The d- and f-Block Elements • Topic 3 of 3

f-Block: Lanthanoids & Actinoids

The f-block elements are the two rows placed below the main body of the periodic table. The lanthanoids (Ce–Lu, following lanthanum) involve the filling of the $4f$ orbitals, and the actinoids (Th–Lr, following actinium) involve the filling of the $5f$ orbitals. They are also called the inner transition elements because the differentiating electron enters an inner $(n-2)f$ subshell.

Electronic configuration

The lanthanoids have the general configuration $[\text{Xe}]4f^{1-14}\,5d^{0-1}\,6s^2$, and the actinoids $[\text{Rn}]5f^{1-14}\,6d^{0-1}\,7s^2$. Because the $4f$ and $5d$ (or $5f$ and $6d$) levels are very close in energy, minor irregularities occur, but the dominant change along each row is the filling of the f subshell.

Oxidation states

The characteristic and most stable oxidation state of the lanthanoids is +3. A few show +2 or +4 when these lead to empty, half-filled or full f subshells (e.g. $\text{Ce}^{4+}$ is $4f^0$; $\text{Eu}^{2+}$ is $4f^7$). Actinoids show a much wider range of oxidation states (+3 to +6 and even +7), because the $5f$, $6d$ and $7s$ levels are similar in energy and the $5f$ electrons are more loosely held.

Lanthanoid contraction

Across the lanthanoid series the atomic and ionic radii steadily decrease. This is the lanthanoid contraction. As each electron is added to the inner $4f$ subshell, it shields the outer electrons from the nucleus very poorly (f orbitals are diffuse and non-penetrating). The effective nuclear charge felt by the outer electrons therefore rises steadily, pulling them inward.

Consequences of lanthanoid contraction

Its effects extend beyond the f-block: (i) the second (4d) and third (5d) transition series have almost the same atomic radii (e.g. Zr ≈ Hf, Nb ≈ Ta), which makes separating these pairs very difficult; (ii) the basicity of $\text{Ln(OH)}_3$ decreases from La to Lu as the cation gets smaller; (iii) the third-row transition metals have unusually high densities.

Colour and magnetic behaviour

Many lanthanoid ions are coloured, owing to f–f transitions; ions with $f^0$ ($\text{La}^{3+}$, $\text{Ce}^{4+}$) and $f^{14}$ ($\text{Lu}^{3+}$) are colourless. Most $\text{Ln}^{3+}$ ions with unpaired f electrons are paramagnetic, though the spin-only formula does not apply well because the orbital angular momentum of f electrons contributes significantly.

Lanthanoids versus actinoids

Both are mostly +3, electropositive and show contraction (the actinoid contraction is even more pronounced). But actinoids show a far greater variety of oxidation states, are all radioactive, and the elements beyond uranium are man-made (transuranium elements). Actinoid chemistry is more complex because the $5f$ electrons are less effectively shielded and more available for bonding than the $4f$ electrons of lanthanoids.

Lanthanoid contraction: ionic radius of Ln3+ decreasing across the seriesIonic radius of Ln3+ (pm)La to Lu (increasing Z)LaLuradius decreases steadily (contraction)
Solved Examples
1
Worked Example
Write the general electronic configuration of the lanthanoids and actinoids.
Solution
  1. Lanthanoids fill the $4f$ subshell after lanthanum.
  2. Actinoids fill the $5f$ subshell after actinium.
  3. Outer $6s$/$7s$ and an occasional $5d$/$6d$ electron complete the configuration.

Answer: Lanthanoids: $[\text{Xe}]4f^{1-14}\,5d^{0-1}\,6s^2$; Actinoids: $[\text{Rn}]5f^{1-14}\,6d^{0-1}\,7s^2$.

2
Worked Example
Why is $\text{Ce}^{4+}$ a common and stable oxidation state of cerium?
Solution
  1. Cerium is $[\text{Xe}]4f^1\,5d^1\,6s^2$ (4 outer electrons available).
  2. Losing four electrons leaves $\text{Ce}^{4+}$ with the configuration $[\text{Xe}]4f^0$.
  3. An empty f subshell ($4f^0$) is especially stable.

Answer: $\text{Ce}^{4+}$ is stable because it attains the extra-stable empty $4f^0$ configuration (it is also a good oxidising agent, reverting to $\text{Ce}^{3+}$).

3
Worked Example
Define lanthanoid contraction and state its cause.
Solution
  1. It is the steady decrease in atomic and ionic radii from La to Lu across the lanthanoid series.
  2. Each added electron goes into the inner $4f$ subshell.
  3. The diffuse $4f$ orbitals shield the outer electrons poorly, so the effective nuclear charge rises and pulls them in.

Answer: Lanthanoid contraction is the gradual decrease in size across the lanthanoids, caused by the poor shielding of the outer electrons by the inner $4f$ electrons.

4
Worked Example
Why is it difficult to separate zirconium from hafnium?
Solution
  1. Zr (4d series) and Hf (5d series) lie in the same group.
  2. Because of the lanthanoid contraction, Hf does not become much larger than Zr despite its higher atomic number.
  3. Their atomic and ionic radii are therefore almost identical, giving very similar chemical properties.

Answer: Lanthanoid contraction makes the radii of Zr and Hf nearly equal, so their chemistry is almost the same and they are very hard to separate.

5
Worked Example
Give two ways in which actinoids differ from lanthanoids.
Solution
  1. Compare oxidation states: lanthanoids are mostly +3; actinoids range from +3 to +6 (even +7).
  2. Compare stability: all actinoids are radioactive, and elements beyond uranium are synthetic.
  3. This wider variability is because $5f$ electrons are more loosely held and closer in energy to $6d/7s$.

Answer: Actinoids show many oxidation states (up to +6/+7) and are all radioactive, whereas lanthanoids are predominantly +3 and (except Pm) non-radioactive.

6
Worked Example
Which lanthanoid ions are colourless and why?
Solution
  1. Colour in lanthanoid ions arises from f–f transitions, which need unpaired f electrons with vacant f orbitals.
  2. $\text{La}^{3+}$ and $\text{Ce}^{4+}$ are $4f^0$ (no f electrons).
  3. $\text{Lu}^{3+}$ is $4f^{14}$ (completely filled f subshell).

Answer: $\text{La}^{3+}$, $\text{Ce}^{4+}$ ($f^0$) and $\text{Lu}^{3+}$ ($f^{14}$) are colourless because no f–f transition is possible.

Key Points

  • f-block (inner transition) elements fill $(n-2)f$ orbitals: lanthanoids $[\text{Xe}]4f^{1-14}5d^{0-1}6s^2$, actinoids $[\text{Rn}]5f^{1-14}6d^{0-1}7s^2$.
  • The characteristic oxidation state is +3; +2/+4 appear where they give $f^0$, $f^7$ or $f^{14}$ (e.g. $\text{Ce}^{4+}$, $\text{Eu}^{2+}$); actinoids show +3 to +6/+7.
  • Lanthanoid contraction: steady decrease in size across La–Lu, due to poor shielding by the inner $4f$ electrons.
  • Consequences: 4d and 5d elements have nearly equal radii (Zr ≈ Hf, hard to separate), $\text{Ln(OH)}_3$ basicity falls, high density of 5d metals.
  • Colour comes from f–f transitions ($f^0$ and $f^{14}$ ions colourless); actinoids are all radioactive and chemically more complex than lanthanoids.
Quick Check — 4 MCQs
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Q1.The most common oxidation state of the lanthanoids is:
Explanation: The +3 state is characteristic and the most stable across the lanthanoid series.
Q2.Lanthanoid contraction is caused by:
Explanation: The diffuse $4f$ electrons shield poorly, so effective nuclear charge rises and the size shrinks steadily.
Q3.Which pair is difficult to separate due to lanthanoid contraction?
Explanation: Zr (4d) and Hf (5d) have almost identical radii because of the contraction, so their chemistry is very similar.
Q4.Compared with lanthanoids, actinoids:
Explanation: Loosely held $5f$ electrons give actinoids many oxidation states (up to +6/+7), and all actinoids are radioactive.
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