f Block Chemistry

Why are they called the rare earths if they are more abundant than some other elements?

Because they are difficult to extract and purify.

Explain the trends depicted in this graph.

Lanthanides generally get rarer as they get heavier due to decreasing nuclear stability.

This is because of the increase in coulombic repulsion in nuclei with more protons.

The odd/even pattern is due to certain nuclei numbers being more stable.

Pm is so radioactive that it has a very short half life and hence a very low abundance.

How are the different lanthanides separated?

On an ion exchange column.

Different ligands which coordinate to different lanthanides are fixed to a column.

What is HEX?

HEX is UF₆.

HEX is used in ²³⁵U separation (from ²³⁸U) by diffusion.

What does Paul mean by ‘lanthanides are lumpy’?

They have high angular nodality

Explain the Relativistic Effect in relation to lanthanides.

A trend of decreasing Ln³⁺ size across the row is observed.

This was previously thought to be just due to the increase in effective nuclear charge (Z_eff).

However, it is now thought to be partially due to relativistic effects:

• as an electron approaches the nucleus, it speeds up considerably

• at higher speeds, the mass of the electron increases to around 1.35x its usual mass

• therefore there is higher electron density near the nucleus, resulting in the observed nuclear contraction

What is the consequence of f block elements having high angular nodality?

They have high corrdination numbers.

They also behave like spheres when it comes to bonding.

Are f orbitals u or g?

They are ungerade.

Draw a radial distribution plot to demonstrate the differences between the 4f, 5f and 6s orbitals and describe the key features.

• 4f orbital has no radial nodes and electron density is closer to the nucleus than 6s so is “core-like”

• 5f has one node and is less core-like - 5f electrons can therefore interact with things like ligands (“the outside world”)

Approximately how far from the nucleus do ligands sit? How does this explain the bonding of lanthanides?

Ligands sit ~2.5 Å away from the nucleus.

The 4f orbital is far too contracted to interact with the ligands in any covalent way.

Therefore, all of the bonding in lanthanides is ionic.

(There is a little bit of covalency in actinides because the 5f orbital is less core-like)

Explain the different magnitudes of significance of the relativistic effect on 4f and 6s orbitals.

The 6f orbital is more significantly impacted by relativistic effects than 4f orbitals.This is because the 6s electrons can pass right through the nucleus, so reach higher speeds, hence increasing electron density closer to the nucleus.

Why are Ln-ligand bond lengths longer than Ac-ligand bonds?

Actanides are larger than lanthanides (due to occupitation of 5f orbitals).

(f-block - ligand bonding is mostly ionic)

What is Lanthanide Contraction?

The trend of decreasing size across the f block rows caused by relativistic effects (which impact the 6s orbital significantly).

Explain why the 5d elements are smaller than expected?

5d transition metals are contracted due to the relativistic effect shrinking orbitals, such as the 5s orbital.

This is why Ti → Zr → Hf doesn’t really increase in size much, even though theres a significant increase in atomic number. Hence, 4d and 5d elements have similar chemistry (and are therefore difficult to separate).

What does the Aufbau Principle tell us about the ionisation of lanthanides and actanides?

That f electrons are expected to be the first to be removed because they sit at higher energies.

Why is 3+ the most common oxidation state for f-block elements?

The 4f/5f orbitals are the highest in energy so are the first to be ionised (Aufbau Principle).

However, once an f electron is removed, the f orbitals drop in energy and the 6s or 5d orbitals become the highest in energy.

Therefore, the lanthanide loses one f electron, then 6s or 5d electrons until it gets to 3+.

It is difficult to ionise beyond 3+.

Valence electrons in a 3+ lanthanide ion are in f orbitals, which explains their chemistry.

What is the general trend of metallic radius in lanthanides? Explain why it is observed.

General trend of a very slight decrease in size.

This similarity in size is observed because of the nature of metallic bonding - positive cations in a sea of electrons. Because all lanthanide ions are 3+, the strength of metallic bonding is the same for each so little change is observed.

Eu and Yb are outliers - they have some stability in a +2 state so can exist in a weaker bonding metal system.

What makes Eu and Yb different from the other lanthanides?

Eu²⁺: f⁷

Yb²⁺: f¹⁴

While other lanthanides are stable at 3+ oxidation states, Eu and Yb can also exist as 2+ ions because of their electron configurations (↑). Their half filled (f⁷) and fully filled (f¹⁴) sublevels in the +2 state make this state more stable and disfavours the third ionisation:

What notable oxidation state is known for Ce and why is it important?

Ce can exist as Ce⁴⁺, which is used as a strong oxidising agent.

How do actinide oxidation states differ from lanthanides?

Actinide oxidation states are much more complex than lanthanides, which generally just occupy +3.

Actinides occupy a variety of oxidation states due to their 5f orbitals being less core-like:

These oxidation states exhibit different solubilities (filled circle is aqueous, unfilled is solid) which can cause issues for radioactive waste handling.

Explain the trend exhibited in the Ligand Field column, specifying the difference between the two circle values

Larger metal ions experience greater splitting by ligands because the valence electrons are held closer to the ligand electrons.

The 4f orbital is core-like and its electron density is held close to the nucleus, so ligands have very little impact on lanthanide energy levels. Molecules which bind lanthanides behave as if the lanthanide is barely bonded at all and it acts like a gas phase ion.

The 5f orbital is less core-like, and holds some electron density further out from the nucleus. Therefore, a larger (but still fairly small) impact is felt on the actinide electrons by the ligand electrons.

Explain the trend exhibited in the e-e repulsion column, specifying the difference between the two circle values

3d → 4d → 5d: orbitals are getting bigger so there is less electron-electron repulsion

4f orbital is core-like so electrons are held close together near the nucleus, hence significant increase in repulsion.

5f orbital is larger and has a radial node so less electron-electron repulsion, but still fairly significant due to the contracted nature of f orbitals.

Explain the trend exhibited in the spin-orbit coupling column.

3d → 4d → 5d → 4f → 5f: electrons have higher angular momentum, hence experience more spin-orbit coupling.

This is denoted by the electrons attraction to the nucleus (distance and size).

What is the Russell Saunders coupling scheme?

The simplification of spin-orbit coupling by summing all of the spin (s) and angular momentum (l) components:

s₁ + s₂ + s₃ … = S

l₁ + l₂ + l₃ … = L

J = S + L

What is the formula for making a term symbol?

What are the rules for J values?

What are Hund’s Rules which define the ground state term symbol?

1) Largest S (maximum spin multiplicity) is the ground state

2) If there are two states with the same S, the ground (/lower) state is the one with the largest L

3) For >half filled states, the maximum value of J is the ground state.

Determine the ground state term symbol for Ce³⁺

Ce³⁺ = 4f¹

S = (+) ½

(2S+1 = 2)
L = + 3

(L = 0 (S), L = 1 (P), L = 2 (D), L = 3 (F), L = 4 (G), …)

Subshell is < half full so minimum J value is ground state.

J = 3.5, 2.5

Term symbol = ²F_2.5

Determine the ground state term symbol for Dy³⁺

Dy³⁺ = 4f⁹

S = (+) 5/2

(2S+1 = 6)
L = + 5

(L = 0 (S), L = 1 (P), L = 2 (D), L = 3 (F), L = 4 (G), L = 5 (H), … )

Subshell is > half full so maximum J value is ground state.

J = 7.5, 6.5, 5.5, 4.5, 3.5, 2.5

Term symbol = ⁶H_7.5