CSIR-NET Lanthanides and Actinides — Contraction, States and Magnetism

By Aniket Bhardwaj · 15 September 2026 · CSIR-NET Chemistry

The f-block is where a lot of otherwise well-prepared candidates lose easy marks. Everything learned about the d-block — crystal field splitting, spin-only magnetic moments, broad d–d bands — quietly stops working, and if you apply it out of habit you will get the wrong answer with complete confidence. This article sets out what actually changes, why it changes, and the arithmetic you need for the magnetic-moment questions that appear most often.

Configurations: why the 4f electrons behave differently

The lanthanides run from La (Z = 57) to Lu (Z = 71) and fill the 4f shell. The general configuration is [Xe] 4fⁿ 5d^0–1 6s², with the familiar half-filled and filled-shell preferences at Gd ([Xe] 4f⁷ 5d¹ 6s²) and Lu ([Xe] 4f¹⁴ 5d¹ 6s²). Ce, being at the start, also takes a 5d electron.

The chemically decisive fact is that 4f orbitals are buried. They lie inside the filled 5s and 5p shells, so they barely overlap with ligand orbitals. Three consequences follow, and almost every f-block question is one of them in disguise:

Ligand-field splitting is tiny — of the order of a few hundred cm⁻¹, against tens of thousands of cm⁻¹ for a transition metal. Bonding is essentially ionic and electrostatic, and coordination numbers are high (8, 9, 10) because they are set by size, not by orbital directionality.
f–f absorption bands are sharp and weak, and their positions hardly shift when the ligand changes — unlike the broad, ligand-sensitive d–d bands of the 3d series.
Spin–orbit coupling is large compared with the ligand field, so J — not just S — is the good quantum number, and the spin-only formula fails.

Actinides fill 5f. Those orbitals are more radially extended than 4f, and the 5f, 6d and 7s levels lie close together. The early actinides therefore behave far more like transition metals, with covalency, a wide range of oxidation states and strongly ligand-dependent chemistry.

The lanthanide contraction

Across the series the ionic radius of Ln³⁺ falls steadily. Using Shannon radii for six-coordination, La³⁺ is about 1.03 Å and Lu³⁺ about 0.86 Å — a contraction of roughly 0.17 Å over fourteen elements. The reason is poor shielding: each added 4f electron screens the growing nuclear charge inefficiently, so the effective nuclear charge felt by the outer electrons rises steadily and the ion shrinks.

Consequence	What it looks like in a question
Zr and Hf are nearly identical in size	Zr⁴⁺ ≈ 0.72 Å and Hf⁴⁺ ≈ 0.71 Å (CN 6), so the two are chemically almost inseparable and occur together in nature
Basicity of Ln(OH)₃ falls La → Lu	Higher charge density on the smaller ion polarises O–H more, so the hydroxide becomes less basic
Stability of Ln³⁺ complexes rises La → Lu	Smaller, more charge-dense ion binds a hard donor more strongly — the basis of ion-exchange separation
Second- and third-row d-block pairs behave alike	Nb/Ta, Mo/W: the third-row element is squeezed back to the size of the second-row one

There is an analogous actinide contraction across the 5f series, from the same shielding argument.

Oxidation states — the +3 rule and its exceptions

The lanthanides are dominated by +3, which is unusual and worth stating plainly: the third ionisation energy is repaid by lattice or hydration energy for all of them. The exceptions are entirely explained by proximity to f⁰, f⁷ and f¹⁴:

+4: Ce⁴⁺ (reaches f⁰) — the basis of cerimetric titrations; also Pr⁴⁺ and Tb⁴⁺ (Tb reaches f⁷).
+2: Eu²⁺ (reaches f⁷) and Yb²⁺ (reaches f¹⁴); Sm²⁺ also exists and is a powerful reductant.

The actinides are quite different. Early in the series the 5f electrons are available for bonding, so U shows +3, +4, +5 and +6, with the linear uranyl ion UO₂²⁺ dominating aqueous chemistry; Np and Pu show a similar spread. Beyond about americium the +3 state reasserts itself and the later actinides begin to resemble the lanthanides.

Magnetic moments — the calculation that decides marks

For the lanthanides, spin–orbit coupling is far larger than the ligand field, so the ground state is a single J level and the moment must be calculated from J, not from the number of unpaired electrons.

μ_eff = g_J √[J(J+1)] BM, where
g_J = 1 + [J(J+1) + S(S+1) − L(L+1)] / [2J(J+1)]

To get J: apply Hund's rules to the fⁿ configuration — maximise S, then maximise L, then take J = |L − S| if the shell is less than half full and J = L + S if it is more than half full.

Worked example 1 — Tb³⁺, f⁸.

Seven f orbitals take one electron each (all parallel), then the eighth pairs up.
Unpaired electrons = 6, so S = 6 × ½ = 3.
L = the m_l value of the doubly occupied orbital = 3.
More than half filled, so J = L + S = 3 + 3 = 6. Ground term ⁷F₆.

J(J+1) = 6 × 7 = 42; S(S+1) = 3 × 4 = 12; L(L+1) = 3 × 4 = 12.
g_J = 1 + (42 + 12 − 12) ÷ (2 × 42) = 1 + 42/84 = 1.50.
μ = 1.50 × √42 = 1.50 × 6.4807 = 9.72 BM.

Compare with spin-only: √[6(6+2)] = √48 = 6.93 BM — about 30% too low. Measured moments for Tb³⁺ compounds sit near the Van Vleck value, not the spin-only one. This contrast is exactly what the question is testing.

Worked example 2 — Ce³⁺, f¹.

S = ½, L = 3, less than half filled so J = |3 − ½| = 5/2. Ground term ²F_5/2.

J(J+1) = (5/2)(7/2) = 8.75; S(S+1) = (½)(3/2) = 0.75; L(L+1) = 12.
g_J = 1 + (8.75 + 0.75 − 12) ÷ (2 × 8.75) = 1 + (−2.5 ÷ 17.5) = 1 − 0.1429 = 0.857 (= 6/7).
μ = 0.857 × √8.75 = 0.857 × 2.9580 = 2.54 BM, in good agreement with measured values for Ce³⁺.

Worked example 3 — Eu³⁺, f⁶: the honest exception.

Six parallel electrons give S = 3; L = 3 + 2 + 1 + 0 + (−1) + (−2) = 3; less than half filled so J = |3 − 3| = 0. Ground term ⁷F₀.

The formula therefore predicts μ = 0. Measured moments for Eu³⁺ are around 3.4 BM instead. The reason is that the spin–orbit splitting to the ⁷F₁ level is comparable to kT at room temperature, so excited J levels are thermally populated and contribute. Sm³⁺ deviates for the same reason. Learn these two as the standard exceptions — a question that lists Eu³⁺ is usually asking whether you know why the simple formula fails.

Spectra and separation

f–f transitions are Laporte-forbidden and only become weakly allowed by vibronic coupling, so molar absorptivities are small and the colours of Ln³⁺ salts are pale. Because the 4f electrons are shielded, the bands are narrow — line-like rather than the broad envelopes seen for d–d transitions — and a few so-called hypersensitive transitions do respond noticeably to the coordination environment. Eu³⁺ and Tb³⁺ are strongly luminescent when sensitised by an organic ligand that absorbs light and transfers energy to the metal, which is the basis of lanthanide luminescent probes.

Because all Ln³⁺ ions are chemically so alike, separation depends entirely on the small, monotonic size difference: ion-exchange chromatography with a chelating eluent, and solvent extraction with organophosphorus extractants. The heaviest (smallest) ion forms the strongest complex and elutes first in the classic ion-exchange scheme.

Lanthanides versus actinides at a glance

Property	4f — lanthanides	5f — actinides
Orbital extension	Buried inside 5s, 5p	More extended; available for bonding early in the series
Common oxidation states	+3 dominant; +2/+4 near f⁰, f⁷, f¹⁴	+3 to +6 for the early members; +3 dominant later
Covalency	Very little; essentially ionic	Appreciable, especially Th–Am
Ligand-field splitting	Very small (~10² cm⁻¹)	Larger than 4f, still small versus 3d
Absorption bands	Sharp, weak, ligand-insensitive	Broader and more intense than 4f
Radioactivity	Essentially all stable (Pm is the exception)	All radioactive
Magnetism	Van Vleck g_J√[J(J+1)] works well	Poorer agreement; ligand field quenches part of the orbital contribution

Mistakes that cost marks

Using the spin-only formula for Ln³⁺. It is right for most 3d ions and wrong for nearly every 4f ion. Decide which block you are in before you write anything.
Getting J the wrong way round. J = |L − S| for a less-than-half-filled shell, J = L + S for more than half filled. Half-filled exactly (f⁷) gives L = 0, so J = S and both rules agree.
Predicting 0 BM for Eu³⁺ and stopping there. State the ⁷F₀ result and then note the thermal population of ⁷F₁.
Applying crystal field theory to the f-block. Δ is far too small to decide spin state; there is no high-spin/low-spin question for lanthanides.
Blaming the lanthanide contraction on relativistic effects alone. Poor 4f shielding is the main cause; relativistic contraction contributes but is not the whole story.
Assuming actinides copy lanthanides. Uranium's +6 uranyl chemistry has no lanthanide analogue at all.

Where this appears in the paper

Sub-topic	Typical question form
Configurations	Write the fⁿ count for a given Ln³⁺; identify the ground term
Contraction	Order a set of ions by radius; explain the Zr/Hf similarity
Oxidation states	Which lanthanide shows +2 or +4, and why
Magnetism	Compute μ_eff from g_J√[J(J+1)]; explain a deviation
Spectra	Why f–f bands are sharp and weak; luminescence and the antenna effect
Separation	Ion-exchange elution order; choice of extractant
Actinides	Uranyl geometry; contrast of 5f with 4f behaviour

Use that as a map of the sub-topics, not as a claim about how many marks each carries. The paper pattern is set by the official notification for the session you are sitting, and that is the only source worth trusting on it.

Get the fⁿ count right before you start. Every magnetic-moment question above begins with the correct electron configuration — one slip there and the whole answer is lost. The Electron Configuration tool writes out the full and noble-gas-core configuration for any element or ion, so you can confirm the fⁿ count in a second.

Open the Electron Configuration Tool →

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