electrons were to be removed, they would be bonding in nature, thus reducing the nuclear-electron attraction, shown in the decrease in bond order, so weaker bond

Question 22

How do bond dissociation enthalpies change across a period for the same bond order?

Answer 22

Same bond order, then the BDE increases
Increased Zeff and so electronegativity across the period, resulting in greater electron-nuclear attraction, and smaller bond lengths, so stronger bonds

Question 23

What is sp mixing and its implications?

Answer 23

For G3-G5, the s and the p orbitals are close enough in energy that they mix, effectively forming sp hybrid orbitals
This occurs with the p z orbital only as they have the same molecular axis symmetry
This results in the s σ bonding and antibonding decreasing in energy, whilst the p σ increasing in energy
As a result, the p σ (2σg) is higher in energy than the p π (1πg)
So π bonds filled before σ

Question 24

Why does sp mixing only occur before g6?

Answer 24

The energy gap between the 2s and 2p orbital increases across period This results in the difference in energy being too large to enable effective orbital mixing, preventing sp mixing Across the period, the Zeff increases, resulting in greater nuclear attraction experienced by the electrons As the 2s is more effective at penetrating the core region than 2p, it will experience a greater more stabilisation, and thus the energy difference between the two increases

Question 25

What does the MO diagram for N2 look like? How do the diagrams and BDE differ down G5 and why?

Answer 25

When drawing the MO, for both the sp mixed bonding and antibonding, show sp hybrids, but opposite sides overlapping Same MO style down G5, but relative, energies would be higher The size of the atoms increases, meaning the bonding electrons are further away from the nuclei This reduces the electrostatic nuclear-electron attraction, decreasing bond strength / BDE and increased bond length

Question 26

What happens to BDE down the halogens and why?

Answer 26

FluorineBromine Cl>Br explained as before, larger atom size F

Question 27

What are the general steps for drawing an MO diagram for 2 different species forming a bond?

Answer 27

Consider the relative energies of the AOs and put on the diagram See if sp mixing occurs, if so the π will go below σ Generally the S will still mix, as will pz as a σ and the other ps as a π Fill in electrons Special cases: NF: sp mixing does not occur Bonding with H: will result in non-bonding with the p x/y and likely s

Question 28

What does the MO diagram for HF look like?

Answer 28

1s of H higher is energy than 2s of F due to Zeff

Question 29

What does the MO diagram for CO look like?

Answer 29

sp mixing still occurs

Question 30

What does the MO diagram for NF look like

Answer 30

sp mixing does not occur enough to enable the σ to switch with the π

Question 31

How does the difference in energy between AOs affect bonding?

Answer 31

The closer in energy between the orbitals, the smaller the HOMO-LUMO gap in general, the larger the stabilisation of bond formation

Question 32

What does the MO diagram for HB look like?

Answer 32

Sp mixing occurs: The lone sp orbital is effectively non bonding, as as the px and py This is because there is no similar energy p in H to enable bonding, so non-bonding The antibonding is higher in energy than the non-bonding

Question 33

How do MO diagrams explain the existence of polar bonds?

Answer 33

There is an unequal contribution to the bonding MO when considering heteroatomic bonds The species with a lower energy AO, and so greater electronegativity, has a greater orbital coefficient for the bonding MO Therefore, the electron density has a greater probability to be found closer to this more electronegative species, resulting in a polar

Question 34

What does it mean to be paramagnetic and diamagnetic? What gives molecules these characteristics?

Answer 34

Paramagnetic: attracted into a magnetic field Diamagnetic: (slightly) repelled by a magnetic field Paramagnetism is caused by a species containing unpaired electrons

Question 35

Why does the BDE of N2 and CO differ despite being isoelectronic?

Answer 35

The isoelectronic species contain 5 electrons pairs There are 3 total electron pairs in C/O compared to 2 for N/N There is a greater reduction in inter-electron repulsion forming N/N stabilising this species, and acting as a driving force to break the bond, resulting in the N2 bond being weaker than CO Alternatively, N/N contain more electrons with parallel spins, so contains greater exchange energy When forming the molecules, this exchange energy must be overcome, thus reducing the net energy released from bond formation as offset, so a weaker bond

Question 36

What does the MO diagram for the terminal B-H-B in borane look like? How would you describe the bond?

Answer 36

3-centre, 2-electron bond

Question 37

What does the MO diagram for the HF2- look like? How would you describe the bond?

Answer 37

3-centre, 2 electron bond Non-bonding electrons do not contribute to the bond type

Question 38

What does the MO diagram XeF2 look like, assuming no d interaction? How would you describe the bond?

Answer 38

3 centre, 2 electron bond Non-bonding electrons do not contribute to the bond type

Question 39

How can the dz2 orbital interact in XeF2? Why is this a weak interaction?

Answer 39

There is some overlap between the dz2 and non-bonding orbitals, resulting in weak stabilisation This is weak because the 5dz2 is far higher in energy than the 2p of the fluorines The large energy gaps results in very poor orbital overlaps, so weak stabilisation

Question 40

Why does the 1s bond with the 2px in HF? Why does nothing else occur?

Answer 40

1s is similar in energy to the 2px The 2s of F is too low in energy for any significant overlap And as the 2p of H is far too high in energy for overlap, they do not form pi bonds, and instead non-bonding

Question 41

How would you approach an MO diagram for elements with the same period? And different periods?

Answer 41

Same period: there will be normal orbital overlap s-s, p-p, potential sp mixing Different period: i.e with H, overlap the orbitals with the same symmetry and similar in energy It could mean non-bonding, if no orbitals are of similar energy, or different bondings i.e s to p, with or without sp mixing

Question 42

How does sp mixing vary down a group?

Answer 42

The effect of sp mixing decreases This is because the energy gap between the np and ns increases with n This is because as the s orbital is more penetrating, the electrons can penetrate the core region and experience the full, greater nuclear charge more than p As nuclear charge increases down the group, the s is stabilised more than p, resulting in larger s-p gaps, reducing the impact of sp mixing

Question 43

How is exchange energy calculated?

Answer 43

For each subshell = k (n spin up ( n spin up -1))/2 + same for spin down

Question 44

Why is the BDE for SiS, P2, and AlCl different despite having the same electron configuration?

Answer 44

First consider exchange energies The exchange energy of P is greater than for Si/S=Al/Cl Therefore, more energy is required to overcome this exchange energy in the process of pairing, resulting in a weaker bond, and so a driving force for the bond breaking, so SiS stronger The orbital overlap between Si and S is stronger Al/Cl as there is a smaller difference in orbital energy This arises from Zeff increasing across the period