JEE Mains Chemistry · The d- and f-Block Elements

Lanthanoids and Actinoids

The lanthanoids Ce to Lu fill the buried 4f subshell and are almost all +3, with Ce⁴⁺ and Tb⁴⁺ as oxidants and Eu²⁺ and Yb²⁺ as reductants; the actinoids fill 5f, which is less buried, so they bond more and show more oxidation states.

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Why this matters

Thirty-one PYQs, twenty-nine of them multiple choice, and two from 2026. Thirteen write a lanthanoid configuration or count the 4f electrons of an ion, often to decide its colour or magnetism; thirteen ask which ions leave the +3 state and whether that makes them oxidants or reductants; five compare the actinoids with the lanthanoids. Nearly every answer is one 4f count away.

Concept 1 of 3: Lanthanoid configurations and 4f counts

Across the lanthanoids the new electron goes into 4f, deep inside the atom. Most atoms are [Xe]4fⁿ6s². Gadolinium and lutetium keep one 5d electron, because that leaves a half-filled 4f⁷ and a full 4f¹⁴. Forming the +3 ion always removes the two 6s electrons and then one more, so every Ln³⁺ is simply [Xe]4fⁿ with n = Z − 57.

Definition

The lanthanoids are the 14 elements Ce (58) to Lu (71). La (57) is the reference element, and actinoids such as Cm are not lanthanoids.
Atoms: [Xe]4fⁿ6s², except Ce 4f¹5d¹6s², Gd 4f⁷5d¹6s², Lu 4f¹⁴5d¹6s² (La is 5d¹6s²).
Ln³⁺: 4f electrons $= Z - 57$ . So Gd³⁺ is 4f⁷ and Lu³⁺ is 4f¹⁴.
Half-filled 4f⁷: the atoms Eu and Gd; the ions Eu²⁺, Gd³⁺ and Tb⁴⁺.
Colourless and diamagnetic: 4f⁰ (La³⁺, Ce⁴⁺) and 4f¹⁴ (Lu³⁺, Yb²⁺). Other Ln³⁺ ions are coloured by f–f transitions.
Isoelectronic ions have the same value of Z minus the charge: Eu³⁺ and Sm²⁺ both have 60 electrons, so they are isoelectronic.
Spin-only moments of f ions: Gd³⁺ (7 unpaired) 7.94 BM, Eu³⁺ (6) 6.93 BM, Ce⁴⁺ 0.

Element (Z)	Atom	M³⁺ ion	Other common ion
La (57)	[Xe]5d¹6s²	4f⁰, colourless	Shows only +3
Ce (58)	[Xe]4f¹5d¹6s²	4f¹	Ce⁴⁺, 4f⁰
Pr (59)	[Xe]4f³6s²	4f²	Pr⁴⁺, 4f¹
Nd (60)	[Xe]4f⁴6s²	4f³	Nd²⁺ 4f⁴; Nd⁴⁺ 4f²
Pm (61)	[Xe]4f⁵6s²	4f⁴	Shows only +3
Sm (62)	[Xe]4f⁶6s²	4f⁵	Sm²⁺, 4f⁶
Eu (63)	[Xe]4f⁷6s²	4f⁶	Eu²⁺, 4f⁷ Eu²⁺ and Gd³⁺ are the two 4f⁷ ions.
Gd (64)	[Xe]4f⁷5d¹6s²	4f⁷	Shows only +3
Tb (65)	[Xe]4f⁹6s²	4f⁸	Tb⁴⁺, 4f⁷
Dy (66)	[Xe]4f¹⁰6s²	4f⁹	Dy⁴⁺, 4f⁸
Ho (67)	[Xe]4f¹¹6s²	4f¹⁰	Shows only +3
Er (68)	[Xe]4f¹²6s²	4f¹¹	Shows only +3
Tm (69)	[Xe]4f¹³6s²	4f¹²	Tm²⁺, 4f¹³
Yb (70)	[Xe]4f¹⁴6s²	4f¹³	Yb²⁺, 4f¹⁴
Lu (71)	[Xe]4f¹⁴5d¹6s²	4f¹⁴, colourless	Shows only +3

The +3 ion always has Z − 57 electrons in 4f; the ions in the last column reach 4f⁰, 4f⁷ or 4f¹⁴, or come close.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q30Moderate

Example 1 · The d- and f-Block Elements · Lanthanoids and Actinoids

Lanthanoid ions with

4f^{7}

configuration are: (A)

Eu^{2 +}

(B)

Gd^{3 +}

(C)

Eu^{3 +}

(D)

Tb^{3 +}

(E)

Sm^{2 +}

Choose the correct answer from the options given below :

The 5d electron in Gd and Lu does not change the ion

Gd is 4f⁷5d¹6s², but Gd³⁺ is still 4f⁷: the three electrons lost are 6s², 5d¹. Use Z − 57 for every +3 ion and the atom's quirks drop out.

Isoelectronic means the same total, Z minus charge

Compare Z − charge, not the charge.

\mathrm{Tb^{2+}}

has 63 electrons and

\mathrm{Tm^{4+}}

has 65, so they are not isoelectronic even though both look like 'lanthanoid ions near 4f⁷'.

Drill 12 more on lanthanoid configurations and 4f counts

Concept 2 of 3: Lanthanoid ions outside the +3 state

+3 is the home state of every lanthanoid. An ion in another state exists because it reaches, or nearly reaches, 4f⁰, 4f⁷ or 4f¹⁴. But reaching a tidy shell does not make it stable in water: the ion still wants to return to +3. So a +4 ion takes an electron and is an oxidant, and a +2 ion gives one away and is a reductant.

Definition

+3 is the most common and most stable state for all lanthanoids.
+4 (oxidants): Ce⁴⁺ (4f⁰) and Tb⁴⁺ (4f⁷). E°(Ce⁴⁺/Ce³⁺) = +1.74 V: it could oxidise water, but slowly, so Ce(IV) is a good analytical reagent. Tb⁴⁺ is an even stronger oxidant.
Pr, Nd, Tb and Dy also show +4, but only in solid oxides $\mathrm{MO_2}$ (with Ce, which forms $\mathrm{CeO_2}$ ). Yb forms no $\mathrm{MO_2}$ .
+2 (reductants): Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴); also Sm²⁺. Aqueous $\mathrm{EuSO_4}$ is a strong reducing agent.
Eu and Yb have the highest third ionisation enthalpies: their third electron would come out of 4f⁷ or 4f¹⁴. So they are the easiest to hold at +2.
Ce leaves the +3 state most easily (to +4).
$\mathrm{CeO_2}$ is used as an oxidant in organic chemistry, for example on aldehydes and ketones.

Ion	4f configuration	Why it exists	Behaviour
Ce⁴⁺	4f⁰	Noble-gas (Xe) core	Strong oxidant; E° = +1.74 V back to Ce³⁺ The noble-gas core favours forming Ce⁴⁺, but Ce³⁺ is still the more stable state in water.
Tb⁴⁺	4f⁷	Half-filled 4f	Stronger oxidant than Ce⁴⁺; found in $\mathrm{TbO_2}$
Pr⁴⁺, Nd⁴⁺, Dy⁴⁺	4f¹, 4f², 4f⁸	Stabilised only in the solid oxide	Found only as $\mathrm{MO_2}$ ; oxidants
Eu²⁺	4f⁷	Half-filled 4f after losing 6s²	Strong reductant; turns into Eu³⁺
Yb²⁺	4f¹⁴	Full 4f after losing 6s²	Reductant; diamagnetic
Sm²⁺	4f⁶	Close to 4f⁷	Reductant
Ln³⁺ (all)	4f¹ to 4f¹⁴	Loss of 6s² and one more electron	The stable state of every lanthanoid

+4 ions are oxidants and +2 ions are reductants, because each tends to return to +3.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q47Moderate

Example 2 · The d- and f-Block Elements · Lanthanoids and Actinoids

Strong reducing and oxidizing agents among the following, respectively, are

A noble-gas core does not make Ce⁴⁺ the stable state

The Xe core favours FORMING Ce⁴⁺, but in water Ce⁴⁺ returns to Ce³⁺, which is why it is an oxidant. 'Ce is more stable as Ce⁴⁺ than as Ce³⁺' is false.

4f⁷ does not stop Eu²⁺ reducing

Eu²⁺ has a half-filled 4f⁷, yet it is a strong reductant. The +3 state is still preferred, so the special configuration does not protect it.

Drill 12 more on lanthanoid ions outside the +3 state

Concept 3 of 3: Actinoids compared with lanthanoids

The actinoids fill 5f instead of 4f. The 5f orbitals reach further out and are shielded less, so their electrons take part in bonding. That gives the actinoids many more oxidation states and a richer chemistry. The poor shielding by 5f electrons also makes the actinoid contraction larger, element to element, than the lanthanoid contraction.

Definition

The actinoids are Th (90) to Lr (103). All are radioactive.
5f, 6d and 7s are close in energy, so configurations are irregular: Np $[\mathrm{Rn}]\,5f^{4}6d^{1}7s^{2}$ , Am $5f^{7}7s^{2}$ , Cm $5f^{7}6d^{1}7s^{2}$ , Es $5f^{11}7s^{2}$ .
Unpaired electrons: Am 7, Cm 8 (seven in 5f and one in 6d).
Oxidation states: +3 is common, but the early actinoids go higher, up to +7 for Np.
Actinoid contraction: the size of M³⁺ falls along the series, by more per element than in the lanthanoids. So $\mathrm{Bk^{3+}}$ is smaller than $\mathrm{Np^{3+}}$ .

Property	Lanthanoids	Actinoids
Subshell being filled	4f, deeply buried	5f, less buried, reaches further out
f electrons in bonding	Very little	To a far greater extent
Oxidation states	Mostly +3; a few +2 and +4	+3 common; up to +7 (Np) in the first half
Contraction along the series	Lanthanoid contraction	Actinoid contraction: larger from element to element
Radioactivity	Only Pm	All of them
Example configuration	Gd [Xe]4f⁷5d¹6s²	Cm [Rn]5f⁷6d¹7s²

Almost every actinoid difference traces back to 5f orbitals being less buried than 4f.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 12 Apr 2023 · Q33Moderate

Example 3 · The d- and f-Block Elements · Lanthanoids and Actinoids

Given below are two statement: one is labelled as Assertion

A

and the other is labelled as Reason

R

Assertion A:

5f

electrons can participate in bonding to a far greater extent than

4f

electrons Reason R:

5f

orbitals are not as buried as

4f

orbitals In the light of the above statements, choose the correct answer from the options given below

Name the right contraction

The shrinking of

\mathrm{M^{3+}}

across Th to Lr is the ACTINOID contraction. A reason that calls it the lanthanoid contraction is wrong for actinoid ions such as

\mathrm{Bk^{3+}}

and

\mathrm{Np^{3+}}

Cm has eight unpaired electrons

Am and Cm both have 5f⁷, but curium adds a 6d electron. 'Cm and Am have seven unpaired electrons' is false.

Drill 4 more on actinoids compared with lanthanoids

Summary — formulas & gotchas at a glance

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Reference tables (3)

Lanthanoid configurations and 4f counts15 rows

Element (Z)	Atom	M³⁺ ion	Other common ion
La (57)	[Xe]5d¹6s²	4f⁰, colourless	Shows only +3
Ce (58)	[Xe]4f¹5d¹6s²	4f¹	Ce⁴⁺, 4f⁰
Pr (59)	[Xe]4f³6s²	4f²	Pr⁴⁺, 4f¹
Nd (60)	[Xe]4f⁴6s²	4f³	Nd²⁺ 4f⁴; Nd⁴⁺ 4f²
Pm (61)	[Xe]4f⁵6s²	4f⁴	Shows only +3
Sm (62)	[Xe]4f⁶6s²	4f⁵	Sm²⁺, 4f⁶
Eu (63)	[Xe]4f⁷6s²	4f⁶	Eu²⁺, 4f⁷ Eu²⁺ and Gd³⁺ are the two 4f⁷ ions.
Gd (64)	[Xe]4f⁷5d¹6s²	4f⁷	Shows only +3
Tb (65)	[Xe]4f⁹6s²	4f⁸	Tb⁴⁺, 4f⁷
Dy (66)	[Xe]4f¹⁰6s²	4f⁹	Dy⁴⁺, 4f⁸
Ho (67)	[Xe]4f¹¹6s²	4f¹⁰	Shows only +3
Er (68)	[Xe]4f¹²6s²	4f¹¹	Shows only +3
Tm (69)	[Xe]4f¹³6s²	4f¹²	Tm²⁺, 4f¹³
Yb (70)	[Xe]4f¹⁴6s²	4f¹³	Yb²⁺, 4f¹⁴
Lu (71)	[Xe]4f¹⁴5d¹6s²	4f¹⁴, colourless	Shows only +3

The +3 ion always has Z − 57 electrons in 4f; the ions in the last column reach 4f⁰, 4f⁷ or 4f¹⁴, or come close.

Lanthanoid ions outside the +3 state7 rows

Ion	4f configuration	Why it exists	Behaviour
Ce⁴⁺	4f⁰	Noble-gas (Xe) core	Strong oxidant; E° = +1.74 V back to Ce³⁺ The noble-gas core favours forming Ce⁴⁺, but Ce³⁺ is still the more stable state in water.
Tb⁴⁺	4f⁷	Half-filled 4f	Stronger oxidant than Ce⁴⁺; found in $\mathrm{TbO_2}$
Pr⁴⁺, Nd⁴⁺, Dy⁴⁺	4f¹, 4f², 4f⁸	Stabilised only in the solid oxide	Found only as $\mathrm{MO_2}$ ; oxidants
Eu²⁺	4f⁷	Half-filled 4f after losing 6s²	Strong reductant; turns into Eu³⁺
Yb²⁺	4f¹⁴	Full 4f after losing 6s²	Reductant; diamagnetic
Sm²⁺	4f⁶	Close to 4f⁷	Reductant
Ln³⁺ (all)	4f¹ to 4f¹⁴	Loss of 6s² and one more electron	The stable state of every lanthanoid

+4 ions are oxidants and +2 ions are reductants, because each tends to return to +3.

Actinoids compared with lanthanoids6 rows

Property	Lanthanoids	Actinoids
Subshell being filled	4f, deeply buried	5f, less buried, reaches further out
f electrons in bonding	Very little	To a far greater extent
Oxidation states	Mostly +3; a few +2 and +4	+3 common; up to +7 (Np) in the first half
Contraction along the series	Lanthanoid contraction	Actinoid contraction: larger from element to element
Radioactivity	Only Pm	All of them
Example configuration	Gd [Xe]4f⁷5d¹6s²	Cm [Rn]5f⁷6d¹7s²

Almost every actinoid difference traces back to 5f orbitals being less buried than 4f.

Watch out for (6)

The 5d electron in Gd and Lu does not change the ion
→ Lanthanoid configurations and 4f counts
Isoelectronic means the same total, Z minus charge
→ Lanthanoid configurations and 4f counts
A noble-gas core does not make Ce⁴⁺ the stable state
→ Lanthanoid ions outside the +3 state
4f⁷ does not stop Eu²⁺ reducing
→ Lanthanoid ions outside the +3 state
Name the right contraction
→ Actinoids compared with lanthanoids
Cm has eight unpaired electrons
→ Actinoids compared with lanthanoids

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