ChemistryISC Class 12 · d- and f-Block Elements

Lanthanoids and Lanthanoid Contraction: ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Lanthanoids are the 14 elements after lanthanum in which 4f orbitals fill. Their common oxidation state is +3. Lanthanoid contraction is the steady fall in atomic and ionic radii from La to Lu, caused by poor shielding by 4f electrons. It makes Zr and Hf almost the same size and lowers basicity.

Notice each ion is smaller than the last: the nucleus grows but 4f electrons shield poorly, so the outer electrons are pulled in.

Illustration drawn for QuestionBuddy.

Understand Lanthanoids and Lanthanoid Contraction

Electronic configuration. The lanthanoids (cerium, Z = 58, to lutetium, Z = 71) follow lanthanum and are usually studied with it (symbol Ln). Their atoms have $6 s^{2}$ in common with a variable number of 4f electrons, and La, Ce, Gd and Lu also have one 5d electron (Gd: [Xe] $4 f^{7} 5 d^{1} 6 s^{2}$ , keeping a half-filled 4f set). All their tripositive ions have the regular form $4 f^{n}$ , with $n$ = 0 (La³⁺) to 14 (Lu³⁺).

Oxidation states. +3 is the characteristic and most stable state. Other states appear when they give an empty, half-filled or filled 4f set: Ce⁴⁺ ( $4 f^{0}$ ) is a strong oxidant ( $E^{\circ}$ Ce⁴⁺/Ce³⁺ = +1.74 V) but reacts with water so slowly that cerium(IV) is a useful analytical reagent; Tb⁴⁺ ( $4 f^{7}$ ) is an oxidant; Eu²⁺ ( $4 f^{7}$ ) and Yb²⁺ ( $4 f^{14}$ ) are strong reductants that revert to +3. Sm also shows +2. Pr, Nd, Tb and Dy show +4 only in oxides MO₂.

Lanthanoid contraction. From La to Lu the atomic and especially the ionic (Ln³⁺) radii decrease steadily (La³⁺ 106 pm to Yb³⁺ 86 pm). As each proton is added to the nucleus, the extra electron enters the 4f subshell. One 4f electron shields another very poorly (worse than d electrons do), because of the diffuse shape of f orbitals, so the effective nuclear charge felt by the outer electrons rises and pulls the whole atom in. The decrease in atomic radii is less regular than for the ions; europium (199 pm) and ytterbium are larger because they use only two electrons in metallic bonding.

Consequences. (i) Similar radii in the 4d and 5d series: the expected increase from the second to the third transition series is cancelled, so Zr (160 pm) and Hf (159 pm), Nb and Ta, Mo and W have almost identical sizes and properties; Zr and Hf occur together and are very hard to separate. (ii) Decrease in basicity: as Ln³⁺ gets smaller, the Ln–OH bond becomes more covalent, so La(OH)₃ is the most basic and Lu(OH)₃ the least. (iii) Difficulty in separating the lanthanoids: neighbours have nearly the same size and the same +3 state. (iv) Increased density and ionisation enthalpy of the third transition series elements.

Colour and magnetism. Many Ln³⁺ ions are coloured in solids and in solution because of f-f transitions; their absorption bands are narrow. La³⁺ ( $4 f^{0}$ ) and Lu³⁺ ( $4 f^{14}$ ) have no electron to excite within the f level and are colourless. All ions except the $f^{0}$ type (La³⁺, Ce⁴⁺) and $f^{14}$ type (Yb²⁺, Lu³⁺) have unpaired electrons and are paramagnetic.

Ionisation enthalpies and reactivity. First ionisation enthalpies are about 600 kJ mol⁻¹ and second about 1200 kJ mol⁻¹, similar to calcium. The third ionisation enthalpies of La, Gd and Lu are abnormally low because removing the third electron leaves an empty, half-filled or full 4f set. $E^{\circ}$ for Ln³⁺/Ln lies between −2.2 and −2.4 V (Eu −2.0 V), so the metals are reactive: early members resemble calcium, later ones aluminium. They are silvery-white soft metals that tarnish in air. Reactions: they burn in oxygen to give Ln₂O₃; with water they give Ln(OH)₃ and H₂; they liberate H₂ from dilute acids; they combine with hydrogen on gentle heating; with halogens they give LnX₃; heated with sulphur, Ln₂S₃; heated with nitrogen, LnN; and with carbon at 2773 K, carbides Ln₃C, Ln₂C₃ and LnC₂. The hydroxides are basic, like alkaline earth hydroxides.

Uses. Alloy steels; mischmetall (about 95% lanthanoid metal and 5% iron, with traces of S, C, Ca and Al), used in Mg-based alloys for bullets, shells and lighter flints; mixed oxides as catalysts in petroleum cracking; some oxides as phosphors in television screens.

Actinoids compared. Actinoids (Th to Lr) fill 5f orbitals. Because 5f, 6d and 7s levels are close in energy, they show a wider range of oxidation states (up to +7 for Np and Pu), though +3 is general. The actinoid contraction from element to element is greater than the lanthanoid contraction, because 5f electrons shield even more poorly. All actinoids are radioactive, and their chemistry is more complex.

Lanthanoids and Lanthanoid Contraction: formulas and key facts

Configuration of Ln³⁺ ions: $L n^{3 +} : [X e] 4 f^{n}, n = 0 (L a^{3 +}) to 14 (L u^{3 +})$; Regular for all tripositive ions.
Stable non-+3 states: $C e^{4 +} (4 f^{0}), T b^{4 +} (4 f^{7}) oxidants; E u^{2 +} (4 f^{7}), Y b^{2 +} (4 f^{14}) reductants$; Empty, half-filled or filled 4f subshell.
Radius trend: $L a^{3 +} (106 pm) > E u^{3 +} (95) > G d^{3 +} (94) > Y b^{3 +} (86)$; Steady fall: the lanthanoid contraction.
Basicity of hydroxides: $L a (O H)_{3} > \dots > Lu (O H)_{3}$; Smaller Ln³⁺ makes the Ln–OH bond more covalent.
Electrode potential: $E_{L n^{3 +} / L n}^{\circ} \approx - 2.2 to - 2.4 V (E u : - 2.0 V)$; Reactive metals: displace H₂ from water and dilute acids.

How to solve Lanthanoids and Lanthanoid Contraction questions

Lanthanoid questions in ISC are short reasoning items and assertion-reason pairs. Answer with a configuration and a cause-effect line.

1Write the configuration of the ion asked ([Xe] $4 f^{n}$ for Ln³⁺; count $n$ = Z − 57 for Ln³⁺).
2For unusual oxidation states, show that the ion reaches $4 f^{0}$ , $4 f^{7}$ or $4 f^{14}$ ; then say whether it is an oxidant (above +3) or reductant (below +3).
3For colour or magnetism, check for unpaired f electrons: $f^{0}$ and $f^{14}$ ions are colourless and diamagnetic.
4For lanthanoid contraction, give the cause in one line (poor shielding by 4f electrons, rising effective nuclear charge) and then the consequence asked.
5For Zr/Hf-type questions, name both facts: almost identical radii, hence similar properties, hence they occur together and are hard to separate.
6For lanthanoid versus actinoid questions, contrast oxidation states, contraction, radioactivity and complex formation.

Lanthanoids and Lanthanoid Contraction in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Cerium(IV) is a strong but slow-acting oxidant

Suppose an analyst titrates an iron(II) solution with a standard cerium(IV) sulphate solution instead of permanganate. 20.0 mL of the iron(II) solution needs 16.0 mL of 0.050 M Ce⁴⁺. The reaction is $C e^{4 +} + F e^{2 +} \to C e^{3 +} + F e^{3 +}$ .

Find the molarity of Fe²⁺, and explain why the cerium(IV) solution stays usable even though $E^{\circ}$ (Ce⁴⁺/Ce³⁺) = +1.74 V is high enough to oxidise water.

Try it first, then show the solutionSolution

Moles of Ce⁴⁺ $= 0.050 \times 0.0160 = 8.0 \times 1 0^{- 4}$ mol; the ratio is 1 : 1, so Fe²⁺ $= 8.0 \times 1 0^{- 4}$ mol.
Molarity of Fe²⁺ $= \frac{8.0 \times 1 0 ^{- 4}}{0.0200} = 0.040$ M.
Ce⁴⁺ ( $4 f^{0}$ ) is a strong oxidant that tends to return to the stable +3 state.
Its reaction with water is extremely slow, so the solution keeps its strength and is a good analytical reagent.

Answer: [Fe²⁺] = 0.040 M. Ce(IV) could oxidise water thermodynamically, but the reaction is very slow, so the reagent is stable in practice.

Memory hook: Ce⁴⁺ wants to go back to +3, but it is in no hurry with water.

Tricks to remember Lanthanoids and Lanthanoid Contraction

Short memory aids. Say each one aloud once, then use it on a question the same day.

0, 7, 14 are the lucky numbers
Any lanthanoid ion with $4 f^{0}$ , $4 f^{7}$ or $4 f^{14}$ is extra stable: Ce⁴⁺ (0), Eu²⁺, Gd³⁺, Tb⁴⁺ (7), Yb²⁺, Lu³⁺ (14).
Use it when: Explaining +2 and +4 states.
Above 3 grabs, below 3 gives
An ion above +3 (Ce⁴⁺, Tb⁴⁺) grabs electrons: oxidant. An ion below +3 (Eu²⁺, Yb²⁺, Sm²⁺) gives electrons: reductant.
Use it when: Oxidising and reducing nature.
Zirconium and hafnium are twins
Lanthanoid contraction squeezes Hf so much that it ends up the same size as Zr: 160 and 159 pm.
Use it when: The most-asked consequence.

Lanthanoids and Lanthanoid Contraction: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spot the oxidant · 1 mark · MCQBoard level

Which of the following lanthanoid ions is a strong oxidising agent?
(a) Eu²⁺
(b) Yb²⁺
(c) Ce⁴⁺
(d) La³⁺

Try it first, then show the solutionSolution

Ions above the common +3 state tend to gain an electron: oxidants.
Ce⁴⁺ ( $4 f^{0}$ ) changes readily to Ce³⁺; Eu²⁺ and Yb²⁺ are reductants; La³⁺ is stable.

Answer: (c) Ce⁴⁺.

Example 2: Configuration of gadolinium · 2 marks · Short answerBoard level

Write the electronic configurations of Gd (Z = 64) and Gd³⁺. Why does the Gd atom have an electron in the 5d orbital rather than an eighth 4f electron?

Try it first, then show the solutionSolution

Gd: [Xe] $4 f^{7} 5 d^{1} 6 s^{2}$ ; Gd³⁺: [Xe] $4 f^{7}$ (1 mark).
Placing the eighth electron in 5d keeps the 4f subshell exactly half-filled ( $4 f^{7}$ ), which is extra stable (1 mark).

Answer: Gd: [Xe] $4 f^{7} 5 d^{1} 6 s^{2}$ ; Gd³⁺: [Xe] $4 f^{7}$ . The half-filled 4f set is extra stable, so the eighth electron goes to 5d.

Example 3: Radii in order, and the europium exception · 3 marks · Short answerHarder board level

(a) Arrange La³⁺, Gd³⁺, Yb³⁺ and Eu³⁺ in decreasing order of ionic radius, giving the reason. (b) The atomic (metallic) radius of europium is unusually large compared with its neighbours. Suggest why.

Try it first, then show the solutionSolution

(a) La³⁺ > Eu³⁺ > Gd³⁺ > Yb³⁺ (106, 95, 94, 86 pm) (1 mark).
Across the series the nuclear charge rises while added 4f electrons shield each other poorly, so the ions contract steadily: the lanthanoid contraction (1 mark).
(b) Eu ( $4 f^{7} 6 s^{2}$ ) keeps its half-filled 4f set and uses only two 6s electrons in metallic bonding, so the bonding is weaker and the metal atom is larger (1 mark).

Answer: (a) La³⁺ > Eu³⁺ > Gd³⁺ > Yb³⁺, due to lanthanoid contraction. (b) Eu uses only two electrons in metallic bonding.

Example 4: General reactions of a lanthanoid · 3 marks · Short answerHarder board level

Write balanced equations, using Ln for a lanthanoid metal in the +3 state, for its reaction with (a) oxygen, (b) water and (c) sulphur on heating.

Try it first, then show the solutionSolution

(a) $4 L n + 3 O_{2} \to 2 L n_{2} O_{3}$ (1 mark).
(b) $2 L n + 6 H_{2} O \to 2 L n (O H)_{3} + 3 H_{2}$ (1 mark).
(c) $2 L n + 3 S \to L n_{2} S_{3}$ (1 mark).

Answer: $4 L n + 3 O_{2} \to 2 L n_{2} O_{3}$ ; $2 L n + 6 H_{2} O \to 2 L n (O H)_{3} + 3 H_{2}$ ; $2 L n + 3 S \to L n_{2} S_{3}$ .

Example 5: Lanthanoids against actinoids · 5 marks · Long answerHOTS / competency

(a) Why is the contraction from one element to the next greater in the actinoids than in the lanthanoids? [1]
(b) Why do actinoids show a much wider range of oxidation states than lanthanoids? Give one actinoid that reaches +7. [2]
(c) Using configurations, explain why Eu shows +2 and Tb shows +4 in addition to +3. (Z: Eu 63, Tb 65) [2]

Try it first, then show the solutionSolution

(a) 5f electrons shield the outer electrons even more poorly than 4f electrons, so the effective nuclear charge rises faster across the series (1 mark).
(b) In actinoids the 5f, 6d and 7s levels are close in energy, so varying numbers of electrons can be used in bonding (1 mark); Np (or Pu) reaches +7 (1 mark).
(c) Eu: [Xe] $4 f^{7} 6 s^{2}$ ; losing the two 6s electrons gives Eu²⁺ with a half-filled $4 f^{7}$ set, so +2 is stable (but Eu²⁺ is a reductant) (1 mark).
Tb: [Xe] $4 f^{9} 6 s^{2}$ ; losing four electrons gives Tb⁴⁺ ( $4 f^{7}$ ), also half-filled, so +4 occurs (Tb⁴⁺ is an oxidant) (1 mark).

Answer: (a) Poorer shielding by 5f electrons. (b) 5f, 6d and 7s energies are close; Np or Pu shows +7. (c) Eu²⁺ and Tb⁴⁺ both reach a half-filled $4 f^{7}$ configuration.

Where students lose marks in Lanthanoids and Lanthanoid Contraction

The slips examiners see most often, and what to write instead.

Mistake: Writing only 'lanthanoid contraction' for Zr and Hf
ISC examiners noted that many candidates named lanthanoid contraction but did not say that Zr and Hf therefore have similar size and properties (2024).
Do this: Write the chain: lanthanoid contraction → nearly identical radii → similar properties → hard to separate.
Usually costs: 1 mark of 2.
Mistake: Writing 'actinoid contraction' for the Zr-Hf similarity
Examiners recorded this confusion in 2024.
Do this: Hf follows the lanthanoids (4f), so it is the lanthanoid contraction that affects it.
Usually costs: 1 mark.
Mistake: Confusing d-d and f-f transitions, or vacant and partly filled f orbitals
In 2024, explanations of colourless La³⁺ and Lu³⁺ mixed up d-d with f-f transitions and did not state 'empty' or 'completely filled' 4f.
Wrong working
La³⁺ is colourless because it has no d-d transition.
Right working
La³⁺ ( $4 f^{0}$ ) and Lu³⁺ ( $4 f^{14}$ ) have empty or filled 4f orbitals, so no f-f transition occurs.
Do this: Lanthanoid colours come from f-f transitions; quote the 4f configuration.
Usually costs: 1 mark.
Mistake: Choosing the wrong assertion-reason option
In a 2025 assertion-reason item on Zr-Hf radii, many candidates chose 'both true but R not the explanation'.
Do this: Lanthanoid contraction is exactly the reason the Zr-Hf radii are equal; check whether R explains A, not just whether both are true.
Usually costs: 1 mark.

Exam tips

Always quote a configuration ( $4 f^{0}$ , $4 f^{7}$ , $4 f^{14}$ ) when explaining stability, colour or magnetism.
For 'consequences of lanthanoid contraction', give Zr/Hf similarity, decreasing basicity of hydroxides, and difficulty in separating lanthanoids.
Use the word 'shielding' and 'effective nuclear charge' in the cause.
Learn the general reactions as a set: O₂ → Ln₂O₃, H₂O → Ln(OH)₃ + H₂, X₂ → LnX₃, S → Ln₂S₃, N₂ → LnN, C (2773 K) → carbides.
In lanthanoid/actinoid comparisons, give at least two clear points: oxidation states and radioactivity are the safest.

Practice questions on Lanthanoids and Lanthanoid Contraction

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The most common oxidation state of the lanthanoids is:
(a) +2
(b) +3
(c) +4
(d) +6
Hint
Think of Ln₂O₃ and LnX₃.
Answer
(b) +3.
Q2 · 2 marks · Short answerBoard level
Explain why Nd³⁺ (Z = 60) is coloured, while La³⁺ (Z = 57) and Lu³⁺ (Z = 71) are colourless.
Hint
Write the 4f configuration of each ion.
Answer
Nd³⁺ is $4 f^{3}$ ; it has unpaired f electrons that absorb visible light in f-f transitions. La³⁺ ( $4 f^{0}$ ) and Lu³⁺ ( $4 f^{14}$ ) have an empty or completely filled 4f set, so no f-f transition is possible and they are colourless.
Q3 · 2 marks · Short answerHarder board level
The $E^{\circ}$ value of Ce⁴⁺/Ce³⁺ is +1.74 V, enough to oxidise water. Why, then, is cerium(IV) used as an analytical reagent in aqueous solution?
Hint
Thermodynamics versus rate.
Answer
Although the oxidation of water by Ce⁴⁺ is feasible, it is extremely slow, so cerium(IV) solutions keep their strength and can be used to oxidise other substances quantitatively.
Q4 · 3 marks · Short answerHarder board level
State the number of unpaired electrons in Ce³⁺ (Z = 58), Gd³⁺ (Z = 64), Lu³⁺ (Z = 71) and Yb²⁺ (Z = 70). Which of these ions are diamagnetic?
Hint
Ce³⁺ $4 f^{1}$ , Gd³⁺ $4 f^{7}$ , Lu³⁺ and Yb²⁺ $4 f^{14}$ .
Answer
Ce³⁺: 1; Gd³⁺: 7; Lu³⁺: 0; Yb²⁺: 0. Lu³⁺ and Yb²⁺ ( $4 f^{14}$ ) are diamagnetic; Ce³⁺ and Gd³⁺ are paramagnetic.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The basic strength of the hydroxides decreases from La(OH)₃ to Lu(OH)₃.
Reason (R): The size of Ln³⁺ ions decreases from La³⁺ to Lu³⁺, so the covalent character of the Ln–OH bond increases.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
A smaller cation polarises the O–H group more.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
The third ionisation enthalpies of La, Gd and Lu are abnormally low compared with their neighbours. Explain, using configurations of the M²⁺ and M³⁺ ions.
Hint
What 4f arrangement does each M³⁺ have?
Answer
Removing the third electron gives La³⁺ ( $4 f^{0}$ ), Gd³⁺ ( $4 f^{7}$ ) and Lu³⁺ ( $4 f^{14}$ ): an empty, half-filled and completely filled 4f set. These arrangements have extra stability (exchange energy), so less energy is needed to reach them.
Q7 · 3 marks · Short answerHOTS / competency
Explain why it is very difficult to separate individual lanthanoids from one another by ordinary chemical methods, and predict, with a reason, whether the radius of Pm³⁺ (Z = 61) lies between those of Nd³⁺ (99 pm) and Sm³⁺ (96 pm).
Hint
Size and oxidation state decide chemical behaviour.
Answer
All lanthanoids show the same +3 state and neighbouring ions differ in radius by only 1 to 3 pm, so their compounds have almost identical solubilities and reactions. Yes: the contraction is steady for the ions, so Pm³⁺ (Z = 61) lies between Nd³⁺ (60) and Sm³⁺ (62), at about 98 pm.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a geologist analyses a zircon sample and finds that it always contains a little hafnium. Zirconium (Z = 40) is in the 4d series and hafnium (Z = 72) in the 5d series of the same group. Their atomic radii are 160 pm and 159 pm, yet hafnium is about twice as dense as zirconium.
(i) Normally the radius increases down a group. Why does it not increase from Zr to Hf? [1]
(ii) Why do the two elements occur together and resist separation? [1]
(iii) Explain why Hf is so much denser although the atoms are the same size. [1]
Hint
Count what lies between La and Hf in the periodic table.
Answer
(i) Fourteen 4f electrons are added between La and Hf; their poor shielding raises the effective nuclear charge and cancels the expected increase (lanthanoid contraction). (ii) Same size and same valence configuration give almost identical chemical properties, so they substitute for each other in minerals and are hard to separate. (iii) Hf atoms are nearly twice as heavy (72 protons against 40, with more neutrons) but take up the same volume, so the density is nearly double.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define lanthanoid contraction.Show answer
The steady decrease in atomic and ionic radii from lanthanum to lutetium.
2.Cause of lanthanoid contraction?Show answer
Imperfect shielding of one 4f electron by another, so the effective nuclear charge increases.
3.Two lanthanoid ions that are reductants?Show answer
Eu²⁺ and Yb²⁺ (also Sm²⁺).
4.What is mischmetall?Show answer
An alloy of about 95% lanthanoid metal and 5% iron with traces of S, C, Ca and Al, used in lighter flints.
5.Range of $E^{\circ}$ for Ln³⁺/Ln?Show answer
About −2.2 to −2.4 V (Eu −2.0 V).
6.Why do actinoids show more oxidation states?Show answer
5f, 6d and 7s orbitals are close in energy.

Lanthanoids and Lanthanoid Contraction: frequently asked questions

What are the consequences of lanthanoid contraction?

Similar radii and properties of 4d and 5d elements of the same group (Zr and Hf), decrease in basicity of hydroxides from La to Lu, and difficulty in separating the lanthanoids.

Why is +3 the common oxidation state of lanthanoids?

Removing the two 6s electrons and one 4f (or 5d) electron gives ions with a regular 4f^n configuration whose high lattice and hydration energies make the +3 state the most stable.

Why are lanthanoid ions coloured?

Unpaired 4f electrons absorb visible light in f-f transitions; La³⁺ and Lu³⁺ have no such transitions and are colourless.

What is the difference between lanthanoids and actinoids?

Lanthanoids fill 4f orbitals, mainly show +3 and are mostly not radioactive; actinoids fill 5f orbitals, show a wider range of oxidation states, are all radioactive and show a larger contraction.

Why are Zr and Hf difficult to separate?

Lanthanoid contraction makes their radii almost identical (160 and 159 pm), so their chemical properties are nearly the same.