Actinoids

Where they sit and what fills. Actinoids follow actinium (Ac, $Z=89$) and occupy the second f-block row, with the general configuration

where: $[\text{Rn}]$ is the radon core, the superscript ranges on $5f$ and $6d$ show that electrons are distributed between these two close-lying subshells depending on the element (an irregular filling pattern, just as lanthanoids show occasional 5d dips), and $7s^2$ is the fixed outermost pair removed first on ionisation. Every actinoid is radioactive; only thorium and uranium occur in nature in workable quantities, with protactinium and actinium present only as trace decay products, and everything from neptunium onward is made artificially by nuclear reactions.

Oxidation states — the syllabus's central comparison. Lanthanoids show $+3$ almost exclusively, with $+2$/$+4$ as rare exceptions, because the 4f orbitals are buried under filled $5s^25p^6$ shells and barely participate once the outer $6s$ (and occasionally one more) electron is stripped. Actinoids instead spread across $+3$ to $+7$: uranium alone shows $+3, +4, +5, +6$, and neptunium, plutonium and americium can reach $+7$. The reason is orbital spacing, not orbital identity — 5f orbitals are more diffuse (spatially larger) than 4f orbitals, so 5f, 6d and 7s sit closer together in energy than the well-separated 4f/5d/6s stack in lanthanoids. Electrons from both 5f and 6d become chemically available, not just the outermost $7s$ pair, so removing progressively more of them stays energetically feasible across a wider range.

Radii, complexation and colour track the same shielding logic. Just as poor shielding by 4f electrons drives the steady lanthanoid contraction, poor shielding by 5f electrons drives an analogous, generally somewhat larger and more irregular actinoid contraction across Th to Lr. Because 5f and 6d orbitals extend further outward than 4f, they overlap better with ligand orbitals, so actinoids form complexes more readily than lanthanoids. Basicity of hydroxides falls steadily across both series as the metal ion shrinks and its charge density rises. Both series give coloured ions from f–f transitions, but actinoid ions are often more intensely coloured, another consequence of the same closer orbital spacing that widens their oxidation-state range.

Actinoid chemistry is purely descriptive at this syllabus level, so the reasoning below explains why the wide oxidation-state range appears rather than deriving a numeric formula, and it is worth walking through carefully because it is the single fact examiners test most on this topic.

Step 1. Compare the two f-block rows at the orbital-energy level. In lanthanoids, the 4f orbitals lie well below 5d and 6s and are shielded from the chemical environment by the filled $5s^25p^6$ shells beneath them. Once the $6s$ electrons (and, for $+3$, one more) are removed, pulling a further electron out of the buried 4f set costs far more energy than the bonding gained back — so ionisation stops at $+3$ for almost every lanthanoid.

Step 2. In actinoids, the 5f orbitals are more diffuse (spatially larger, less core-shielded) than 4f orbitals, and they sit close in energy to both 6d and 7s rather than well below them. This closeness means electrons from 5f, 6d, and 7s are all in similar energetic reach of the bonding electrons around them.

Step 3. Because three subshells' worth of electrons are chemically accessible rather than just one, an actinoid atom can lose progressively more electrons — three, four, five, even six or seven — while each successive ionisation still pays for itself through bond formation, unlike in lanthanoids where the 4f wall stops the process at $+3$.

Step 4. The same poor shielding by the diffuse 5f electrons that opens up this wider ionisation range also causes the steady radius decrease across the series — the actinoid contraction — for the identical reason poor 4f shielding causes lanthanoid contraction. Wider oxidation-state range and stronger radii contraction are two faces of the same orbital story, not two separate facts to memorise.

Problem. Uranium shows oxidation states $+3, +4, +5, +6$, while its rough f-block counterpart, the lanthanoid europium, shows almost exclusively $+3$ (with $+2$ as a minor exception). Explain the difference in terms of orbital energies, and name the uranium species corresponding to its most stable and its highest common oxidation state.

Step 1. In europium, the $4f$ orbitals sit well below $5d/6s$ and are shielded by the filled $5s^25p^6$ core, so once the outer $6s^2$ pair is removed to reach $+3$ (occasionally stopping at $+2$), further ionisation from the buried 4f set is too costly to occur under normal chemical conditions.

Step 2. In uranium, the $5f$, $6d$ and $7s$ orbitals are far closer together in energy and less shielded than europium's $4f/5d/6s$ set, because 5f orbitals are more diffuse than 4f orbitals.

Step 3. Electrons from both the 5f and 6d subshells of uranium are therefore chemically accessible for bonding, not just the outermost $7s$ pair, so removing progressively more electrons from these closely spaced orbitals stays feasible.

Step 4. This gives uranium its observed range: $+3$ as $\text{U}^{3+}$, $+4$ as $\text{UO}_2$ (its most thermodynamically stable oxide), $+5$, and $+6$ as the uranyl ion $\text{UO}_2^{2+}$, the most common form of uranium in aqueous and industrial chemistry. Answer: the gap between europium's buried 4f set and uranium's closely spaced 5f/6d/7s set explains the difference; uranium's most stable form is $\text{UO}_2$ ($+4$) and its highest common form is $\text{UO}_2^{2+}$ ($+6$).

Competitive stretch

A question may ask you to state two similarities and one difference between the series in a single answer. Similarities: (1) both show a steady radii contraction across the row from poor f-electron shielding — lanthanoid contraction and actinoid contraction are the same mechanism applied to 4f and 5f respectively; (2) both favour $+3$ as an important state and give coloured ions through f–f transitions. Difference: all actinoids are radioactive, many (from neptunium onward) entirely synthetic, while lanthanoids are stable and naturally occurring — a distinction with real consequences, such as uranium and plutonium's role in nuclear fission, that has no lanthanoid counterpart.

• Assuming actinoids show only $+3$ like the lanthanoid "default" — the syllabus specifically wants the wider, more variable range acknowledged.
• Forgetting that most actinoids are synthetic — only Th and U occur in nature in workable amounts, with Pa and Ac present merely as trace decay products.
• Mixing up the cause of lanthanoid versus actinoid contraction — both come from poor f-electron shielding, but citing "different causes" loses marks; the difference is only that 5f is more diffuse than 4f.
• Treating the f-block as one uniform family instead of stating the specific three-way comparison examiners want: radii trend similar, oxidation-state range different, natural occurrence different.

Ask the tutor: want a quick side-by-side quiz drilling uranium's four oxidation states against their common compounds?