The Lanthanoids

Who they are. The lanthanoids are the 14 elements from cerium (Z = 58) to lutetium (Z = 71), coming right after lanthanum (Z = 57, [Xe]$5d^1 6s^2$). Across the series electrons fill the 4f subshell while the outer $5d^{0-1}6s^2$ arrangement barely changes, which is exactly why their chemistry stays so uniform from element to element. The general configuration is written as:

where: [Xe] is the xenon core (Z = 54), the $4f$ exponent ranges from 1 (Ce) to 14 (Lu), and $5d$ holds 0 or 1 electron depending on the element (La is conventionally grouped with the lanthanoids for chemistry despite having zero 4f electrons).

Oxidation states. The dominant state throughout is +3: removing $6s^2$ plus either $5d^1$ or one 4f electron leaves an ion with an empty, half-filled, or fully-filled 4f subshell, and those configurations carry extra stability. A handful of lanthanoids also show +2 or +4 when that reaches an $f^0$, $f^7$, or $f^{14}$ configuration — Ce$^{4+}$ ($f^0$) and Eu$^{2+}$ ($f^7$) are the two classic examples — but +3 stays dominant for the series as a whole.

General reactivity. Lanthanoids behave like reactive, electropositive metals, broadly similar to the alkaline earths. They tarnish in air, react with cold water slowly (faster hot) to liberate H$_2$, burn readily in halogens, and combine with H$_2$, carbon, sulfur, nitrogen and phosphorus on heating.

Lanthanoid contraction itself. This is the steady, cumulative fall in atomic and ionic radii across the series from La to Lu. The mechanism is shielding: 4f orbitals are diffuse and shield outer electrons from the nucleus poorly, so each new 4f electron only partly cancels the extra proton added alongside it. Effective nuclear charge on the outer shell rises a little at every step, and the electron cloud is pulled in a little at every step. No single contraction is dramatic, but fourteen of them stack up into a real, examinable shrink.

Three consequences of the contraction:

• Ionic radii trend — Ln$^{3+}$ radius decreases smoothly and continuously from La$^{3+}$ down to Lu$^{3+}$, with no reversal partway through.
• Basicity trend — a smaller, more polarising Ln$^{3+}$ holds OH$^-$ more tightly and covalently, so Ln(OH)$_3$ basic strength falls from La(OH)$_3$ (most basic) to Lu(OH)$_3$ (least basic).
• 4d/5d twinning — the accumulated contraction across the 14 lanthanoids almost exactly cancels the normal radius increase expected on descending a group from period 5 to period 6, so pairs such as Zr/Hf, Nb/Ta and Mo/W end up nearly the same size and are notoriously hard to separate chemically. Separating the lanthanoids from one another is difficult for the same reason: their sizes change only gradually across the series.

Lanthanoid contraction has no separate "formula" to derive — it falls straight out of comparing shielding by inner versus outer electrons, and the reasoning is worth walking through carefully because exams reward the mechanism, not just the trend.

Step 1. Start from the shielding rule. Electrons in inner, compact orbitals shield outer electrons from nuclear charge efficiently; electrons in diffuse, spread-out orbitals shield poorly. Among orbitals of similar principal quantum number, the ordering of "how well an electron shields others" roughly follows how close it stays to the nucleus on average — s and p electrons shield well, d electrons shield less well, and f electrons — the most diffuse and penetrating of all at that shell — shield worst.

Step 2. Apply it to the lanthanoid series. Moving from Ce to Lu, each successive proton added to the nucleus is accompanied by one more electron entering the 4f subshell. Because 4f shields so poorly, that new electron cancels only a fraction of the extra nuclear charge. The net pull felt by the outer $5d/6s$ electrons — the effective nuclear charge — therefore climbs steadily across all 14 elements.

Step 3. Translate rising effective nuclear charge into a shrinking radius. A larger net pull draws the outer electron cloud in closer to the nucleus. Since this happens at every single step from La to Lu, the radius does not jump — it contracts smoothly and continuously, which is exactly the shape seen in the figure above.

Step 4. Read off the numbers. Ionic radius falls from about $r(\text{La}^{3+}) \approx 106\ \text{pm}$ to $r(\text{Lu}^{3+}) \approx 85\ \text{pm}$:

where: $\Delta r$ is the total contraction accumulated across the series, and the fall happens despite the nuclear charge rising by 14 units — proof that shielding, not charge alone, is what governs ionic size here.

Step 5. Extend the logic to 4d/5d twinning. Descending any group normally increases radius, because a whole new principal shell is added. But Hf (period 6, 5d) sits immediately after the lanthanoid series, so the 21 pm of accumulated contraction is subtracted from the radius increase Hf would otherwise have shown over Zr (period 5, 4d). The two effects very nearly cancel, leaving Zr$^{4+}$ and Hf$^{4+}$ almost the same size — the reasoning chain closes exactly where the Pass-1 worked example leaves it.

Explain why Ce (Z = 58) commonly shows a +4 oxidation state in addition to the usual +3, using electronic configuration, and then use the same reasoning to compare Zr$^{4+}$ and Hf$^{4+}$ radii.

Step 1. Cerium (Z = 58) has configuration [Xe]$4f^1 5d^1 6s^2$.

Step 2. Ce$^{3+}$, formed by removing $6s^2$ and $5d^1$, is [Xe]$4f^1$ — one electron remains in the 4f subshell.

Step 3. Ce$^{4+}$, formed by removing that remaining 4f electron too, is [Xe]$4f^0$ — the stable xenon core with an empty 4f subshell. An empty ($f^0$), half-filled ($f^7$), or fully-filled ($f^{14}$) subshell carries extra stability, so cerium readily gives up that fourth electron, making +4 accessible alongside the dominant +3. This is why ceric ammonium sulfate/nitrate is used as a mild oxidising agent — Ce$^{4+}$ is readily reduced back to the more stable Ce$^{3+}$.

Step 4 (stretch). Given $r(\text{La}^{3+}) \approx 106\ \text{pm}$ and $r(\text{Lu}^{3+}) \approx 85\ \text{pm}$, the accumulated contraction of $\Delta r \approx 21\ \text{pm}$ across the 14 lanthanoids offsets the normal period-5-to-period-6 radius increase for Group 4. So Zr$^{4+}$ and Hf$^{4+}$, though one shell apart, land at nearly identical radii of about 79–80 pm rather than Hf$^{4+}$ being clearly larger.

Answer: Ce shows +4 because Ce$^{4+}$ reaches the extra-stable $f^0$ configuration, and this same shielding-driven contraction is why Zr$^{4+}$ and Hf$^{4+}$ end up almost the same size, making Zr and Hf chemically hard to separate.

• Thinking lanthanoid contraction means radii increase across the series — it is a steady decrease from La to Lu, not a rise.
• Assuming every lanthanoid shows only +3 — Ce$^{4+}$ ($f^0$) and Eu$^{2+}$ ($f^7$) are genuine exceptions, but +3 remains dominant overall.
• Blaming the wrong electrons for the shielding — it is poor shielding by inner 4f electrons, not by 5d or 6s electrons; mixing this up wrecks the basicity and radius explanations.
• Quoting the 4d/5d twinning (Zr–Hf, Nb–Ta, Mo–W) as the definition of lanthanoid contraction when asked to define it — it is a consequence, and the definition itself is about the radius decrease across the lanthanoid series.

Ask the tutor: Want to work through why Eu shows +2 the same way Ce shows +4, or practice ranking a set of Ln(OH)$_3$ compounds by basicity?