f – Block Elements

ChemistryThe d- And f-Block ElementsFor JEE aspirants

f-block elements are the elements in which the differentiating electron enters the anti-penultimate $(n - 2) f$ subshell: the 4f lanthanoids (Ce to Lu, studied with La) and the 5f actinoids (Th to Lr, studied with Ac). Also called inner transition elements, the f-block elements show a dominant +3 state, a steady shrinking of size called lanthanoid contraction, coloured paramagnetic ions and, for actinoids, radioactivity and many oxidation states. Every trend here is drawn as a figure. The topic is asked every year in NEET and JEE Main.

On this page1Position2Configuration3Ln oxidation states4Lanthanoid contraction5Ln properties6Actinoids7Ln vs An8Solved examples

Key Formulas - Quick Reference

★ Must learnGeneral configuration $(n - 2) f^{1 - 14} (n - 1) d^{0 - 1} n s^{2}$ : lanthanoids $[Xe] 4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}$ , actinoids $[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ .
★ Must learn $4 f$ electrons in $L n^{3 +}$ $= Z - 57$ : from $L a^{3 +}$ ( $f^{0}$ ) to $L u^{3 +}$ ( $f^{14}$ ).
★ Must learnLanthanoids: +3 is typical; $C e^{4 +}$ ( $f^{0}$ , oxidant, $E^{\circ} = + 1.74$ V), $T b^{4 +}$ ( $f^{7}$ ), $E u^{2 +}$ ( $f^{7}$ ) and $Y b^{2 +}$ ( $f^{14}$ , reductants).
★ Must learnLanthanoid contraction: $L n^{3 +}$ radius falls about 17 pm (103 pm $\to$ 86 pm) because $4 f$ electrons shield poorly.
Basic strength: $La (OH)_{3} > \dots > Lu (OH)_{3}$ ; Zr (160 pm) $\approx$ Hf (159 pm).
Shielding power: $s > p > d > f$ .
★ Must learnActinoids: highest state Th +4, Pa +5, U +6, Np and Pu +7; actinoid contraction is larger than lanthanoid contraction.
Mischmetall $\approx$ 95% lanthanoid metals + 5% Fe (traces of S, C, Ca, Al).

1. What Are f-Block Elements?

In the f-block elements the differentiating electron (the last electron, which makes an element different from the one before it) enters the anti-penultimate shell, the $(n - 2) f$ subshell. Because they form a series within the transition series, they are called inner transition elements; the lanthanoids were once called rare earth elements. Depending on whether the electron enters $4 f$ or $5 f$ orbitals, they form two series.

Lanthanoids (4f series): cerium (Ce, 58) to lutetium (Lu, 71). They come immediately after lanthanum, which gives the series its name, and lanthanum is studied with them as the reference element.
Actinoids (5f series): thorium (Th, 90) to lawrencium (Lr, 103). They come immediately after actinium, which is studied with them.

Figure 1: The two f-block rows are pulled out below the table to keep it narrow. The 4f row (lanthanoids) belongs after La in period 6 and the 5f row (actinoids) after Ac in period 7; electrons enter the anti-penultimate

(n - 2) f

subshell.

Why the names end in "-oid": IUPAC prefers lanthanoid and actinoid (like lanthanum, actinium) over the older "lanthanide" and "actinide", because "-ide" usually means a negative ion (chloride, oxide). Both names are accepted in exams.

2. Electronic Configuration

The general electronic configuration of the f-block elements is

(n - 2) f^{1 - 14} (n - 1) d^{0 - 1} n s^{2}

so the lanthanoids are $[Xe] 4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}$ and the actinoids $[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ . All lanthanoid atoms have $6 s^{2}$ . The $4 f$ and $5 d$ levels are so close in energy that only La, Ce, Gd and Lu keep an electron in $5 d$ ; the others place all their electrons beyond the xenon core in $4 f$ . The trivalent ions, however, fill regularly from $4 f^{0}$ to $4 f^{14}$ .

Element	Z	Atom ( $[Xe]$ +)	$L n^{3 +}$	$L n^{3 +}$ radius / pm
La	57	$5 d^{1} 6 s^{2}$	$4 f^{0}$	103
Ce	58	$4 f^{1} 5 d^{1} 6 s^{2}$	$4 f^{1}$	101
Pr	59	$4 f^{3} 6 s^{2}$	$4 f^{2}$	99
Nd	60	$4 f^{4} 6 s^{2}$	$4 f^{3}$	98
Pm	61	$4 f^{5} 6 s^{2}$	$4 f^{4}$	97
Sm	62	$4 f^{6} 6 s^{2}$	$4 f^{5}$	96
Eu	63	$4 f^{7} 6 s^{2}$	$4 f^{6}$	95
Gd	64	$4 f^{7} 5 d^{1} 6 s^{2}$	$4 f^{7}$	94
Tb	65	$4 f^{9} 6 s^{2}$	$4 f^{8}$	92
Dy	66	$4 f^{10} 6 s^{2}$	$4 f^{9}$	91
Ho	67	$4 f^{11} 6 s^{2}$	$4 f^{10}$	90
Er	68	$4 f^{12} 6 s^{2}$	$4 f^{11}$	89
Tm	69	$4 f^{13} 6 s^{2}$	$4 f^{12}$	88
Yb	70	$4 f^{14} 6 s^{2}$	$4 f^{13}$	87
Lu	71	$4 f^{14} 5 d^{1} 6 s^{2}$	$4 f^{14}$	86

Figure 2: Atoms fill

4 f

irregularly (La, Ce, Gd and Lu keep one

5 d

electron), but the

L n^{3 +}

ions run smoothly from

4 f^{0}

(La) to

4 f^{14}

(Lu). Eu and Gd (

4 f^{7}

) and Yb and Lu (

4 f^{14}

) show the stability of half-filled and full sets.

Exam Trick

Z minus 57. The number of $4 f$ electrons in any $L n^{3 +}$ ion is $Z - 57$ : $N d^{3 +}$ ( $Z = 60$ ) is $4 f^{3}$ , $E r^{3 +}$ ( $Z = 68$ ) is $4 f^{11}$ . The atoms are irregular, the ions never are.

Key idea

Only four lanthanoid atoms (La, Ce, Gd, Lu) carry a

5 d

electron; in the ions, the

4 f

count is simply

Z - 57

3. Oxidation States of Lanthanoids

Lanthanoids show one typical, stable oxidation state: +3. It is reached by losing the two $6 s$ electrons and one $5 d$ (or $4 f$ ) electron, and it is the common state in solution and in solids (oxides $L n_{2} O_{3}$ , halides $Ln X_{3}$ ). A few elements also show +2 or +4, especially when this gives an empty, half-filled or completely filled f subshell:

Reason	Example	Behaviour
noble gas core, empty $4 f^{0}$	$C e^{4 +}$ ( $4 f^{0}$ )	strong oxidant, $E^{\circ} (C e^{4 +} /C e^{3 +}) = + 1.74$ V; used in volumetric analysis
half-filled $4 f^{7}$	$E u^{2 +}$ , $T b^{4 +}$	$E u^{2 +}$ is a strong reductant; $T b^{4 +}$ is an oxidant
completely filled $4 f^{14}$	$Y b^{2 +}$ ( $4 f^{14}$ )	reductant, changes back to +3

Figure 3: The unusual states reach

f^{0}

f^{7}

f^{14}

C e^{4 +}

(

f^{0}

) and

T b^{4 +}

(

f^{7}

E u^{2 +}

(

f^{7}

) and

Y b^{2 +}

(

f^{14}

). Because +3 is the normal state,

C e^{4 +}

is an oxidant and

E u^{2 +}

a reductant.

Stability of these configurations is only part of the story: $S m^{2 +}$ ( $4 f^{6}$ ) and $T m^{2 +}$ ( $4 f^{13}$ ) exist, and Pr, Nd and Dy form +4 oxides ( $M O_{2}$ ). Whenever a lanthanoid is found in +2 or +4, remember that +3 is "home": a +4 ion tends to gain an electron (oxidant) and a +2 ion tends to lose one (reductant).

Key idea

+3 is home: +4 lanthanoid ions are oxidants (

C e^{4 +}

), +2 ions are reductants (

E u^{2 +}

Y b^{2 +}

4. Lanthanoid Contraction

Across the lanthanoid series, atomic and ionic radii decrease steadily with increasing atomic number. This regular decrease is called the lanthanoid contraction. The $L n^{3 +}$ radius falls from about 103 pm for $L a^{3 +}$ to 86 pm for $L u^{3 +}$ , a total of only about 17 pm over fourteen elements, but the effect accumulates.

Figure 4: (A) Metallic radii shrink slowly, except Eu and Yb, which keep

f^{7}

and

f^{14}

and release only two electrons to the metallic bond, so their atoms are larger. (B) The

L n^{3 +}

radius falls steadily by about 17 pm from

L a^{3 +}

L u^{3 +}

: the lanthanoid contraction.

Cause. With each step the nuclear charge rises by one and one electron enters the $4 f$ subshell. The $4 f$ orbitals are diffuse and shield the other electrons very poorly (shielding power $s > p > d > f$ ). The effective nuclear charge felt by the outer electrons therefore increases, and the whole electron cloud is pulled in. The atomic (metallic) radii of Eu and Yb are exceptions: these atoms keep their stable $4 f^{7}$ and $4 f^{14}$ sets and give only two electrons to the metallic bond, so their atoms are larger and the metals softer.

4.1 Consequences of Lanthanoid Contraction

Similar chemistry, difficult separation. Neighbouring lanthanoid ions differ very little in size, so their chemical properties are almost identical and it is very difficult to obtain them pure (ion-exchange chromatography is used).
4d and 5d twins. The third transition series (5d) elements that follow the lanthanoids are almost the same size as the 4d elements above them: Zr 160 pm and Hf 159 pm, Nb and Ta, Mo and W. Such pairs occur together in nature and are hard to separate.
Basicity of hydroxides decreases. As $L n^{3 +}$ gets smaller from La to Lu, it polarises the hydroxide ion more, so the covalent character of the M-OH bond increases and basic strength decreases: $La (OH)_{3}$ is the most basic and $Lu (OH)_{3}$ the least basic.
Higher density and IE of 5d metals. Because 5d atoms are no bigger than 4d atoms but much heavier, the 5d metals (Os, Ir, Pt) are very dense, and their ionisation enthalpies are higher.

Figure 5:

4 f

electrons shield worst of all, so

Z_{eff}

climbs across the series. Results: Zr and Hf are twins, the lanthanoids are hard to separate, and

Ln (OH)_{3}

becomes less basic from

La (OH)_{3}

Lu (OH)_{3}

as the smaller ion polarises O-H bonds more.

Exam Trick

Same size, same chemistry. Any question that pairs Zr/Hf, Nb/Ta or Mo/W, or asks why lanthanoids are hard to separate, has one answer: lanthanoid contraction.

Key idea

Poor 4f shielding shrinks the lanthanoids by about 17 pm; this makes them hard to separate, makes 5d metals twins of the 4d metals, and lowers basicity from La to Lu.

Quick Recall: tap to check

Which is more basic,

La (OH)_{3}

Lu (OH)_{3}

La (OH)_{3}

; the larger

L a^{3 +}

ion gives a more ionic M-OH bond.

Why are the metallic radii of Eu and Yb unusually large?

They keep

4 f^{7}

and

4 f^{14}

and use only two electrons in metallic bonding.

Configuration of

G d^{3 +}

[Xe] 4 f^{7}

(

Z - 57 = 7

5. General Characteristics of Lanthanoids

5.1 Physical Properties

Lanthanoids are silvery-white, soft metals that tarnish rapidly in air. Hardness increases with atomic number; samarium is as hard as steel. Melting points lie between about 1000 K and 1200 K (samarium 1623 K). They conduct heat and electricity well. Their first three ionisation enthalpies add up to fairly low values, which is why the +3 state forms so easily.

5.2 Colour and Magnetic Properties

Many trivalent lanthanoid ions are coloured in the solid state and in solution. The colour comes from f-f transitions: electrons jump between $4 f$ levels. Ions with $f^{n}$ and $f^{14 - n}$ electrons tend to have similar colours. $L a^{3 +}$ ( $f^{0}$ ) and $L u^{3 +}$ ( $f^{14}$ ) are colourless, and so are $C e^{3 +}$ , $G d^{3 +}$ and $Y b^{3 +}$ in visible light. Because the $4 f$ orbitals are buried inside the $5 s$ and $5 p$ shells, the absorption bands are very sharp and hardly change with the ligand.

Figure 6:

L n^{3 +}

colours come from f-f transitions. Ions with

f^{n}

and

f^{14 - n}

look alike (Pr green and Tm pale green; Sm and Dy yellow), while

f^{0}

f^{7}

and

f^{14}

ions are colourless.

All lanthanoid ions except the $f^{0}$ ones ( $L a^{3 +}$ , $C e^{4 +}$ ) and the $f^{14}$ ones ( $Y b^{2 +}$ , $L u^{3 +}$ ) are paramagnetic. Their magnetic moments include a large contribution from orbital motion, so the spin-only formula is only a rough guide here (it fits $G d^{3 +}$ , $4 f^{7}$ , well: about 7.9 BM).

Figure 7: For

L n^{3 +}

4 f

electrons

= Z - 57

G d^{3 +}

(

Z = 64

) is

4 f^{7}

with 7 unpaired electrons;

L a^{3 +}

and

L u^{3 +}

are colourless and diamagnetic.

5.3 Complex Formation

Lanthanoids have little tendency to form complexes. Their $L n^{3 +}$ ions are large, so their charge density is low, and the $4 f$ orbitals are too deeply buried to take part in bonding. The tendency to form complexes and the stability of the complexes increase slightly from La to Lu as the ions get smaller.

5.4 Chemical Behaviour

The first few members (La, Ce, Pr) are quite reactive, rather like calcium; as the atomic number rises the behaviour approaches that of aluminium. $E^{\circ} (L n^{3 +} /Ln)$ values lie between about $- 2.2$ and $- 2.4$ V (Eu about $- 2.0$ V), so all are strong reducing agents. Their main reactions are:

They combine with hydrogen on gentle heating, form carbides ( $L n_{3} C$ , $L n_{2} C_{3}$ , $Ln C_{2}$ ) when heated with carbon at 2773 K, and burn in halogens to give $Ln X_{3}$ .
They liberate hydrogen from dilute acids, and from water (slowly in the cold, faster on heating).
They burn in oxygen to form oxides $M_{2} O_{3}$ and hydroxides $M (OH)_{3}$ , which are basic, like alkaline earth metal oxides and hydroxides. On heating they also combine with nitrogen (LnN) and sulphur ( $L n_{2} S_{3}$ ).

2Ln + 6HCl 2LnC l_{3} + 3 H_{2} ↑

2Ln + 6 H_{2} O 2Ln (OH)_{3} + 3 H_{2} ↑

4Ln + 3 O_{2} 2L n_{2} O_{3}

Figure 8: Lanthanoids behave like reactive metals of the +3 state: they burn to

L n_{2} O_{3}

, form

LnN

L n_{2} S_{3}

and

Ln X_{3}

, liberate

H_{2}

from water and dilute acids, and give carbides at 2773 K.

Key idea

Lanthanoids are reactive, trivalent, paramagnetic metals whose colours come from sharp f-f transitions.

6. The Actinoids

6.1 Configuration and Size

Actinoids have the configuration $[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ . The $5 f$ and $6 d$ energies are even closer than $4 f$ and $5 d$ , so early actinoids often keep $6 d$ electrons (thorium has none in $5 f$ at all):

Element	Th	Pa	U	Np	Pu	Am	Cm
Z	90	91	92	93	94	95	96
$[Rn]$ +	$6 d^{2} 7 s^{2}$	$5 f^{2} 6 d^{1} 7 s^{2}$	$5 f^{3} 6 d^{1} 7 s^{2}$	$5 f^{4} 6 d^{1} 7 s^{2}$	$5 f^{6} 7 s^{2}$	$5 f^{7} 7 s^{2}$	$5 f^{7} 6 d^{1} 7 s^{2}$

Actinoid atoms and ions also shrink across the series. This actinoid contraction is greater from element to element than the lanthanoid contraction, because $5 f$ electrons shield the outer electrons even more poorly than $4 f$ electrons do.

6.2 Oxidation States

The dominant oxidation state is +3, as for the lanthanoids, but +4 is also common, and the early actinoids show still higher states. The maximum oxidation state first increases up to the middle of the series and then decreases: +4 for Th, +5 for Pa, +6 for U and +7 for Np and Pu, falling back to +3 (or +2 for No) for the later members. High states often occur as oxocations such as $U O_{2}^{2 +}$ , $Pu O_{2}^{2 +}$ and $Np O_{2}^{+}$ .

Figure 9: Early actinoids use their

5 f

6 d

and

7 s

electrons freely, so the highest state rises to +7 at Np and Pu. From Am onwards the

5 f

electrons are held more tightly and +3 (or +2 for No) dominates, as in the lanthanoids.

JEE Advanced

Why actinoids show many more oxidation states. The $5 f$ , $6 d$ and $7 s$ levels are very close in energy, and the $5 f$ orbitals extend further from the nucleus than $4 f$ orbitals do (they are less buried inside filled shells). So $5 f$ electrons can be removed or used in bonding much more easily. In the second half of the series the growing nuclear charge stabilises the $5 f$ electrons, and the chemistry becomes lanthanoid-like (+3). The same greater exposure of $5 f$ orbitals explains why actinoids form more complexes and more covalent compounds than lanthanoids.

6.3 General Characteristics and Chemical Behaviour

The ability of actinoids to exist in many oxidation states makes their chemistry more complicated. Moreover, all of them are radioactive, and the later ones have short half-lives, so their chemistry is difficult to study in the laboratory. They are silvery metals with a wide range of structures, and their ionisation enthalpies are lower than those of the early lanthanoids because $5 f$ electrons are more effectively shielded from the nucleus.

In the finely divided state they react with boiling water to give a mixture of oxide and hydride.
They combine with most non-metals at moderate temperatures.
All actinoid metals are attacked by hydrochloric acid, but nitric acid has very little effect because a protective oxide layer forms on the surface. Alkalis have no action.

Figure 10: Only Th, Pa and U are found in useful amounts; the later actinoids are made artificially and are so short-lived (Fm-257 about 100 days) that their chemistry is studied on tiny amounts. This is why actinoid chemistry is harder to study.

Key idea

Actinoids are all radioactive, reach +7 at Np and Pu, contract more than lanthanoids and form more complexes.

Quick Recall: tap to check

Which actinoids show the +7 state?

Neptunium and plutonium.

Why does

HN O_{3}

hardly attack actinoid metals?

A protective oxide layer forms on the metal surface.

Which is larger per step, lanthanoid or actinoid contraction?

Actinoid contraction, because

5 f

electrons shield even more poorly.

7. Lanthanoids vs Actinoids

Property	Lanthanoids	Actinoids
Filling subshell	$4 f$	$5 f$
Oxidation states	mainly +3; a few +2, +4	+3 to +7 (+4, +5, +6 common early)
Radioactivity	only promethium	all are radioactive
Complex formation	weak	much stronger
Contraction per element	smaller	larger (5f shields worse)
Oxocations	none	$U O_{2}^{2 +}$ , $Pu O_{2}^{2 +}$ , $Np O_{2}^{+}$
Hydroxides	basic	less basic
Magnetic behaviour	easy to explain	more complex
Occurrence	all occur naturally except Pm	only Th, Pa, U in useful amounts

Similarities

Both are inner transition series with +3 as the common state, both show a contraction in size, and both give coloured, paramagnetic ions.

Key difference

$5 f$ orbitals are less buried than $4 f$ , so actinoids use them in bonding: more oxidation states, more complexes, more covalent compounds.

Exam Trick

Lanthanoids are loyal, actinoids are adventurous. Lanthanoids stay at +3 (a few step to +2 or +4); actinoids climb to +7 at Np and Pu. If an option offers a lanthanoid above +4, it is wrong.

8. Uses of f-Block Elements

The best-known alloy is mischmetall, about 95% lanthanoid metals and 5% iron with traces of S, C, Ca and Al. A magnesium alloy containing about 3% mischmetall is used in bullets, shells and lighter flints. Mixed lanthanoid oxides are catalysts in petroleum cracking, and some lanthanoid oxides are phosphors in television screens and fluorescent materials. Among actinoids, uranium and plutonium are nuclear fuels, and thorium is a potential nuclear fuel.

Figure 11: Mischmetall (about 95% lanthanoid metals, 5% Fe) is alloyed with Mg for bullets, shells and lighter flints. Lanthanoid oxides are cracking catalysts and phosphors; U and Pu are nuclear fuels.

9. Revision Map

The whole f-block chapter in one picture.

Figure 12: The f-block on one page: configuration explains the +3 state and the contraction, and the contraction explains the similar chemistry of the lanthanoids.

10. Solved Examples

Solved Example 1

Why is

C e^{4 +}

a good oxidising agent, whereas

E u^{2 +}

is a good reducing agent?

Solution:

+3 is the most stable oxidation state of lanthanoids. $C e^{4 +}$ ( $4 f^{0}$ ) readily gains an electron to become $C e^{3 +}$ , so it is an oxidant ( $E^{\circ} = + 1.74$ V). $E u^{2 +}$ ( $4 f^{7}$ ) readily loses an electron to become $E u^{3 +}$ , so it is a reductant. Their unusual states survive only because $4 f^{0}$ and $4 f^{7}$ are relatively stable.

Solved Example 2

Which of the following ions is colourless?
(A)

N d^{3 +}

(B)

S m^{3 +}

(C)

L u^{3 +}

(D)

P r^{3 +}

Solution:

Answer: (C). $L u^{3 +}$ ( $Z = 71$ ) has $71 - 57 = 14$ f electrons: $4 f^{14}$ , so no f-f transition is possible. The others are $4 f^{3}$ (lilac), $4 f^{5}$ (yellow) and $4 f^{2}$ (green).

Solved Example 3

Calculate the number of unpaired electrons and the spin-only magnetic moment of

G d^{3 +}

(Z = 64).

Solution:

$4 f$ electrons $= 64 - 57 = 7$ : $4 f^{7}$ , all seven unpaired. $μ = 7 \times 9 = 63 = 7.94$ BM, close to the measured value (about 7.9 BM) because a half-filled set has no orbital contribution.

Solved Example 4

Because of lanthanoid contraction, which pair of elements has almost identical atomic radii?
(A) Ti and Zr
(B) Zr and Hf
(C) V and Nb
(D) Sc and Y

Solution:

Answer: (B). Zr (4d) is 160 pm and Hf (5d) 159 pm. The 14 lanthanoids placed before Hf cancel the size gain of the extra shell. In the other pairs the lower element is clearly larger.

Solved Example 5

Arrange

La (OH)_{3}

Gd (OH)_{3}

and

Lu (OH)_{3}

in decreasing order of basic strength and explain.

Solution:

$La (OH)_{3} > Gd (OH)_{3} > Lu (OH)_{3}$ . The ionic radius falls $La > Gd > Lu$ (lanthanoid contraction). A smaller, more polarising cation makes the M-OH bond more covalent, so the hydroxide releases $O H^{-}$ less easily.

Solved Example 6

Why do actinoids show a wider range of oxidation states than lanthanoids?

Solution:

In actinoids the $5 f$ , $6 d$ and $7 s$ orbitals are very close in energy, and $5 f$ orbitals are less shielded (less buried) than $4 f$ orbitals. Electrons from all three can take part in bonding, giving states up to +7 (Np, Pu). In lanthanoids the $4 f$ electrons are held deep inside, so +3 dominates.

Solved Example 7

The highest oxidation state shown by neptunium is
(A) +4
(B) +5
(C) +6
(D) +7

Solution:

Answer: (D). The maximum oxidation state of actinoids rises Th +4, Pa +5, U +6, Np +7 (Pu also +7), then falls. Np uses all its $5 f^{4} 6 d^{1} 7 s^{2}$ electrons.

Practice Questions

Why is $La (OH)_{3}$ more basic than $Lu (OH)_{3}$ ?Answer: $L a^{3 +}$ is larger (lanthanoid contraction makes $L u^{3 +}$ smaller), so the La-OH bond is more ionic.
Name a lanthanoid that shows +4 and one that shows +2 in solution.Answer: $C e^{4 +}$ ( $4 f^{0}$ ); $E u^{2 +}$ ( $4 f^{7}$ ) or $Y b^{2 +}$ ( $4 f^{14}$ ).
Why are zirconium and hafnium difficult to separate?Answer: Lanthanoid contraction makes their radii almost equal (160 and 159 pm), so their chemistry is nearly identical.
Why is it difficult to separate the lanthanoids from one another?Answer: Their ions differ very little in size, so their properties are very similar; ion-exchange methods are needed.
Why is the chemistry of actinoids harder to study than that of lanthanoids?Answer: All actinoids are radioactive, many are short-lived and available only in tiny amounts, and they show many oxidation states.
What is the spin-only magnetic moment of $C e^{3 +}$ ( $Z = 58$ )?Answer: $4 f^{1}$ , $n = 1$ , $μ = 3 = 1.73$ BM.
What is mischmetall and one use of it?Answer: An alloy of about 95% lanthanoid metals and 5% Fe (traces of S, C, Ca, Al); with Mg it makes lighter flints, bullets and shells.

Common Mistakes to Avoid

Watch out

Writing Gd as $4 f^{8} 6 s^{2}$ : it is $4 f^{7} 5 d^{1} 6 s^{2}$ , keeping a half-filled set.
Counting $4 f$ electrons in $L n^{3 +}$ from the atom's configuration and getting it wrong; use $Z - 57$ .
Calling $C e^{4 +}$ a reducing agent: it is a strong oxidant ( $E^{\circ} = + 1.74$ V).
Writing the basicity order as $Lu (OH)_{3} > La (OH)_{3}$ : basicity falls from La to Lu.
Assuming all atomic radii fall smoothly: the metallic radii of Eu and Yb jump up.
Applying the spin-only formula to lanthanoid ions as if it were exact: orbital contribution is large ( $G d^{3 +}$ is the exception).
Giving lanthanoids states up to +7: only actinoids (Np, Pu) reach +7.
Thinking actinoids dissolve easily in nitric acid: an oxide film protects them; HCl attacks them.

Frequently Asked Questions

What are f-block elements?

f-Block elements are the elements in which the last electron enters the anti-penultimate (n-2)f subshell. They form two series: the 4f lanthanoids (Ce to Lu) and the 5f actinoids (Th to Lr). They are called inner transition elements and are placed in two rows below the main periodic table.

What is lanthanoid contraction and what causes it?

Lanthanoid contraction is the steady decrease in atomic and ionic radii from La to Lu, about 17 pm for the trivalent ions. It happens because each added $4 f$ electron shields the outer electrons poorly, so the effective nuclear charge rises and pulls the electron cloud inwards.

What are the consequences of lanthanoid contraction?

Lanthanoids have nearly identical chemistry and are hard to separate; 4d and 5d elements such as Zr and Hf have almost equal radii; basic strength of hydroxides falls from $La (OH)_{3}$ to $Lu (OH)_{3}$ ; and 5d metals are very dense with high ionisation enthalpies.

Why is +3 the most common oxidation state of lanthanoids?

Losing the two 6s electrons and one 5d or 4f electron needs a fairly low total energy, and the resulting ions are well stabilised by hydration and lattice energy. Other states appear only when they give $f^{0}$ , $f^{7}$ or $f^{14}$ , as in $C e^{4 +}$ , $E u^{2 +}$ and $Y b^{2 +}$ .

Why are lanthanoid ions coloured?

Their colour comes from f-f transitions of electrons within the 4f subshell. Ions with $f^{n}$ and $f^{14 - n}$ electrons have similar colours, while $f^{0}$ and $f^{14}$ ions such as $L a^{3 +}$ and $L u^{3 +}$ are colourless. The bands are sharp because 4f orbitals are shielded by 5s and 5p electrons.

How are actinoids different from lanthanoids?

Actinoids are all radioactive, show many more oxidation states (up to +7 for Np and Pu), form more complexes, show a larger contraction per element and form oxocations such as $U O_{2}^{2 +}$ . Lanthanoids stay mainly at +3 and only promethium is radioactive.

Which f-block questions are common in NEET?

NEET mostly asks direct NCERT statements: lanthanoid contraction and its consequences, the basicity order of hydroxides, the stable +3 state with $C e^{4 +}$ as an oxidant and $E u^{2 +}$ as a reductant, colourless ions, and the comparison of lanthanoids with actinoids.

How is the f-block tested in JEE Main?

JEE Main asks one or two questions per paper from this area: identifying ions with $f^{0}$ , $f^{7}$ or $f^{14}$ configurations, magnetic moment calculations, pairs with equal radii such as Zr and Hf, the maximum oxidation states of actinoids, and statements about lanthanoid contraction.

Previous year questions on f – Block Elements

7 questions from past papers, each with a step-by-step solution.

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