Class 12 Chemistry Notes

Chapter 4: The d- and f-Block Elements

Part 1: Introduction, Position in the Periodic Table & Electronic Configuration

4.1 Introduction

The d-block elements occupy the middle portion of the periodic table. In these elements, electrons are progressively filled into the (n−1)d orbitals.

The f-block elements are placed separately at the bottom of the periodic table. In these elements, electrons enter the 4f and 5f orbitals. These are called inner transition elements.

Classification of d- and f-Block Elements

1. Transition (d-block) Elements

These elements belong to Groups 3–12 of the periodic table.

Four Transition Series

Series	Orbitals Filled	Elements
First Transition Series	3d	Sc → Zn
Second Transition Series	4d	Y → Cd
Third Transition Series	5d	La, Hf → Hg
Fourth Transition Series	6d	Ac, Rf → Cn

2. Inner Transition (f-block) Elements

The f-block consists of two series.

Lanthanoids

4f series
Ce → Lu

Actinoids

5f series
Th → Lr

These elements are placed separately below the periodic table to keep the table compact.

IUPAC Definition of Transition Elements

A transition element is a metal whose atom or one of its common ions possesses an incomplete d-subshell.

Why are Zn, Cd and Hg NOT Transition Elements?

Although these elements belong to Group 12, they are not considered transition elements because:

Their atoms have completely filled d-orbitals.
Their common ions also have completely filled d-orbitals.

Element	Electronic Configuration	Transition Element?
Zn	3d¹⁰4s²	No
Cd	4d¹⁰5s²	No
Hg	5d¹⁰6s²	No

Their chemistry is usually studied along with transition metals because they occur at the end of the transition series.

Why are Transition Elements Important?

Transition elements differ from s- and p-block elements because of their partially filled d-orbitals.

They generally show:

Variable oxidation states
Formation of coloured ions
Complex compound formation
Catalytic activity
Paramagnetism
Alloy formation
High melting and boiling points

4.2 Position in the Periodic Table

The d-block lies between the s-block and p-block.

s-block     d-block               p-block

1 2 | 3 4 5 6 7 8 9 10 11 12 | 13 14 15 16 17 18

The d-orbitals of the penultimate shell are progressively filled. This gives rise to four transition series.

General Electronic Configuration

The general electronic configuration of transition elements is

(n − 1)d¹–¹⁰ ns¹–²

where

(n−1)d orbitals contain 1–10 electrons.
ns orbitals contain 1 or 2 electrons.

This general configuration has a few important exceptions.

Why Do Exceptions Occur?

The energy difference between (n−1)d and ns orbitals is very small.

Half-filled and completely filled d-orbitals are more stable.

Therefore, one electron from the ns orbital shifts to the d-orbital in certain cases.

Important Exceptions

Chromium (Cr)

Expected configuration

3d⁴4s²

Actual configuration

3d⁵4s¹

Reason:

The 3d⁵ configuration is half-filled and therefore more stable.

Copper (Cu)

Expected configuration

3d⁹4s²

Actual configuration

3d¹⁰4s¹

Reason:

The 3d¹⁰ configuration is completely filled and therefore more stable.

Electronic Configuration of the First Transition Series

Element	Atomic Number	Electronic Configuration
Sc	21	3d¹4s²
Ti	22	3d²4s²
V	23	3d³4s²
Cr	24	3d⁵4s¹
Mn	25	3d⁵4s²
Fe	26	3d⁶4s²
Co	27	3d⁷4s²
Ni	28	3d⁸4s²
Cu	29	3d¹⁰4s¹
Zn	30	3d¹⁰4s²

Why is Sc a Transition Element but Zn is Not?

Scandium

Electronic configuration

3d¹4s²

It has an incomplete d-orbital, so it is a transition element.

Zinc

Electronic configuration

3d¹⁰4s²

Its atom and common ion Zn²⁺ both have completely filled d-orbitals.

Therefore, Zn is not a transition element.

Important Board Points

✔ Transition elements belong to Groups 3–12.

✔ f-block elements are called inner transition elements.

✔ Lanthanoids = 4f series (Ce–Lu).

✔ Actinoids = 5f series (Th–Lr).

✔ General electronic configuration:
(n−1)d¹–¹⁰ ns¹–²

✔ Exceptions:

Cr → 3d⁵4s¹
Cu → 3d¹⁰4s¹

✔ Zn, Cd and Hg are not transition elements because they possess completely filled d-subshells in their atoms and common ions.

Quick Revision

Transition elements: Groups 3–12 with incomplete d-subshells.
Inner transition elements: Lanthanoids (4f) and Actinoids (5f).
General configuration: (n−1)d¹–¹⁰ ns¹–².
Exceptions: Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹).
Zn, Cd, Hg: Not transition elements due to fully filled d-orbitals.

Part 2: General Properties of Transition Elements (4.3)

4.3 General Properties of Transition Elements

Transition elements possess partially filled d-orbitals, which give them several unique physical and chemical properties.

These properties include:

Metallic character
High melting and boiling points
High enthalpy of atomisation
Variable oxidation states
Formation of coloured ions
Paramagnetism
Complex formation
Catalytic activity
Alloy formation
Formation of interstitial compounds

4.3.1 Physical Properties

Almost all transition metals show typical metallic properties.

Characteristics

Hard and strong
High tensile strength
Malleable
Ductile
Good conductors of heat
Good conductors of electricity
Metallic lustre
Low volatility

Exceptions: Zn, Cd, Hg and Mn differ in some physical properties.

High Melting and Boiling Points

Most transition metals have high melting and boiling points.

Reason

Both (n−1)d electrons and ns electrons participate in metallic bonding.

More electrons take part in bonding, making the metallic bonds stronger.

Hence,

Strong metallic bonding
High melting point
High boiling point

Across a transition series, melting points generally increase up to the middle of the series and then decrease.

Enthalpy of Atomisation

Definition

The energy required to convert one mole of a solid metal into gaseous atoms.

Trend

Transition metals have high enthalpy of atomisation.

Reason

Strong metallic bonding
Large number of unpaired electrons
Greater participation of d-electrons in bonding

Greater the number of unpaired electrons, stronger is the metallic bond and higher is the enthalpy of atomisation.

Board Question

Why do transition elements have high enthalpy of atomisation?

Answer:
Because a large number of unpaired d-electrons participate in metallic bonding, producing strong interatomic attraction.

4.3.2 Atomic and Ionic Sizes

Trend Across a Series

Atomic radius decreases only slightly from left to right.

Reason

Nuclear charge increases.
New electrons enter d-orbitals.
d-electrons shield nuclear charge poorly.
Effective nuclear charge increases.

Hence, atoms become slightly smaller across the series.

Comparison of 3d, 4d and 5d Series

Normally, atomic size should increase down a group.

However,

4d elements are larger than 3d elements.
5d elements have almost the same size as 4d elements.

Example:

Zr = 160 pm
Hf = 159 pm

This is due to lanthanoid contraction.

Lanthanoid Contraction

Definition

The gradual decrease in the size of lanthanoid atoms and ions with increasing atomic number is called lanthanoid contraction.

Cause

Poor shielding by 4f electrons.

As nuclear charge increases, electrons are pulled closer to the nucleus.

Effects

4d and 5d elements have nearly equal sizes.
Zr and Hf show very similar chemical properties.
Density of heavier transition metals increases.

4.3.3 Ionisation Enthalpy

Definition

The energy required to remove the outermost electron from an isolated gaseous atom.

Trend

Ionisation enthalpy generally increases across the transition series.

However, the increase is small and irregular.

Reasons

Increase in nuclear charge
Poor shielding by d-electrons
Stability of half-filled and completely filled d-orbitals

Why is the Trend Irregular?

Three factors influence ionisation enthalpy:

Nuclear attraction
Electron–electron repulsion
Exchange energy

Half-filled (d⁵) and completely filled (d¹⁰) configurations are especially stable because of higher exchange energy.

Important Points

ns electrons are removed before d-electrons during ion formation.
Second and third ionisation enthalpies increase more sharply than the first.
Cr⁺ (d⁵), Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) are particularly stable.

4.3.4 Oxidation States

One of the most important properties of transition elements is the variable oxidation state.

Reason

Both ns and (n−1)d electrons participate in chemical bonding.

Therefore, different numbers of electrons can be lost.

Common Oxidation States

Element	Common Oxidation States
Sc	+3
Ti	+2, +3, +4
V	+2, +3, +4, +5
Cr	+2, +3, +6
Mn	+2, +3, +4, +5, +6, +7
Fe	+2, +3
Co	+2, +3
Ni	+2
Cu	+1, +2
Zn	+2

Trend in Oxidation States

The maximum number of oxidation states is shown by elements near the middle of the series.
Manganese exhibits oxidation states from +2 to +7.
Early elements have fewer d-electrons.
Late elements have almost filled d-orbitals, making higher oxidation states less common.

Stability of Oxidation States

Up to manganese, the highest oxidation state approximately equals the total number of s and d electrons.
After manganese, higher oxidation states become less stable.

Examples:

Ti(IV)
V(V)
Cr(VI)
Mn(VII)

After Mn, the +2 and +3 oxidation states become more common.

Important Board Question

Which transition element does not show variable oxidation states?

Answer: Scandium (Sc), which commonly exhibits only the +3 oxidation state.

Quick Revision

Reasons for High Melting Point

Strong metallic bonding
d-electrons participate in bonding

Reasons for High Enthalpy of Atomisation

Large number of unpaired electrons
Strong interatomic attraction

Atomic Size

Slight decrease across a series
Due to increasing effective nuclear charge

Lanthanoid Contraction

Caused by poor shielding of 4f electrons
Makes 4d and 5d elements nearly equal in size

Ionisation Enthalpy

Generally increases
Trend is irregular due to exchange energy and stable d⁵/d¹⁰ configurations

Variable Oxidation States

Due to participation of both ns and d electrons
Mn shows the maximum number of oxidation states
Sc shows only +3

Part 3: Electrode Potentials, Reactivity, Magnetic Properties, Coloured Ions & Complex Formation

4.3.5 Trends in Standard Electrode Potentials

The standard electrode potential (E°) helps to understand the tendency of a metal to:

Lose electrons (oxidation)
Gain electrons (reduction)

For transition metals, the important reaction is: $M^{2+} + 2e^- \rightarrow M$ M2++2e−→M

Variation of E°(M²⁺/M)

Across the first transition series:

E° values generally become less negative from left to right.
This indicates a decrease in the tendency to form M²⁺ ions.

Reason

The value depends on:

Enthalpy of atomisation
Ionisation enthalpy
Hydration enthalpy of ions

The overall trend is not regular because these factors vary irregularly.

Important Exceptions

Manganese (Mn)

Mn has a more negative E° value because:

Mn²⁺ has a stable half-filled d⁵ configuration.
It is difficult to convert Mn²⁺ into Mn³⁺.

Zinc (Zn)

Zn has a negative E° value because:

Zn²⁺ has a stable d¹⁰ configuration.
Formation of Zn²⁺ is favourable.

Copper (Cu)

Copper has a positive E° value: $Cu^{2+}+2e^- \rightarrow Cu$ Cu2++2e−→Cu

E° = +0.34 V

Reason

Copper has:

High enthalpy of atomisation
High ionisation enthalpy
Low hydration enthalpy

Therefore, Cu does not easily form Cu²⁺ ions.

4.3.6 Trends in M³⁺/M²⁺ Standard Electrode Potentials

The stability of M³⁺ and M²⁺ ions depends on electronic configuration.

Important Examples

Scandium

Sc³⁺ is stable because it has noble gas configuration: $Sc^{3+} = [Ar]$ Sc3+=[Ar]

Manganese

Mn²⁺ is very stable because: $Mn^{2+}=3d^5$ Mn2+=3d5

(half-filled configuration)

Iron

Fe³⁺ is stable because: $Fe^{3+}=3d^5$ Fe3+=3d5

Zinc

Zn²⁺ is stable because: $Zn^{2+}=3d^{10}$ Zn2+=3d10

(fully filled configuration)

4.3.7 Stability of Higher Oxidation States

Transition metals show higher oxidation states because both:

ns electrons
(n−1)d electrons

can participate in bonding.

Stability of Higher Oxidation States

Higher oxidation states are generally stabilised by:

Oxygen
Fluorine

because they form strong bonds with metals.

Examples:

Oxidation State	Compound
+4	TiO₂
+5	V₂O₅
+6	CrO₃
+7	Mn₂O₇

Role of Oxygen

Oxygen stabilises high oxidation states by forming multiple bonds.

Examples:

VO₂⁺ → V(+5)
CrO₄²⁻ → Cr(+6)
MnO₄⁻ → Mn(+7)

Higher oxides become more acidic as oxidation state increases.

Example: $Mn_2O_7$ Mn2O7

is acidic and forms permanganic acid.

4.3.8 Chemical Reactivity and E° Values

Transition metals show different chemical reactivities.

General Behaviour

Many transition metals react with mineral acids.
Some metals are noble and resist attack by acids.

Examples:

Copper does not liberate hydrogen from dilute acids.
Titanium and vanadium become passive in dilute acids.

Strong Reducing Agents

Some transition metal ions act as reducing agents.

Examples:

Cr²⁺

Cr²⁺ is a strong reducing agent.

Reaction: $2Cr^{2+}+2H^+ \rightarrow 2Cr^{3+}+H_2$ 2Cr2++2H+→2Cr3++H2

Reason:

Cr²⁺ changes from d⁴ to the more stable d³ configuration.

Strong Oxidising Agents

Mn³⁺ and Co³⁺ are strong oxidising agents.

Reason:

They easily get reduced to more stable states.

4.3.9 Magnetic Properties

Transition elements show magnetic behaviour because of unpaired electrons.

Types of Magnetism

1. Diamagnetism

Substances are repelled by magnetic field.
All electrons are paired.

Example:

Zn²⁺ (d¹⁰)

2. Paramagnetism

Substances are attracted by magnetic field.
Contains unpaired electrons.

Most transition metal ions are paramagnetic.

Magnetic Moment Formula

$\mu=\sqrt{n(n+2)}$ μ=n(n+2)

where:

μ = magnetic moment in Bohr Magneton (BM)
n = number of unpaired electrons

Examples

Ion	Configuration	Unpaired Electrons
Sc³⁺	d⁰	0
Ti³⁺	d¹	1
V²⁺	d³	3
Mn²⁺	d⁵	5
Fe²⁺	d⁶	4
Cu²⁺	d⁹	1
Zn²⁺	d¹⁰	0

4.3.10 Formation of Coloured Ions

Many transition metal ions are coloured.

Reason

The d-orbitals split into different energy levels.

When light falls on the ion:

Electrons absorb certain wavelengths.
They jump from lower d-level to higher d-level.
The remaining transmitted/reflected light gives colour.

This is called d-d transition.

Important Colours

Ion	Colour
Ti³⁺	Purple
V²⁺	Violet
V³⁺	Green
Cr³⁺	Violet
Mn²⁺	Pink
Fe³⁺	Yellow
Fe²⁺	Green
Co²⁺	Pink
Ni²⁺	Green
Cu²⁺	Blue
Zn²⁺	Colourless

4.3.11 Formation of Complex Compounds

Definition

Complex compounds are compounds in which metal ions are surrounded by ions or molecules called ligands.

Examples: $[Fe(CN)_6]^{3-}$ [Fe(CN)6]3− $[Cu(NH_3)_4]^{2+}$ [Cu(NH3)4]2+ $[PtCl_4]^{2-}$ [PtCl4]2−

Why Do Transition Metals Form Complexes?

Because they have:

Small size
High ionic charge
Availability of vacant d-orbitals

Quick Revision

Electrode Potential

Depends on atomisation enthalpy, ionisation enthalpy and hydration enthalpy.
Cu has positive E° due to high energy requirement.

Higher Oxidation States

Stabilised by oxygen and fluorine.
Mn shows +7 oxidation state.

Magnetic Properties

Due to unpaired electrons.
Formula:

$\mu=\sqrt{n(n+2)}$

Coloured Ions

Due to d-d transition.
d⁰ and d¹⁰ ions are generally colourless.

Complex Formation

Due to small size, high charge and vacant d-orbitals.

Part 4: Catalytic Properties, Interstitial Compounds, Alloys & Important Compounds

4.3.12 Catalytic Properties of Transition Elements

Transition metals and their compounds are widely used as catalysts.

Why do Transition Elements Act as Catalysts?

Transition metals show catalytic activity because of:

Variable oxidation states
Ability to form intermediate compounds
Presence of vacant d-orbitals
Ability to adsorb reactant molecules on their surface

Examples of Catalytic Activity

1. Iron Catalyst

Used in the Haber process: $N_2 + 3H_2 \rightarrow 2NH_3$ N2+3H2→2NH3

Iron helps in breaking the strong N≡N bond.

2. Vanadium Pentoxide (V₂O₅)

Used in the Contact process: $2SO_2 + O_2 \rightarrow 2SO_3$ 2SO2+O2→2SO3

V₂O₅ acts as a catalyst.

3. Nickel

Nickel catalyses hydrogenation of vegetable oils.

Example: $Vegetable\ oil + H_2 \rightarrow Saturated\ fat$ Vegetable oil+H2→Saturated fat

4.3.13 Formation of Interstitial Compounds

Transition metals form compounds by trapping small atoms like:

Hydrogen (H)
Carbon (C)
Nitrogen (N)

inside the empty spaces of their crystal lattice.

These are called interstitial compounds.

Examples

Iron Carbide

$Fe_3C$ Fe3C

Titanium Carbide

$TiC$ TiC

Tungsten Carbide

$WC$ WC

Properties of Interstitial Compounds

They generally show:

High hardness
High melting point
High chemical stability
Metallic conductivity

Uses

Tungsten Carbide (WC)

Used for:

Cutting tools
Hard materials

because of its extreme hardness.

4.3.14 Alloy Formation

Transition metals easily form alloys with each other.

Reason

Transition metals have:

Similar atomic sizes
Similar crystal structures
Similar metallic properties

Therefore, atoms of one metal can replace atoms of another metal in the crystal lattice.

Important Alloys

Alloy	Composition
Stainless steel	Fe + Cr + Ni
Brass	Cu + Zn
Bronze	Cu + Sn
Nichrome	Ni + Cr + Fe
Duralumin	Al + Cu + Mg

Importance of Alloys

Alloys generally have:

Greater strength
Greater hardness
Better corrosion resistance

than pure metals.

4.4 Some Important Compounds of Transition Elements

The most important compounds in this chapter are:

Potassium dichromate (K₂Cr₂O₇)
Potassium permanganate (KMnO₄)

4.4.1 Potassium Dichromate (K₂Cr₂O₇)

Preparation

Potassium dichromate is prepared from chromite ore (FeCr₂O₄).

The preparation involves three steps:

Conversion of chromite ore into sodium chromate
Conversion of sodium chromate into sodium dichromate
Conversion of sodium dichromate into potassium dichromate

Step 1: Formation of Sodium Chromate

Chromite ore is heated with sodium carbonate in the presence of air.

Reaction: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$ 4FeCr2O4+8Na2CO3+7O2→8Na2CrO4+2Fe2O3+8CO2

Sodium chromate is formed.

Step 2: Conversion of Sodium Chromate to Sodium Dichromate

Sodium chromate solution is treated with sulphuric acid.

Reaction: $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ 2Na2CrO4+H2SO4→Na2Cr2O7+Na2SO4+H2O

Step 3: Formation of Potassium Dichromate

Sodium dichromate is treated with potassium chloride.

Reaction: $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$ Na2Cr2O7+2KCl→K2Cr2O7+2NaCl

Potassium dichromate crystallises because it is less soluble than sodium dichromate.

Properties of Potassium Dichromate

Physical Properties

Orange crystalline solid
Soluble in water
Strong oxidising agent

Oxidising Action of K₂Cr₂O₇

In acidic medium: $Cr_2O_7^{2-}+14H^+ +6e^- \rightarrow 2Cr^{3+}+7H_2O$ Cr2O72−+14H++6e−→2Cr3++7H2O

Chromium changes from: $Cr^{+6} \rightarrow Cr^{+3}$ Cr+6→Cr+3

Oxidation of Iodide Ion

$Cr_2O_7^{2-}+14H^+ +6I^- \rightarrow 2Cr^{3+}+3I_2+7H_2O$ Cr2O72−+14H++6I−→2Cr3++3I2+7H2O

Oxidation of Fe²⁺

$Cr_2O_7^{2-}+14H^+ +6Fe^{2+} \rightarrow 2Cr^{3+}+6Fe^{3+}+7H_2O$ Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2O

Uses of Potassium Dichromate

Laboratory oxidising agent
Manufacture of dyes
Leather industry
Preparation of chromium compounds

4.4.2 Potassium Permanganate (KMnO₄)

Preparation

KMnO₄ is prepared from pyrolusite ore (MnO₂).

Preparation occurs in two stages:

Formation of potassium manganate (K₂MnO₄)
Conversion of manganate into permanganate

Step 1: Formation of Potassium Manganate

MnO₂ is fused with KOH in presence of air or oxidising agent.

Reaction: $2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O$ 2MnO2+4KOH+O2→2K2MnO4+2H2O

Step 2: Formation of Potassium Permanganate

Potassium manganate is converted into KMnO₄.

Reaction: $2K_2MnO_4+Cl_2 \rightarrow 2KMnO_4+2KCl$ 2K2MnO4+Cl2→2KMnO4+2KCl

Properties of KMnO₄

Physical Properties

Dark purple crystalline solid
Soluble in water
Powerful oxidising agent

Oxidising Action of KMnO₄

In Acidic Medium

$MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O$ MnO4−+8H++5e−→Mn2++4H2O

In Neutral or Alkaline Medium

$MnO_4^-+2H_2O+3e^- \rightarrow MnO_2+4OH^-$ MnO4−+2H2O+3e−→MnO2+4OH−

Oxidation of Oxalate Ion

$2MnO_4^-+5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+}+10CO_2+8H_2O$ 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O

Uses of KMnO₄

Strong oxidising agent
Used in organic oxidation reactions
Used as disinfectant
Used in water treatment

Quick Revision

Catalysts

Fe → Haber process
V₂O₅ → Contact process
Ni → Hydrogenation

Interstitial Compounds

Small atoms occupy spaces in metal lattice.
Hard, high melting and stable.

Alloys

Formed due to similar atomic sizes.
Stronger than pure metals.

K₂Cr₂O₇

Orange solid
Strong oxidising agent
Cr changes +6 → +3

KMnO₄

Purple solid
Strong oxidising agent
Mn changes +7 → +2 (acidic medium)

Part 5: The f-Block Elements — Lanthanoids and Actinoids

4.5 The f-Block Elements

The elements in which the last electron enters the f-orbital are called f-block elements.

They are also known as inner transition elements because the differentiating electron enters an inner shell.

The f-block contains two series:

Lanthanoids (4f series)
Actinoids (5f series)

4.5.1 Lanthanoids

Definition

The elements in which the 4f orbitals are progressively filled are called lanthanoids.

They extend from: $Ce (58) \rightarrow Lu (71)$ Ce(58)→Lu(71)

(Usually La is also included because it resembles lanthanoids.)

Electronic Configuration of Lanthanoids

General electronic configuration: $[Xe]4f^{1-14}5d^{0-1}6s^2$ [Xe]4f1−145d0−16s2

where:

[Xe] represents xenon core
Electrons are gradually filled in 4f orbitals

Important Electronic Configurations

Element	Configuration
La (57)	[Xe] 5d¹6s²
Ce (58)	[Xe] 4f¹5d¹6s²
Gd (64)	[Xe] 4f⁷5d¹6s²
Lu (71)	[Xe] 4f¹⁴5d¹6s²

Oxidation States of Lanthanoids

The most common oxidation state is: $+3$ +3

Reason:

Lanthanoids lose:

Two 6s electrons
One 5d or 4f electron

to form Ln³⁺ ions.

Other Oxidation States

Some lanthanoids also show:

+2 State

Examples:

Sm²⁺
Eu²⁺
Yb²⁺

+4 State

Examples:

Ce⁴⁺
Tb⁴⁺

Stability of Oxidation States

Ce⁴⁺

Stable because: $Ce^{4+}=4f^0$ Ce4+=4f0

It has an empty f-subshell.

Eu²⁺

Stable because: $Eu^{2+}=4f^7$ Eu2+=4f7

It has a half-filled f-subshell.

Yb²⁺

Stable because: $Yb^{2+}=4f^{14}$ Yb2+=4f14

It has a completely filled f-subshell.

Physical Properties of Lanthanoids

Lanthanoids are:

Silvery white metals
Soft metals
Good conductors of heat and electricity
Highly electropositive
Reactive metals

Their hardness increases across the series.

Lanthanoid Contraction

Definition

The gradual decrease in the atomic and ionic size of lanthanoids with increasing atomic number is called lanthanoid contraction.

Cause

The main reason is:

Poor shielding effect of 4f electrons

The 4f electrons do not effectively shield the outer electrons from increasing nuclear charge.

Therefore:

Nuclear attraction increases.
Atomic size decreases.

Consequences of Lanthanoid Contraction

1. Similarity between Zr and Hf

Zirconium (Zr) and Hafnium (Hf) have almost identical sizes.

Therefore, their properties are very similar.

2. Difficulty in Separation of Lanthanoids

Because lanthanoids have very similar sizes and chemical properties, separating them is difficult.

3. Basic Strength of Hydroxides Decreases

Across the series: $La(OH)_3 > Lu(OH)_3$ La(OH)3>Lu(OH)3

Basic character decreases because ionic size decreases.

Chemical Properties of Lanthanoids

1. Reaction with Oxygen

Lanthanoids form oxides: $4Ln+3O_2 \rightarrow 2Ln_2O_3$ 4Ln+3O2→2Ln2O3

2. Reaction with Water

They react slowly with cold water: $2Ln+6H_2O\rightarrow2Ln(OH)_3+3H_2$ 2Ln+6H2O→2Ln(OH)3+3H2

3. Reaction with Halogens

They form halides: $2Ln+3X_2\rightarrow2LnX_3$ 2Ln+3X2→2LnX3

Examples:

LnCl₃
LnF₃

Colour of Lanthanoid Ions

Many lanthanoid ions are coloured due to:

f-f transitions

Examples:

Ion	Colour
Ce³⁺	Colourless
Pr³⁺	Green
Nd³⁺	Purple
Sm³⁺	Yellow
Eu³⁺	Pink

Magnetic Properties of Lanthanoids

The magnetic behaviour depends on:

Number of unpaired electrons in 4f orbitals

Examples:

La³⁺ → diamagnetic (4f⁰)
Gd³⁺ → highly paramagnetic (4f⁷)

4.5.2 Actinoids

Definition

The elements in which the 5f orbitals are progressively filled are called actinoids.

They range from: $Th(90)\rightarrow Lr(103)$ Th(90)→Lr(103)

Electronic Configuration of Actinoids

General configuration: $[Rn]5f^{1-14}6d^{0-1}7s^2$ [Rn]5f1−146d0−17s2

Important Examples

Element	Configuration
Th	[Rn] 5f⁰6d²7s²
Pa	[Rn] 5f²6d¹7s²
U	[Rn] 5f³6d¹7s²
Pu	[Rn] 5f⁶7s²

Oxidation States of Actinoids

Actinoids show a wider range of oxidation states than lanthanoids.

Common oxidation states: $+3,\ +4,\ +5,\ +6$ +3, +4, +5, +6

Examples

Element	Oxidation States
Th	+4
U	+3, +4, +5, +6
Np	+3 to +7
Pu	+3 to +7

Why Do Actinoids Show More Oxidation States?

Because:

Energy difference between 5f, 6d and 7s orbitals is small.
More electrons can participate in bonding.

Similarities Between Lanthanoids and Actinoids

Both show:

+3 oxidation state
Formation of coloured ions
Complex formation
Magnetic behaviour
Similar chemical properties

Differences Between Lanthanoids and Actinoids

Lanthanoids	Actinoids
4f orbitals filled	5f orbitals filled
Less variable oxidation states	More variable oxidation states
Mostly non-radioactive	All are radioactive
Less complex formation	Greater complex formation
+3 is dominant	+3, +4, +5, +6 common

Important Board Points

Lanthanoids

4f series
General configuration:

$[Xe]4f^{1-14}5d^{0-1}6s^2$ [Xe]4f1−145d0−16s2

Most common oxidation state: +3
Show lanthanoid contraction

Actinoids

5f series
General configuration:

$[Rn]5f^{1-14}6d^{0-1}7s^2$ [Rn]5f1−146d0−17s2

All radioactive
Show many oxidation states

Chapter Complete Revision Summary

d-Block Elements

Transition elements have incomplete d-orbitals.
General configuration:

$(n-1)d^{1-10}ns^{1-2}$ (n−1)d1−10ns1−2

Show:
- Variable oxidation states
- Colour
- Magnetism
- Complex formation
- Catalytic activity