F-Block and Nuclear Chemistry Flashcards
Describe the trend in the radius of the lanthanides.
There is a general decrease in the radius of the lanthanides as we progress from left to right across the period. This is due to the increase in effective nuclear charge (Zeff). However, the decrease is so big that it cannot be fully explained by the increase in Zeff. Itis also due to the relativsitc effect.
What is the relativistic effect?
For highly charged nuclei, as the electron approaches the nucleus, the speed of the electron increases up to ~2/3 of the speed of light. As the electron approaches the speed of light, the theory of relativity says that it increases in mass. This is important because the radii of orbitals have a 1/m dependency.
What are the different kinds of f-orbitals and why do they exist?
The Schrödinger wave equation gives solutions for the f-orbitals, however, like all orbitals higher than s, these solutions have a complex part. Visualising the f-orbitals and ignoring the complex part, we generate two different solutions: the cubic set and the general set.
Describe the general set of f-orbitals.
The f-orbitals have a high number of angular nodes, so lanthanides and actinides have a very high number of coordination geometries and coordination numbers.
The symmetry of the f-orbitals is u, so any electronic transition between f-orbitals is forbidden by the Laporte or symmetry selection rule.
Describe the radial distribution plots of f-orbitals and bonding.
For the 5f-orbitals, the main part of the electron density is quite close to the nucleus. The overlap between 5f and ligand orbitals is small.
The 4f-orbitals have zero nodes and they are highly spatially contracted. Covalent overlap between 4f and ligand orbitals is essentially zero. 4f-orbitals are core-like. Ligand-lanthanide bonding is ionic.
What is the effect of the relativistic effect on radial distribution?
Because Ψ is non-zero at the nucleus for s-orbitals, the electron in an s-orbital can slightly penetrate the nucleus so has a large relativistic effect. Because of this, s-orbitals are highly contracted. Therefore, 4f-orbitals are mostly core-like.
Describe the filling of the f-orbitals.
We see standard filling of the orbitals according to the Aufbau principle, except in cases where we see half-filled and filled f-shells (i.e. f7 and f14). This refelcts the unusual stability of half-filled and filled shells.
What is the most common oxidation state of lanthanides and why?
The core-like nature of the 4f-orbitals in the atomic state means that the electron-electron repulsion in these orbitals causes there to be high energy. As we oxidise, the Zeff increases and the 4f-orbitals feel a large electrostatic attraction from the positive nucleus. Therefore, the energy of the electrons in the 4f-orbitals decreases significantly. At the +3 oxidation state the f-electrons are stable so the fourth ionisation energy is very high for lanthanides.
Modern thinking is that the +2 oxidation state is also common.
What determines the relative stability of the +2 and +3 oxidation states in lanthanides?
The key factor in the value of ΔGm is I3 - the ionisation energy of Ln2+.
Describe the oxidation states of actinides.
For actinides, because the 5f-orbitals are much less core-like, there is a much wider range of oxidation states from +2 to +7.
What determines the energy of electronic states in metal ions?
There are three principle energy terms: electron-electron repulsion in the orbitals, ligand field splitting and spin-orbit coupling.
What is spin-orbit coupling?
The magnetic field generated by the angular momentum increases as the nuclear charge increases.
The orbital angular momentum couples with the spin angular momentum through mixing of the spin and angular momentum wavefunctions. This splits the energies of different electronic configurations.
The splitting gets bigger as the element gets heavier. Spin-orbit coupling is significant for f-elements.
How do we figure out the values of j for a multi-electron system?
For lanthanides we can work out the spin-orbit coupling quantum number using the Russell-Saunders coupling scheme:
How do you determine the term symbol of the ground state using Hund’s rules?
2S+1LJ
The ground state is the one with the maximum spin multiplicity (2S+1). When we have states with the same spin multiplicity, the ground state is the one with the maximum orbital angular momentum (L). When we have states with the same value of S and L, the ground state is the one with the smallest value of J for half-filled shells or less, or the one with the biggest value of J for more than half-filled shells.
How do you calculate the magnetic moment of lanthanides?
We use the Lande formula. We can do this because the spin-orbit coupling is large so J is a good quantum number, and there is very little quenching of the orbital angular momentum by the ligands.
Describe how a Nd YAG laser works and give a general scheme for lanthanide fluorescence.
- Intense white light is used to populate the excited state.
- Relaxation via intersystem crossing occurs.
- This state decays rapidly to the F1 state.
- This transition is slow because of the symmetry or Laporte selection rule, which creates a population inversion. More light is used to cause an intense stimulated emission (fluorescence).
- The F2 state then relaxes back to the ground state via thermal decay.
Give an example of an antenna ligand.
Give the common coordination numbers for lanthanides and the coordination geometry.
How could uranium be extracted?
Uranium is oxo-philic so it will coordinate to ligands via an oxygen atom. Tributylphosphate is a lipophilic ligand, which allows uranium to be selectively extracted into an organic solvent.
Describe the bonding in the uranyl cation.
The U=O bonds in the uranyl cation (UO22+) are very short and strong, therefore they are likely to contain some covalent character.
Describe covalent bonding in lanthanide complexes.
The f-orbital in lanthanides are below the d-orbitals. If we can mix these orbital interactions then f-electron density can be used to stabilise covalent M-L bonding. The degree of mixing depends on the d-f energy gap. It is only the lanthanides with small d-f energy gaos that exhibit stable M-L covalent bonding.
How can lanthanide complexes be used in catalysis?
As catalysts lanthanides act as Lewis acids to polarise substrates including organic molecules. They have high ligand exchange rates, which gives high catalytic rates. They are stable catalyst so their reactions are not complicated by unwanted side redox reactions.
