Question
Compare lanthanoids and actinoids in terms of contraction, oxidation states, and general chemical behaviour. Why do actinoids show a wider range of oxidation states than lanthanoids?
Solution — Step by Step
As we move from La (Z=57) to Lu (Z=71), each added 4f electron shields the nuclear charge very poorly. The 4f orbitals have a diffuse shape and penetrate poorly toward the nucleus. Result: effective nuclear charge increases steadily, pulling the electron cloud inward.
Atomic radii decrease from La (187 pm) to Lu (175 pm) — a total drop of about 12 pm across 14 elements. This steady shrinkage is called lanthanoid contraction.
Similarly, actinoids (Ac to Lr) show contraction as 5f electrons are added. But actinoid contraction is slightly greater than lanthanoid contraction because 5f orbitals shield even more poorly than 4f orbitals.
Lanthanoids predominantly show +3 oxidation state. Why? Removing three electrons (two from 6s and one from 5d or 4f) gives a stable configuration. Exceptions exist:
- Ce: +4 (achieves noble gas core, )
- Eu: +2 (achieves half-filled )
- Yb: +2 (achieves completely filled )
Actinoids show oxidation states from +3 to +7. The 5f, 6d, and 7s orbitals are very close in energy, so electrons from all three subshells can participate in bonding. This is why U shows +3, +4, +5, +6 states and Np even shows +7.
The early actinoids (U, Np, Pu) resemble transition metals more than they resemble lanthanoids in their chemistry.
| Property | Lanthanoids (4f) | Actinoids (5f) |
|---|---|---|
| Predominant oxidation state | +3 | +3 to +7 |
| Contraction magnitude | ~12 pm across series | Slightly greater |
| Shielding of f-electrons | Poor (4f) | Poorer (5f) |
| Radioactivity | Non-radioactive (mostly) | All radioactive |
| Complex formation | Limited | More extensive |
flowchart TD
A["f-block elements"] --> B["Lanthanoids 4f"]
A --> C["Actinoids 5f"]
B --> D["Poor 4f shielding"]
C --> E["Poorer 5f shielding"]
D --> F["Lanthanoid contraction ~12 pm"]
E --> G["Actinoid contraction slightly more"]
B --> H["Mainly +3 state"]
C --> I["+3 to +7 states"]
H --> J["4f electrons not easily removed"]
I --> K["5f, 6d, 7s close in energy"]Why This Works
The root cause is orbital energy spacing. In lanthanoids, the 4f orbitals sit well below the 5d and 6s levels — once you remove the 6s and 5d electrons, pulling out 4f electrons requires significantly more energy. So +3 dominates.
In actinoids, the 5f, 6d, and 7s orbitals overlap in energy. This means removing additional electrons beyond +3 does not cost dramatically more energy, enabling higher oxidation states.
Alternative Method
Think of it through ionisation enthalpy trends. Plot the successive IEs for a lanthanoid like Gd — there is a sharp jump after IE3. For an actinoid like U, the jump is much more gradual, explaining why removing 4th, 5th, even 6th electrons is feasible.
Common Mistake
A frequent error in CBSE boards: students write “lanthanoid contraction is caused by poor shielding of d-electrons.” Wrong — it is caused by poor shielding of 4f electrons. The d-block contraction is a separate phenomenon. In the f-block, it is specifically the diffuse f-orbitals that cannot shield effectively. This distinction has cost students marks in CBSE 2024 board papers.