Chemistry · The d- and f-Block Elements

The Lanthanoids & Lanthanoid Contraction

The lanthanoids are the fourteen elements from cerium to lutetium in which the 4f subshell is progressively filled. NCERT §4.5 treats them as a single, almost monotonous family bound together by one stable +3 state and one defining trend — the lanthanoid contraction. For NEET, this subtopic is a reliable scorer: questions on the Zr/Hf radius coincidence, the Ce⁴⁺/Eu²⁺ exceptions and the f-configurations recur almost every year.

Who are the lanthanoids

The f-block consists of two series — the lanthanoids (the fourteen elements following lanthanum) and the actinoids (the fourteen elements following actinium). Because lanthanum (Z = 57) so closely resembles the elements that follow it, it is conventionally discussed alongside them, and the general symbol Ln is used for the whole family. The 4f orbitals are filled across cerium (Z = 58) to lutetium (Z = 71).

What sets the lanthanoids apart from the ordinary transition metals is their uniformity. They resemble one another far more closely than the members of any single 3d, 4d or 5d series do, because the differentiating electron enters the deeply buried 4f subshell rather than the valence shell. Their chemistry therefore offers a clean window onto the effect of small, steady changes in size and nuclear charge across a row of otherwise near-identical elements.

FeatureWhat NCERT §4.5 says
MembersCe (58) to Lu (71); La (57) included by convention
Differentiating orbital4f (buried inside the [Xe] core)
General symbolLn
Stable oxidation state+3 for all (Ln³⁺ = 4fn)
Mutual resemblanceGreater than within any d-series

Electronic configurations

Every lanthanoid atom carries the common $\ce{6s^2}$ outer pair, with variable occupancy of the 4f level beneath it. The general atomic configuration is [Xe] 4f1–14 5d0–1 6s2. Two members, lanthanum and lutetium (and cerium and gadolinium), keep a single 5d electron, while the rest place all their extra electrons in 4f. The 5d1 appears at La (4f0), Ce, Gd (because 4f7 is a stable half-filled set) and Lu (4f14).

The crucial NEET point is the configuration of the tripositive ion. Once a lanthanoid loses three electrons to give Ln³⁺ — the dominant species — the configuration collapses to the simple regular form $\ce{4f^n}$, with n running from 1 (Ce³⁺) to 14 (Lu³⁺). This is why the Ln³⁺ radii decrease so smoothly, while neutral-atom radii are slightly irregular.

ElementZAtom config (outside [Xe])Ln³⁺ config
Lanthanum (La)575d¹ 6s²4f⁰
Cerium (Ce)584f¹ 5d¹ 6s²4f¹
Europium (Eu)634f⁷ 6s²4f⁶
Gadolinium (Gd)644f⁷ 5d¹ 6s²4f⁷
Terbium (Tb)654f⁹ 6s²4f⁸
Ytterbium (Yb)704f¹⁴ 6s²4f¹³
Lutetium (Lu)714f¹⁴ 5d¹ 6s²4f¹⁴
NEET Trap

Eu, Gd and Tb configurations

NEET 2016 asked the exact configurations of Eu, Gd and Tb. The half-filled-stability rule decides them: Eu stops at $\ce{4f^7 6s^2}$ (no 5d), Gd promotes one electron to give $\ce{4f^7 5d^1 6s^2}$ (keeping 4f⁷), and Tb is $\ce{4f^9 6s^2}$. Aspirants who blindly fill 4f without applying the half-filled rule choose the wrong option.

Half-filled (4f⁷) and filled (4f¹⁴) sets earn a 5d¹ at Gd and Lu — never memorise the row without this rule.

Oxidation states & the +2/+4 exceptions

In the lanthanoids, La(III) and Ln(III) compounds are the predominant species. The +3 state is so dominant because the energy needed to remove the third electron is repaid by lattice or hydration energy, and the resulting 4fn ions are well shielded. Occasionally, however, +2 and +4 ions appear in solution or in solid compounds. This irregularity — mirrored in the third ionisation enthalpies — arises mainly from the extra stability of empty, half-filled or completely-filled f subshells.

Cerium provides the classic +4 case: $\ce{Ce^4+}$ has an empty $\ce{4f^0}$ noble-gas configuration, so its formation is favoured. But Ce(IV) is a strong oxidant that readily reverts to Ce(III); the standard potential $E^\circ$ for $\ce{Ce^4+/Ce^3+}$ is +1.74 V, large enough that it could in principle oxidise water — though the reaction is so slow that Ce(IV) survives as a useful analytical reagent. Pr, Nd, Tb and Dy also show +4, but only in their oxides $\ce{MO2}$.

On the reducing side, $\ce{Eu^2+}$ forms by losing only the two 6s electrons, leaving the half-filled $\ce{4f^7}$ set; it is a strong reducing agent that changes back to the common +3 state. Similarly $\ce{Yb^2+}$ ($\ce{4f^14}$) is a reductant, and samarium behaves much like europium, showing both +2 and +3. Terbium, with a half-filled $\ce{4f^7}$ in its +4 state, is an oxidant.

IonConfigDriving stabilityRedox role
$\ce{Ce^4+}$4f⁰Empty f (noble gas)Strong oxidant → Ce³⁺
$\ce{Tb^4+}$4f⁷Half-filled fOxidant
$\ce{Eu^2+}$4f⁷Half-filled fStrong reductant → Eu³⁺
$\ce{Yb^2+}$4f¹⁴Filled fReductant

The same exchange-energy stability explains an ionisation-enthalpy quirk that NEET loves: lanthanum, gadolinium and lutetium show abnormally low third ionisation enthalpies, because removing the third electron from each leaves a stable 4f0, 4f7 or 4f14 core respectively. Gadolinium's case (NEET 2022) is purely an exchange-enthalpy effect, not a size or electronegativity effect.

The lanthanoid contraction

The single most examined idea in this subtopic is the lanthanoid contraction: the overall, steady decrease in atomic and ionic radii on moving from lanthanum to lutetium with increasing atomic number. The decrease in the Ln³⁺ ionic radii is remarkably regular; the metallic-atom radii decrease too, but slightly less smoothly because of variable 5d/4f occupancies.

Ionic radius of Ln³⁺ decreasing across the 4f series Ln³⁺ ionic radius / pm Atomic number (La → Lu) 117 85 La³⁺ Lu³⁺ radius decreases steadily →
Figure 1. The Ln³⁺ ionic radius falls smoothly from La³⁺ to Lu³⁺ as the nuclear charge rises and each added 4f electron shields it poorly (schematic, NCERT Fig. 4.6).

Why the contraction happens

The cause is the same one that drives the ordinary contraction across a transition series — the imperfect shielding of one electron by another in the same subshell — but it is more pronounced here. As we cross the series, each successive electron is added to the 4f subshell while the nuclear charge rises by one unit at every step. The 4f orbitals are diffuse and poorly directed, so one 4f electron shields another even less effectively than one d electron shields another.

The consequence is that the effective nuclear charge experienced by the outer 5s, 5p and 6s electrons creeps up steadily across the row. The whole electron cloud is pulled inward, and the size of the entire 4fn shell shrinks in a fairly regular fashion with increasing atomic number. Because the change at each step is small, the cumulative effect over fourteen elements is what matters — and it is exactly enough to undo the size increase that would otherwise be expected on going from one principal shell to the next.

It is worth being precise about the comparison NCERT draws. The contraction across the lanthanoids is "of the same kind" as the ordinary contraction seen across a normal transition series — both stem from imperfect intra-subshell shielding — but the 4f orbitals are far poorer screeners than d orbitals because of their shape and their deep penetration toward the nucleus. That is why the contraction in the 4f series, though gentle at each step, accumulates into a size reduction large enough to leave a lasting imprint on the elements that follow. The trend in the regular Ln³⁺ ions is therefore smooth, whereas the metallic (neutral-atom) radii show small humps at Eu and Yb, where the half-filled and filled 4f sets allow only the two 6s electrons into the metallic bonding, slightly enlarging those atoms.

Build the foundation

The same poor-shielding logic underlies size, density and ionisation trends across the d-series. Revisit general properties of the transition elements to see where the contraction fits.

Consequences of the contraction

The lanthanoid contraction has consequences that reach well beyond the 4f block itself. The 4f orbitals are filled before the 5d transition series begins, so the contraction is "stored up" and then delivered to the third transition series. This is the most important examinable consequence.

1. Near-identical radii of the 4d and 5d series

On descending a transition group, the radius normally increases from the 3d to the 4d member. The expected further increase from 4d to 5d is cancelled by the intervening lanthanoid contraction. The result is that second-row and third-row transition metals have almost identical sizes — for example Zr (160 pm) and Hf (159 pm), and likewise the pairs Nb/Ta and Mo/W.

Atomic radii of 3d, 4d and 5d series, with 4d and 5d nearly overlapping Atomic radius / pm across a transition group → 3d 4d 5d 4d ≈ 5d (lanthanoid contraction cancels expected rise) Zr 160 / Hf 159 pm
Figure 2. Atomic radius rises from the 3d to the 4d series, but the 5d series tracks the 4d series almost exactly because the lanthanoid contraction offsets the expected increase (schematic, NCERT Fig. 4.3).

2. Similar physical and chemical properties of 4d/5d pairs

Because their radii match so closely, members of the second and third transition series in the same group show much greater similarity in physical and chemical behaviour than ordinary family relationships would predict. This is why Zr and Hf occur together in nature and are notoriously difficult to separate — their compounds behave almost identically.

3. Difficulty of separating the lanthanoids themselves

Within the 4f series, the radii change so gradually that adjacent Ln³⁺ ions differ only minutely in size. Their salts have nearly identical solubilities and complexation behaviour, which makes separation of individual lanthanoids by classical fractional methods extremely tedious — modern separation relies on ion-exchange and solvent-extraction techniques exploiting these tiny radius differences.

4. Decreasing basicity of the hydroxides

As the ionic radius of Ln³⁺ shrinks across the series, the charge density on the cation rises, the $\ce{Ln-OH}$ bond becomes stronger and more covalent, and the hydroxide releases $\ce{OH-}$ less readily. Hence basic strength falls steadily: $\ce{La(OH)3}$ is the most basic and $\ce{Lu(OH)3}$ the least basic of the series.

ConsequenceEffect
4d vs 5d radiiNearly equal: Zr ≈ Hf, Nb ≈ Ta, Mo ≈ W
Properties of pairs5d metals resemble 4d congeners closely
SeparationZr/Hf and adjacent lanthanoids hard to separate
Hydroxide basicityDecreases La(OH)₃ → Lu(OH)₃
Covalent characterIncreases toward the heavier (smaller) ions
NEET Trap

Don't confuse cause with consequence

The cause of the lanthanoid contraction is poor 4f shielding of an increasing nuclear charge. The consequence is the Zr ≈ Hf radius coincidence. NEET 2021 asked why Zr and Hf have similar radii — the answer is "lanthanoid contraction," not "diagonal relationship" or "same group." Read the stem carefully to know whether it wants the reason for the contraction or its downstream effect.

Cause = poor 4f shielding · Consequence = identical 4d/5d radii (Zr ≈ Hf).

General characteristics

All lanthanoids are silvery-white soft metals that tarnish rapidly in air; hardness rises with atomic number, samarium being steel-hard. Melting points lie between 1000 and 1200 K (samarium is an outlier at 1623 K), and they are good conductors of heat and electricity. The decrease in metallic radius combined with rising atomic mass gives a general increase in density across the series, while properties change smoothly except for Eu and Yb (and occasionally Sm and Tm), whose anomalies trace back to the half-filled and filled f-sets.

Many trivalent ions are coloured in both solid and solution, the colour arising from f–f transitions; the absorption bands are narrow. The exceptions are the $\ce{4f^0}$ ions ($\ce{La^3+}$, $\ce{Ce^4+}$) and the $\ce{4f^14}$ ions ($\ce{Yb^2+}$, $\ce{Lu^3+}$), which are colourless and diamagnetic. All Ln ions other than these f⁰ and f¹⁴ types are paramagnetic. Chemically the earlier members behave like calcium and the later ones more like aluminium; the lanthanoids burn in oxygen, react with water to give $\ce{Ln(OH)3}$ and hydrogen, and combine with halogens to give $\ce{LnX3}$.

Worked Example

Which lanthanoid ions are diamagnetic, and why?

Diamagnetism requires no unpaired electrons. Among lanthanoid ions this is satisfied only by the empty-f and full-f configurations: $\ce{Ce^4+}$ and $\ce{La^3+}$ are $\ce{4f^0}$, while $\ce{Yb^2+}$ and $\ce{Lu^3+}$ are $\ce{4f^14}$ — all with zero unpaired electrons. Hence the pair $\ce{Ce^4+}$ and $\ce{Yb^2+}$ is diamagnetic (NEET 2024). Ions such as $\ce{Eu^2+}$ ($\ce{4f^7}$, seven unpaired) or $\ce{Gd^3+}$ ($\ce{4f^7}$) are strongly paramagnetic.

Quick Recap

Lanthanoids & the contraction in one minute

  • Lanthanoids = Ce–Lu (4f series); La included by convention; symbol Ln; differentiating electron enters 4f.
  • General atom config [Xe] 4f1–14 5d0–1 6s²; all Ln³⁺ are simply $\ce{4f^n}$ — the dominant +3 state.
  • Exceptions from f-stability: $\ce{Ce^4+}$ (4f⁰, oxidant), $\ce{Eu^2+}$ (4f⁷, reductant), $\ce{Yb^2+}$ (4f¹⁴), $\ce{Tb^4+}$ (4f⁷).
  • Lanthanoid contraction = steady fall in atomic/ionic radii La → Lu, caused by poor 4f shielding of a rising nuclear charge.
  • Consequences: Zr ≈ Hf (and Nb/Ta, Mo/W) radii equal; 4d/5d pairs hard to separate; lanthanoids hard to separate; basicity of Ln(OH)₃ falls La → Lu.
  • Coloured/paramagnetic except f⁰ (La³⁺, Ce⁴⁺) and f¹⁴ (Yb²⁺, Lu³⁺), which are colourless and diamagnetic.

NEET PYQ Snapshot — The Lanthanoids & Lanthanoid Contraction

Real NEET previous-year questions drawn only from this subtopic — contraction, configurations and oxidation states.

NEET 2021

Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of:

  • (1) Having similar chemical properties
  • (2) Belonging to same group
  • (3) Diagonal relationship
  • (4) Lanthanoid contraction
Answer: (4) Lanthanoid contraction

The cumulative contraction across the 4f series makes third-series radii match the second series; the almost identical radii of Zr (160 pm) and Hf (159 pm) are a direct consequence.

NEET 2024

The pair of lanthanoid ions which are diamagnetic is:

  • (1) $\ce{Ce^4+}$ and $\ce{Yb^2+}$
  • (2) $\ce{Ce^3+}$ and $\ce{Eu^2+}$
  • (3) $\ce{Gd^3+}$ and $\ce{Eu^3+}$
  • (4) $\ce{Pm^3+}$ and $\ce{Sm^3+}$
Answer: (1) Ce⁴⁺ and Yb²⁺

$\ce{Ce^4+}$ is 4f⁰ and $\ce{Yb^2+}$ is 4f¹⁴ — both have zero unpaired electrons, so the pair is diamagnetic. The other ions all have unpaired 4f electrons and are paramagnetic.

NEET 2022

Gadolinium has a low value of third ionisation enthalpy because of:

  • (1) high exchange enthalpy
  • (2) high electronegativity
  • (3) high basic character
  • (4) small size
Answer: (1) high exchange enthalpy

Gd is [Xe] 4f⁷ 5d¹ 6s². The third electron is removed from 5d, leaving the stable half-filled [Xe] 4f⁷ core. Its high exchange energy makes the third electron easy to remove, lowering the third ionisation enthalpy.

FAQs — The Lanthanoids & Lanthanoid Contraction

Concept checks NEET aspirants ask most about the 4f series and the contraction.

What is the lanthanoid contraction?

The lanthanoid contraction is the steady, overall decrease in atomic and ionic radii of the lanthanoids from lanthanum to lutetium as the atomic number increases. It arises because the 4f electrons being progressively added shield the outer electrons from the rising nuclear charge very poorly, so the effective nuclear charge felt by the outer shell increases steadily and the electron cloud is pulled inward.

Why do Zr and Hf have almost identical radii?

The 4f orbitals are filled in the lanthanoids, which sit between the second (4d) and third (5d) transition series. The cumulative lanthanoid contraction across these fourteen elements cancels the expected size increase on going from the 4d to the 5d series. As a result Zr (160 pm) and Hf (159 pm) have almost identical radii, which is why they occur together in nature and are very difficult to separate.

What is the most common oxidation state of the lanthanoids and why?

The +3 state is the most stable and common oxidation state for all lanthanoids, giving Ln3+ ions of the form 4f^n. It corresponds to the loss of the two 6s electrons and one 4f (or 5d) electron, and the resulting trivalent ions account for the close resemblance of the lanthanoids to one another.

Why does cerium show a +4 state and europium a +2 state?

These exceptions are driven by the extra stability of empty, half-filled or fully-filled f subshells. Ce4+ has a noble-gas (4f^0) configuration, so it forms readily — though it is a strong oxidant that reverts to Ce3+. Eu2+ has a half-filled 4f^7 configuration, so europium can keep its two 6s electrons; Eu2+ is a strong reducing agent that changes to the common +3 state.

How does the lanthanoid contraction affect the basicity of Ln(OH)3?

As the ionic radius of Ln3+ decreases across the series, the charge density and the strength of the Ln–OH bond increase, so the hydroxides become less ionic and less basic. Thus La(OH)3 is the most basic and Lu(OH)3 the least basic, with basic strength decreasing steadily from La to Lu.