F-Elements and Nuclear Chemistry

Question 1

What pattern do the lanthanide radii show?

Answer 1

The radii steadily drop as the row goes from left to right, which is also shown by the 3D TM’s.

Question 2

Why do the radii of the lanthanides show a pattern?

Answer 2

The drop in radius with increased atomic number could be due to the increase in Z_eff. However, the drop is very significant and can’t be fully explained by the increase in Z_eff. Calculations by Laerdahl suggested that 30% of the contraction in radius was due to the relativistic effect.

Question 3

What is the relativistic effect?

Answer 3

For highly charged nuclei, as the electron approaches the nucleus it speeds up to 2/3 of the speed of light due to the electrostatic attraction between the electron and the nucleus. As it approaches the speed of light, the elctron also increases in mass. Radii of orbitals have a 1/mass dependency, therefore this increase in mass causes a decrease in radii.

Question 4

What are the different kinds of f-orbitals and why do they exist?

Answer 4

The Schrodinger equation gives solutions for f-orbitals, however, like all orbitals higher than s, these solutions have a complex part. Visualising the f-orbitals, while ignoring the complex part, generates two different solutions - the cubic set and the general set.

Question 5

Describe the general set of f-orbitals.

Answer 5

• high number of angular nodes causing lots of coordination geometries and coordination numbers
• ungerade symmetry (non-centrosymmetric) - any f-f transition is u-u and therefore symmetry-Laporte forbidden

Question 6

Describe the radial distribution plot of the 6s, 7s and 5f orbitals of the actinides and what it suggests about bonding.

Answer 6

• the orbitals move away from the nucleus (greater r) in the order 5f < 6s < 7s
• 5f - one node, 6s - 5 nodes, 7s - 6 nodes
• the main part of electron density of the 5f orbitals is quite close to the nucleus
• overlap between 5f and ligand orbitals is small

Question 7

Describe the radial distribution plot of the 4f, 5s and 6s orbitals of the lanthanides and what it suggests about the bonding.

Answer 7

• the orbitals move away from the nucleus (greater r) in the order 4f < 5s < 6s
• 4f - no nodes, 5s - 4 nodes, 6s - 5 nodes
• the 4f orbitals are highly spatially contracted (close to the nucleus)
• the covalent overlap between ligand orbitals and 4f orbitals is essentially zero
• 4f orbitals are essentially ‘core-like’
• Ln-L bonding is essentially ionic

Question 8

What are the effects of the relativistic effect on the radial distributions?

Answer 8

• s-orbitals have non-zero ψ at the nucleus so the electron can slightly penetrate the nucleus
• s-electron gains enormous speed causing a large relativistic effect and therefore a large increase in electron mass
• as a result, the s-orbitals are highly contracted = direct relativistic contraction and reduced Z_eff
• the reduced Z_eff causes the 4f orbitals to move out (indirect relativistic expansion) causing them to be mostly core-like

Question 9

Describe how the f-orbitals are filled with electrons and relate it to the lanthanides.

Answer 9

There is standard filling of the orbitals according to the Aufbau principle, except for half-filled (f⁷) or fully-filled (f¹⁴) f-orbitals. In the lanthanides, ganolidium is f⁸, one more than half-full, and instead of being in the f-orbitals the extra electron is placed into the above d-orbital to retain the f⁷ configuration. This is because f⁷ is particularly stable, similar to the fully-filled orbital, which is unusual.

Question 10

What does the filling of orbitals for atomic lanthanides suggest?

What happens when we ionise the lanthanides?

Answer 10

It suggests that the 6s orbital is lower in energy than the 4f orbitals as the 6s orbitals are filled before the 4f orbitals. This is counterintuitive as we might expect the 6s orbital to be higher in energy and therefore filled second in preference to the 4f orbitals.

Even more unusual is that upon ionisation, the 6s electrons are the first to be removed even though they’re seemingly lower in energy.

Question 11

Describe the 4f and 6s orbitals in the lanthanides of oxidation state zero.

Answer 11

When the oxidation state is zero, the 4f orbitals are core-like with large electron-electron repulsions causing the orbitals to be high in energy. The 4f orbitals are also small in size which increases the repulsion.

In comparison, the 6s orbital is more diffuse and therefore there is less repulsion between electrons. This means the 6s orbital is lower in energy than the 4f orbitals.

Question 12

Describe the energies of the 4f and 6s orbitals when the lanthanides are oxidised.

Answer 12

• Upon oxidation, Z_eff increases
• The 4f orbitals feel an increase in electrostatic attraction from the nucleus (which is large due to their core-like nature)
• Therefore, the 4f orbitals see a significant drop in energy causing them to be more stabilised
• The 6s orbital doesn’t feel as big of an increase in electrostatic attraction due to its distance from the nucleus so it doesn’t drop in energy as much
• The 4f orbitals become lower in energy than the 6s orbitals

Question 13

What is the most common oxidation state of the lanthanides?

Answer 13

Once the lanthanides have gotten to the +3 oxidation state the f-electrons are stabilised and therefore it’s difficult to remove any more, so the +4 ionisation energy is very high.

This makes the +3 oxidation state the most common for lanthanides. Most textbooks will say this is the only common oxidation state, but modern thinking suggests that the +2 oxidation state is also common.

Question 14

Describe the metallic/atomic radii of both the lanthanides and actinides (before any ionisation).

Answer 14

• there is a general drop in radius of the lanthanides from left to right
• there are two distinct lanthanide exceptions - europium and ytterbium - due to the half-filled and full-filled subshells
• for the actinides there’s much less regularity, due to more complicated orbital structure, but overall the have smaller metallic radii than the lanthnides (more compact)

Question 15

Write the reaction and equation for ΔG_m for the +2 and +3 lanthanide oxidation states.

Define ΔG_m and state what the key factor in its equation is.

Answer 15

Ln²⁺_(aq) + H⁺_(aq) –> Ln⁺³_(aq) + 1/2H_2(g) ΔG_m

ΔG_m = I₃ + ΔH_hyd(M³⁺) - ΔH_hyd(M²⁺) - TΔS + ΔH_H

where H⁺ + e^- –> 1/2H_2(g) ΔH_H

ΔG_m = measure of ox. state stability, with the key factor in it being I₃, the ionisation energy of Ln²⁺.

Question 17

Describe the trends in I₃ and ΔG_m for the lanthanides.

Answer 17

The ionisation energies of Eu and Yb are much higher than the rest of the lanthanides. This means that the ΔG_m values for these two elements are less negative so the +3 oxidation state is favoured less by Eu and Yb.

We can form Eu²⁺ and Yb²⁺ easily due to the relative stability of the f⁷ and f¹⁴ shells in Ln²⁺.

Question 18

Define ΔH_atomisation for the lanthanides and their trend.

Answer 18

ΔH_atomisation is the energy released when condensing an atomic element down to a metal.

The trend in ΔH_atomisation is also anomalous for Eu and Yb, showing the ‘least’ stability in the metallic state.

Question 19

Describe the trend in oxidation states for the actinides.

Answer 19

In the actinides, the 5f orbitals are much less core-like leading to a much wider range of oxidation states because the orbitals are larger so the IE’s are less.

It becomes easier to reduce the ions from left to right due to the increase in Z_eff. Americium and curium are half-filled and full-filled subshell exceptions.

A lot of the actinides can reach oxidation states greater than +3, which U being stable as +6.

Question 20

What are the three principal energy terms?

Answer 20

1. Electron-electron repulsion in the orbital (d-d or f-f)
2. Ligand field splitting - CFSE is an example
3. Spin-orbit coupling

Question 21

Why do 4d orbitals of the TMs have a higher ligand field splitting value than 3d orbitals?

Answer 21

4d orbitals are bigger and overlap more with the ligands, therefore the ligands give a higher degree of splitting.

Question 22

Why do the 4f orbitals of the lanthanides have a large electron-electron repulsion value? Why do they also have a very small ligand field splitting value?

Answer 22

The 4f orbitals are quite compact and core-like, therefore there’s more electron-electron repsulsion. This is also means that there’s much less overlap with ligands so the ligand field splitting is much smaller.

For the first time, the spin-orbit coupling is higher than LF.

Question 23

What would you describe the principal energy term as for 3d, 4d+5d, 4f and 5f orbitals?

Answer 23

• 3d TM - e-e repsulsion and LF
• 4d+5d - LF
• 4f - e-e repsulsion and spin-orbit coupling
• 5f - same as 4f with more spin-orbit coupling contribution

Question 24

What is spin-orbit coupling?

Answer 24

A charged electron is orbiting a nucleus. This orbiting charge creates a magnetic field in the centre of the orbit, therefore the nucleus has a magnetic field. The nucleus magnetic field has orbital angular momentum while the electron has spin angular momentum. These couple to each other.

Question 25

Why does spin-orbit coupling increase down the periodic table?

Answer 25

Can view it as positively charged nucleus orbiting the electron, where the nucleus becomes more charged down the periodic group. The more charge, the larger the magnetic field created at the electron. Aka the magnetic field generated by angular momentum increases as the atomic number increases/higher nuclear charge.

Question 26

How can quantify spin-orbit coupling?

Answer 26

For actinides, we need to turn to j-j coupling to make sense of the spin-orbit coupling.

For lanthanides, we can work out the spin-coupling quantum numbers using the Russell Saunders coupling scheme.

Each electron’s orbital and spin angular momentum can be summed separately and then used to find the total angular momentum:

L = l1 + l2 + l3 …

S = s1 + s2 + s3 …

J = L + S, L + S - 1, L + S - 2, … |L - S|

Which can then give a term symbol.

Question 27

What are Hund’s rules?

Answer 27

1. Ground state is the state with the maximum spin multiplicity (2S+1) value
2. When states have the same spin multiplicity, the ground state is the one with the maximum orbital angular momentum (L)
3. When states have the same S + L, the ground state is:
   
   • > half-filled - minimum J value
   • < half-filled - maximum J value

Question 28

How do you calculate the magnetism of systems with no orbital angular momentum?

Answer 28

E.g. for 3d TMs as there is very little orbital angular momentum due to the motion of the electrons being hindered by the ligands. Therefore, only spin angular momentum is important and the ‘spin-only’ formula can be used.

Question 29

How do you calculate the magnetism of spin-orbit coupled systems?

Answer 29

We can’t use the ‘spin-only’ formula as the orbital angular momentum is important now. We use a different formula, called the Lande formula.

Question 30

Compare calculated and experimental magnetic moments.

Answer 30

There is a very good match between them. This demonstrates how 4f orbitals are core-like as they’re interfered with by ligands. Eu deviates from the calculated formula, possibly becomes the ground state isn’t described well by a single term.

Question 31

Describe the magnetism of the actinides.

Answer 31

Actinide magnetism is much more complicated and the Russell Saunders scheme breaks down. This means you need to turn to j-j coupling to understand the system. The greater spatial extension of the 5f orbitals means that there’s orbital angular momentum quenching and L is no longer a good quantum number,

Question 32

Describe the electronic absorptions in f-elements.

Answer 32

Lanthanides and actinides don’t have vibrations which affect the energies of the f-orbitals. Therefore we can expect the f-f transitions to be sharp as they aren’t broadened out by metal-ligand vibrations.

Question 33

Describe the intensities of f-f transitions of lanthanides.

Answer 33

The symmetry of the f-orbitals is ungerade (uncentrosymmetric) therefore an f-f transition is u -> u which is forbidden by the symmetry/Laporte selection rule. Therefore, the bands of lanthanide +3 salts (visible absorption bands) are weak. The combination of sharp but weak 4f-4f transition is a characteristic of the lanthanides.

Question 34

Describe the absorptions of the actinides.

Answer 34

The actinides have broader bands than lanthanides, due to a degree of orbital-ligand overlap which gives some vibrational fine structure to the bands.

The spectra of actinides are more intense than lanthanides due to 5f-ligand overlap breaking down the ungerade symmetry of the 5f orbital.

Question 35

Describe the how an Nd YAG laser works.

Answer 35

1. Use intense white light to populate excited states
2. Rapid decay of the excited states to the ⁴F_3/2 level f-configuration
3. Transition to the other f-configuration - slow because the transition is forbidden, causing a build-up of the excited state
4. Stimulated emission by shining in more light, intense and sharp due to the build-up of the population in the excited state
5. Decay back to the ground state

Question 36

Describe the general scheme for lanthanide fluorescence.

Answer 36

• excitation into singlet ligand states where we then get isc to triplet state
• decay to first f-configuration where the population inversion/build-up is created
• decay to the second f-configuration gives a sharp transition
• decay back to the ground state

Question 37

What can we use to obtain the high intensity transition from lanthanides?

Answer 37

We can use ligands like antennae to harvest light to feed into the f-levels via isc to give the high intensity transition.

Question 38

Give an example of a lanthanide metal with an ‘antennae’ ligand and how it can be used.

Answer 38

An example is an Ln₄L₆ cage. Each ligand acts as an antennae which you can promote to create an excited state and therefore the population inversion. The void in the centre of the cage can be deteced by fluorescence, and it will fluoresce a different colour when certain molecules are inside. This means lanthanides can be used to detect different molecules.

Question 39

Name two applications of lanthanide fluorescence.

Answer 39

1. Screen phosphors
   
   • high intensit, narrow emission bands which give very intense colours on TVs
2. MRI contrast agents
   
   • lanthanides have very high magnetism
   • Gd³⁺ has maximum unpaired spin (f⁷) so it has a very large magnetic field
   • this magnetic field can be used to magnetically relax protons which are in a magnetically excited state

Question 40

How does Gd act as an MRI contrast agent?

Answer 40

Gd coordinates 8 atoms from DTPA with one vacant space. This allows Gd to coordinate to a water molecule and relax any magnetic excitation it has. The Gd then exchanges it for a different magnetically excited molecule.

DTPA is a strong chelate and doesnt allow free Gd in the body (which is toxic by itself).

Question 41

Describe the bonding and coordination chemistry of the lanthanides.

Answer 41

• core-like nature of the f-orbital means we can expect little to no M-L covalent bonding involving these orbitals
• therefore M-L bonding is ionic
• Ln³⁺ ions are classified as hard ions so they bond well with hard ligands - typically those coordinated through oxygen
• Lanthanides are oxo-philic - they like to coordinate to oxygen atoms
• high coordination numbers are common - 6 to 12 around one ion

Question 42

Describe the stability of lanthanide complexes.

Answer 42

The chelate effect is very important in lanthanide coordination chemistry. There’s a general increase in the stability of Ln-L complexes as the Z_eff of the lanthanide increases (stronger ionic bond).

Europium is unusual because of the stability of the f⁷ configuration.

Question 43

Describe actinide coordination chemistry using UO₂²⁺ as an example.

Answer 43

Also oxo-philic, which can be taken advantage of when wanting to separate actinides.

The U-O bond is stronger than just ionic bonding, it’s likely that the bond has some covalent character.

There are also organometallic actinide complexes. These must require a degree of M-L covalency to be stable so the 5f orbitals are likely to give some covalent overlap.

Question 44

There must be some covalency in the M-L bonds of lanthanides, but the 4f orbitals are core-like. How is this explained?

Answer 44

Use Sc(Cp)₂ complex as an example - Sc is a TM not lanthanide, but we can draw up an MO energy level diagram for M-L bonding. It shows that if we can promote electrons into the d-orbitals then we can get covalent bonding.

Lanthanide f-orbitals are lower in energy than d-orbitals. If we can ‘mix’ the interactions of these orbitals then f-electron density can be used to stabilised covalent M-L bonding.

Question 45

What lanthanides exhibit stable M-L covalent bonding?

Answer 45

The degree of mixing between the lower energy f-orbitals and higher energy d-orbitals depends on the energy gap. A small energy gap means we might expect some f-electron density to occupy the d-orbitals.

This means that only the lanthanides with small d-f energy gaps exhibit stable M-L covalent bonding.

Question 46

How can lanthanides act as catalysts?

Answer 46

• as catalysts, lanthanides basically act as Lewis acids (M³⁺ character) to polarise substrates including organic molecules - organometallic-like chemistry
• high ligand exchange rates giving high catalytic rates
• not complicated by unwanted side redow reactions - catalysts tend to be stable (high yield, good product)