Lanthanoids

Position and configuration. Lanthanoids are the 14 elements Ce (Z = 58) to Lu (Z = 71), following La (Z = 57), across which the 4f subshell is progressively filled: $[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^2$. La plus these 14 are together called the lanthanides, or rare earths.

Why the contraction happens. Every added electron enters the same 4f subshell, an orbital that is compact and diffuse at once — it sits close to the nucleus but has a shape that screens the outer $6s$ electrons very poorly. So while the nuclear charge $Z$ climbs by one at every step, the screening constant $\sigma$ rises by much less than one. The net pull felt by the outermost electrons increases steadily, and the atom shrinks a little at every element. Because this happens 14 times in a row, the cumulative effect is large even though each individual step is small.

Consequences of the contraction. Three effects follow directly: The radii of the second (4d) and third (5d) transition series in the same group converge almost exactly, because the contraction across the intervening 4f series cancels the radius increase normally expected on adding an extra shell — Zr and Hf are the textbook pair, and Nb/Ta and Mo/W show the same effect. Basic strength of the hydroxides $\text{Ln(OH)}_3$ falls steadily from La to Lu, since a shrinking $\text{Ln}^{3+}$ radius means rising charge density and rising covalent character in the Ln–OH bond, which resists releasing $\text{OH}^-$. The general chemical similarity of the lanthanoids — the reason they are so hard to separate — is itself a consequence of their radii changing only slowly and smoothly across the series, rather than in sharp jumps.

Oxidation states, colour, complexes. The dominant oxidation state throughout the series is $+3$; a few elements also show $+2$ or $+4$ when this yields a more stable $4f^0$, $4f^7$, or $4f^{14}$ configuration — $\text{Ce}^{4+}$ is $4f^0$, $\text{Eu}^{2+}$ is $4f^7$. Many $\text{Ln}^{3+}$ ions are coloured, from f–f transitions that are weaker and sharper than the d–d transitions of the d-block, because the 4f orbitals sit shielded beneath the filled $5s$ and $5p$ shells. Complex formation is comparatively limited, since those buried 4f orbitals overlap poorly with ligand orbitals.

Chemical reactivity. Lanthanoids are reactive, electropositive metals: they tarnish in air forming $\text{Ln}_2\text{O}_3$, react slowly with water liberating $\text{H}_2$, react readily with dilute acids liberating $\text{H}_2$, combine with hydrogen to give non-stoichiometric hydrides $\text{LnH}_{2-3}$, and combine with halogens to give $\text{LnX}_3$. On heating they also form sulphides, nitrides and carbides. They alloy readily, most famously as misch metal (about 95% lanthanoid plus iron, used in lighter flints).

Problem: Explain why zirconium (Zr, 4d series) and hafnium (Hf, 5d series) have almost identical atomic radii of about $159\ \text{pm}$ each, even though Hf lies directly below Zr and one full period lower, and state one practical consequence.

Step 1. Normally, descending a group adds a full extra principal shell, so Hf would be expected to be noticeably larger than Zr — the usual pattern seen for lighter groups such as Ti/Zr/Hf's own neighbours before lanthanoid effects are considered.

Step 2. Between the 4d series (ending near Zr) and the 5d series (starting near Hf), the entire 4f series — all 14 lanthanoids, Ce to Lu — is filled first. By Steps 1–5 of the derivation above, this produces a steady, cumulative lanthanoid contraction across those 14 elements.

Step 3. That contraction almost exactly cancels the expected radius increase from adding the extra principal shell on going from period 5 (Zr) to period 6 (Hf), since both effects are comparable in size but act in opposite directions.

Step 4. Net result: $r_{\text{Zr}} \approx r_{\text{Hf}} \approx 159\ \text{pm}$ — essentially identical atomic radii, despite Hf sitting a full period below Zr.

Step 5. Practical consequence: because their radii and chemical behaviour are so alike, Zr and Hf are extremely difficult to separate and are almost always found together in ores; the same lanthanoid-contraction effect makes the Nb/Ta and Mo/W pairs similarly hard to separate.

Competitive stretch

Rank $\text{La(OH)}_3$, $\text{Nd(OH)}_3$, and $\text{Lu(OH)}_3$ in order of decreasing basic strength. Since $\text{La}^{3+}$ (Z = 57) has the largest ionic radius among the three, is the least polarising, and gives the most ionic Ln–OH bond, it is the strongest base; $\text{Lu}^{3+}$ (Z = 71, 4f fully filled) has the smallest radius and is the weakest base; $\text{Nd}^{3+}$, mid-series, falls between them. Order: $\text{La(OH)}_3 > \text{Nd(OH)}_3 > \text{Lu(OH)}_3$.

• Confusing lanthanoid contraction with the smaller, more irregular radius decrease across the 3d transition series — the lanthanoid effect is larger and more regular because it spans 14 elements of poorly-shielding 4f electrons, not d electrons.
• Assuming every lanthanoid shows only the $+3$ state — remember $\text{Ce}^{4+}$ ($4f^0$) and $\text{Eu}^{2+}$ ($4f^7$), both stabilised by reaching an empty or half-filled 4f configuration.
• Mixing up f–f transitions (lanthanoids, weak and sharp, since 4f orbitals are shielded by outer $5s, 5p$ electrons) with d–d transitions (d-block) when explaining coloured ions.
• Treating lanthanoid contraction as a single-element property rather than a cumulative one — its visible consequences, such as Zr/Hf radius convergence, appear only once the whole 14-element series has been crossed.

Ask the tutor: Want to see how effective nuclear charge is estimated with Slater's rules, or work through more Ln$^{3+}$ colour and complex examples?