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NCERT Class 12 Chemistry Chapter 8 Notes The d and f-Block Elements - Download PDF

NCERT Class 12 Chemistry Chapter 8 Notes The d and f-Block Elements - Download PDF

Edited By Shivani Poonia | Updated on Jul 04, 2025 12:42 PM IST

What do the fiery red glow of a ruby, the shimmer of a jet engine, and the flash of a camera have in common? They all owe their magic to the transition and inner transition elements. That glowing red in a ruby? It’s due to Cr³⁺ ions—a d-block element—replacing aluminum in its crystal, creating beauty through chemistry. In the d- and f-block, elements are also known as transition metals (d-block elements) and the inner transition metals (f-block elements). The d-block elements are elements of groups 3 to 12 of the periodic table. The elements of the f-block are those in which the 4f and 5f orbitals are filled.

This Story also Contains
  1. NCERT Notes For Class 12 Chapter 4 : Download PDF
  2. NCERT Notes For Class 12 Chapter 4
  3. The d- and f- block Elements Previous Year Question and Answer
  4. Approach to solve questions of Class 12 Chemistry The d-and f-block elements
  5. CBSE Class 12 Chemistry Chapter-wise Notes
  6. Subject-Wise NCERT Exemplar Solutions
  7. Subject-Wise NCERT Solutions
  8. NCERT Books and Syllabus
NCERT Class 12 Chemistry Chapter 8 Notes The d and f-Block Elements - Download PDF
NCERT Class 12 Chemistry Chapter 8 Notes The d and f-Block Elements - Download PDF

These NCERT Notes of class 12 the d- and f-block elements contains all the important topics that are being asked in the Class 12 board examinations and competitive exams. It explains the variable oxidation states of the transition metals, along with magnetic properties, chemical reactivity, electronic configuration, and the formation of colored compounds. These elements are crucial in fields ranging from metallurgy and catalysis to materials science and medicine. These NCERT notes are based on the latest CBSE Syllabus for Class 12 Chemistry and are beneficial for board examinations. The main topics covered in the d and f-block Elements class 12 notes are the physical properties of d-block elements, some important compounds of transition elements, lanthanoids, actinoids, lanthanide contraction, etc.

Also, students can refer,

NCERT Notes For Class 12 Chapter 4 : Download PDF

These comprehensive notes cover all important concepts of The d- and f- block elements in a simple and exam-friendly format. Students can easily download the PDF by clicking the button given below.

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NCERT Notes For Class 12 Chapter 4

The d- and f-block elements are known as transition and inner-transition metals, respectively. These elements display a wide range of oxidation states, form complex compounds, and play vital roles in catalysis and metallurgy. Below is a structured, topic-wise breakdown of all the concepts covered in the NCERT textbook to help you learn this chapter effectively. These notes are useful for last minute revision as well as exam preparation.

Position of d block in the Periodic Table

In the periodic table, the d-block elements are found in the intermediate region of the s- and p-block elements. Because of its location between s- and p-block components, it was given the term 'transition.'

Electronic Configuration of d Block Elements

(n-1) d1-10ns1-2 is the electronic configuration of d-block elements. They have two outer shells that aren't complete.

Where (n–1) denotes inner d orbitals with electrons ranging from 1 to 10, and ns denotes the outermost orbital, which may have one or two electrons (n-1) The electronic configurations of Zn, Cd, and Hg are represented by d10s2.

D-block elements have a variable valency that varies by one unit.

Table : Electronic configuration of 4th period transition elements

ElementSymbolAtomic Number (Z)Electronic Configuration (Outer Orbitals)
ScandiumSc21[Ar] 3d¹ 4s²
TitaniumTi22[Ar] 3d² 4s²
VanadiumV23[Ar] 3d³ 4s²
ChromiumCr24[Ar] 3d⁵ 4s¹ (exception)
ManganeseMn25[Ar] 3d⁵ 4s²
IronFe26[Ar] 3d⁶ 4s²
CobaltCo27[Ar] 3d⁷ 4s²
NickelNi28[Ar] 3d⁸ 4s²
CopperCu29[Ar] 3d¹⁰ 4s¹ (exception)
ZincZn30[Ar] 3d¹⁰ 4s²

Table : Electronic configuration of 5th period transition elements

ElementSymbolAtomic Number (Z)Electronic Configuration (Outer Orbitals)
YttriumY39[Kr] 4d¹ 5s²
ZirconiumZr40[Kr] 4d² 5s²
NiobiumNb41[Kr] 4d⁴ 5s¹ (exception)
MolybdenumMo42[Kr] 4d⁵ 5s¹ (exception)
TechnetiumTc43[Kr] 4d⁵ 5s²
RutheniumRu44[Kr] 4d⁷ 5s¹ (exception)
RhodiumRh45[Kr] 4d⁸ 5s¹ (exception)
PalladiumPd46[Kr] 4d¹⁰ (exception – no 5s electron)
SilverAg47[Kr] 4d¹⁰ 5s¹
CadmiumCd48[Kr] 4d¹⁰ 5s²

Table : Electronic configuration of 6th period transition elements

ElementSymbolAtomic Number (Z)Electronic Configuration (Outer Orbitals)
HafniumHf72[Xe] 4f¹⁴ 5d² 6s²
TantalumTa73[Xe] 4f¹⁴ 5d³ 6s²
TungstenW74[Xe] 4f¹⁴ 5d⁴ 6s²
RheniumRe75[Xe] 4f¹⁴ 5d⁵ 6s²
OsmiumOs76[Xe] 4f¹⁴ 5d⁶ 6s²
IridiumIr77[Xe] 4f¹⁴ 5d⁷ 6s²
PlatinumPt78[Xe] 4f¹⁴ 5d⁹ 6s¹ (exception)
GoldAu79[Xe] 4f¹⁴ 5d¹⁰ 6s¹ (exception)
MercuryHg80[Xe] 4f¹⁴ 5d¹⁰ 6s²

General Properties of the Transition Elements (d-Block)

Physical Properties

1)Metallic characteristics :

All transition metals have a metallic character. They are excellent heat and electricity conductors. They are abrasive and persistent. They are malleable, ductile, and resonant since they are metal. They combine with other metals to form alloys. They can be found in three different structures: face-centred cubic (fcc), hexagonal close-packed (hcp), and body-centred cubic (bcc). Within the atoms of transition elements, there is both covalent and metallic bonding.

Figure shows the trends in melting points of transition elements.

2)Atomic size:

As the atomic number increases, the atomic radii of the elements in the 3d series decrease. Because of lanthanoid contraction, the atomic radii rise from 3d to 4d, and the atomic radii of the 4d and 5d transition series are quite close. Zirconium and Hafnium, are examples. The rise in the density of elements is caused by a decrease in the metallic radius due to an increase in the atomic mass. As a result, the density of titanium to copper increases.

Figure shows the trends in atomic radii of transition elements.

3)Ionisation enthalpy:

Because transition elements are tiny, they have a high ionisation energy. Since their ionisation potentials lie between S and P-block elements, they are less electropositive than s-block elements. Transition elements form covalent compounds. The ionisation potentials of the d-block elements increase from left to right due to the screening effect of the additional electrons added to the (n-1) d subshell. Because of the stable electronic configuration, the second ionisation energy increases with the increase in atomic number in the first transition series. Ionization energy reduces as moves through the group whereas the amount of ionisation energy increases across the period.

4)Oxidation state:

The small energy difference between the ns and (n -1) d orbitals cause transition elements to have different oxidation states. (n -1) d electrons participate in bonding with ns electrons. Scandium, on the other hand, does not have different oxidation states due to the limited number of electrons available for bonding. Zinc has fewer orbitals accessible for bonding due to the existence of more d electrons, and so does not have a variable oxidation state. The elements in the 8th group have the highest oxidation state among the d-block elements. Manganese, which belongs to the 7th group, has the highest oxidation state among the 3d –series elements. Ruthenium, which belongs to the 8th group, has the highest oxidation state among the 4d-Series elements. Osmium, which belongs to the 8th group, has the highest oxidation state of all the elements in the 5d series.

The table shows Oxidation States of the first row Transition Metal (the most common ones are in bold types)

ElementSymbolCommon Oxidation States
ScandiumSc+3
TitaniumTi+2, +3, +4
VanadiumV+2, +3, +4, +5
ChromiumCr+2, +3, +6
ManganeseMn+2, +3, +4, +6, +7
IronFe+2, +3
CobaltCo+2, +3
NickelNi+2, +3
CopperCu+1, +2
ZincZn+2

5) Trends in the M2+/M Standard Electrode Potentials:

The stability of M2+ ions in aqueous medium depends on three factors:

(i)Enthalpy of atomisation

(ii) Summation of first and second ionization enthalpies
(iii) Hydration enthalpy

An elements in Mn2+ state in aqueous medium is more stabler if the electrode potential (M2+/M) value of more negative. Across the period, the tendency to form M2+ ion decreases. Except copper, all elements of first transition series show negative values of electrode potentials. The exceptional behaviour of the copper due to low enthalpy of atomisation and very high summation of first and second ionization enthalpies which is not compensated by its hydration enthalpy ( Cu2+. Because of positive electrode potential, copper doesnot liberate hydrogen gas from dilute acids and reacts only with oxidizing acids such as nitric acid and hot concentrated sulphuric acid.
The electrode potentials of Mn,Ni and Zn are more negative than expected. However, the electrode potential values of Mn and Zn are lowered because of low second ionization enthalpies while Ni has exceptionally more negative electrode potential due to its high hydration enthalpy.
Across the period M2+/M value decreases because of increase in first and second ionization enthalpies.

Table showing thermochemical data (kJ mol-1) for the first row Transition Elements and the Standard Electrode Potentials for the Reduction of MII to M.

ElementΔₐH⁰ (M) Atomization EnthalpyΔᵢH₁⁰ 1st Ionization EnthalpyΔᵢH₂⁰ 2nd Ionization EnthalpyΔhydH⁰ (M²⁺) Hydration EnthalpyE⁰/V (M²⁺/M)
Ti4696561309–1866–1.63
V5156501414–1895–1.18
Cr3986531592–1925–0.90
Mn2797171509–1862–1.18
Fe4187621561–1998–0.44
Co4277581644–2079–0.28
Ni4317361752–2121–0.25
Cu3397451958–2121+0.34
Zn1309061734–2059–0.76

Figure showing observed and calculated values for the standard electrode potentials (M2+/M°) of the elements Ti to Zn.

6)Trends in the M3+/M2+ Standard Electrode Potentials:

Table showing values of M3+/M2+ Standard Electrode Potentials for First-Row Transition Metals.

ElementIon Pair (M³⁺/M²⁺)E(M3+/M2+) (V)
ScSc³⁺/Sc²⁺–2.03
TiTi³⁺/Ti²⁺–0.37
VV³⁺/V²⁺–0.26
CrCr³⁺/Cr²⁺–0.41
MnMn³⁺/Mn²⁺+1.51
FeFe³⁺/Fe²⁺+0.77
CoCo³⁺/Co²⁺+1.82
NiNi³⁺/Ni²⁺+1.60
CuCu³⁺/Cu²⁺+0.15 (less common)
ZnZn³⁺/Zn²⁺Not commonly observed (Zn³⁺ is unstable)

The low value for scandium indicates the exceptional stability of Sc³⁺, which possesses a noble gas configuration. On the other hand, zinc shows the highest value, as removing an electron from Zn²⁺ will destabilize its stable d¹⁰ configuration. The relatively high value for manganese suggests the exceptional stability of Mn²⁺ (d⁵), while the lower value for iron reflects the additional stability of Fe³⁺ (also d⁵). Similarly, the low value for vanadium is linked to the stability of V²⁺, which has a half-filled t₂g orbital configuration.

7)Trends in Stability of Higher Oxidation States:

Table below shows the stable halides of 3d transition metals. Highest oxidation states are seen in TiX4,VF5, and CrF6Mn7+ does not form simple halides but exists as MnO3 F, and beyond Mn , only FeX3 and CoF3 are common trihalides.

Fluorine stabilizes high oxidation states due to strong lattice energies or bond enthalpies (e.g., VF5,CrF6 ). Lower halides like VX2 and CuX are unstable. Cu2+ does not form CuI2 because it oxidizes Ito I2.

2Cu2++4ICu2I2( s)+I2

However, many copper (I) compounds are unstable in aqueous solution, due to lower hydration energy compared to Cu2+ and undergo disproportionation.

2Cu+Cu2++Cu

In oxides, oxygen stabilizes higher oxidation states better than fluorine. The highest oxidation number matches the group number from Sc2O3 to Mn2O7. Beyond this, stable higher oxides are rare. Oxocations like VO2+,VO2+, and TiO2+ are common. Oxygen's ability to form multiple bonds explains its effectiveness in stabilizing high oxidation states, as seen in tetrahedral oxoanions like [MO4]nfor V,Cr, and Mn .

Oxidation Number+6+5+4+3+2+1
CrF₆VF₅, CrF₅TiX₄, VX₄, CrX₄, MnF₄TiX₃, VX₃, CrX₃, MnF₃, FeX₃¹, CoF₃TiX₂³, VX₂, CrX₂, MnX₂, FeX₂, Co X₂, NiX₂, CuX₂², ZnX₂CuX¹

8) Chemical Reactivity and Eo Values:

Transition metals show a wide range of chemical reactivity. Many are reactive enough to dissolve in mineral acids, though some, known as noble metals, resist attack by simple acids. In the first transition series, all metals except copper are generally more reactive and can be oxidized by 1 M H+, though some like titanium and vanadium are passive in dilute non-oxidising acids at room temperature due to slow reaction rates.

The standard electrode potentials, E(M2+/M), show a decreasing tendency to form M2+ ions across the series. This trend toward less negative E values correlates with increasing first and second ionisation enthalpies. However, Mn,Ni, and Zn deviate from this trend:

  • Mn2+(d5) and Zn2+(d10) are stabilized by half-filled and fully filled d-subshells, respectively.
  • Ni's E is influenced by its highly exothermic hydration enthalpy.

For the redox couple M3+/M2+, ions like Mn3+ and Co3+ act as strong oxidising agents in aqueous solutions, while Ti2+,V2+, and Cr2+ serve as strong reducing agents, capable of liberating hydrogen from dilute acids.

For example:

2Cr2+(aq)+2H+(aq)2Cr3+(aq)+H2( g)

9)Magnetic properties:

When a magnetic field is applied, substances exhibit mainly two types of behavior: diamagnetism (repelled by the field) and paramagnetism (attracted by the field). Materials that are very strongly attracted are classified as ferromagnetic, an extreme form of paramagnetism. In paramagnetic substances, the magnetic behavior arises from unpaired electrons; each unpaired electron contributes a magnetic moment (about 1.73 BM) primarily due to its spin, as the orbital contribution is largely quenched in first-row transition metal compounds. Transition metals and most of their compounds are paramagnetic in nature. The spin only formula is used to compute the magnetic moment of these elements.

μ =n(n-1)

where n is the number of unpaired electrons

Table shows the observed and calculated magnetic moment.

Ion

Configuration

Unpaired Electrons

Calculated Magnetic Moment (BM)

Observed Magnetic Moment (BM)

Sc³⁺3d⁰00.000.00
Ti³⁺3d¹11.731.75
Ti²⁺3d²22.842.76
V²⁺3d³33.873.86
Cr²⁺3d⁴44.904.80
Mn²⁺3d⁵55.925.96
Fe²⁺3d⁶44.905.3 – 5.5
Co²⁺3d⁷33.874.4 – 5.2
Ni²⁺3d⁸22.842.9 – 3.4
Cu²⁺3d⁹11.731.8 – 2.2
Zn²⁺3d¹⁰00.000.00

10)Formation of coloured compounds

There is an excitation of an electron from a lower to a higher level, and the energy of excitation corresponds to the energy difference between the levels. This frequency is usually in the visible range. The existence of unpaired or incomplete d-orbitals gives the transition metal ions their colour. The transition metal cations' absorption of visible light and hence their colourful character is due to the transition of one or more unpaired d-electrons from a lower to a higher level within the same d-subshell. This transition requires a small amount of visible light energy.Sc3+, Cu+, Zn2+, and Ti4+ have either a completely empty or completely filled 3d-orbital, i.e. no unpaired d-electron, and hence appear colourless.

The table shows the colors of some of the First Row (aquated) Transition Metal Ions

IonElectronic ConfigurationColour (in aqueous solution)
Ti³⁺ (aq)3d¹Purple
V²⁺ (aq)3d³Violet
V³⁺ (aq)3d²Green
Cr³⁺ (aq)3d³Violet
Mn²⁺ (aq)3d⁵Pale pink
Fe²⁺ (aq)3d⁶Light green
Fe³⁺ (aq)3d⁵Yellow to brown
Co²⁺ (aq)3d⁷Pink
Ni²⁺ (aq)3d⁸Green
Cu²⁺ (aq)3d⁹Blue
Zn²⁺ (aq)3d¹⁰Colourless

11)Formation of complex compounds

Transition metal cations have a strong tendency to form complexes with several molecules or ions known as ligands. Such complexes are named coordinate complexes because the links involved in their formation are coordinated. The structure of these complex ions might be linear, square, planar, tetrahedral, or octahedral, depending on the nature of the metal ion hybridization. Weak ligands like CO and NO form complexes only when transition metals are in a zero oxidation state. This is because these ligands contain empty orbitals in the donor atom in addition to the lone pair of electrons. Highly electronegative and basic ligands like F-, Cl- can form complexes with transition metals despite being in high oxidation states due to the presence of small, highly charged or neutral ligands with lone pairs of electrons that can form strong sigma bonds by donating a lone pair of electrons. The stability of complexes increases as the atomic number increases in a transition series. The transition metal atom has several oxidation states; the higher the valence, the more stable the complex.

12)Catalytic Properties

Transition metals and their compounds are well-known catalysts, primarily due to their ability to exhibit variable oxidation states and form complexes. Examples include V2O5 in the Contact Process, finely divided Fe in the Haber Process, and Ni in catalytic hydrogenation. Catalysis on a solid surface typically involves the adsorption of reactant molecules onto the catalyst's surface. First-row transition metals utilize their 3d and 4s electrons to form bonds with reactants, which helps to increase their concentration at the surface and weaken internal bonds, thereby lowering the activation energy.

Their variable oxidation states make transition metal ions particularly effective in redox reactions. For instance, Fe3+ catalyzes the reaction between iodide and persulphate ions as follows:

2I+S2O82I2+2SO42

The catalytic mechanism involves these redox steps:

2Fe3++2I2Fe2++I22Fe2++S2O822Fe3++2SO42

13)Formation of interstitial compounds

Interstitial compounds with non-stoichiometric composition are formed when transition elements combine with tiny atoms such as H, B, C, N, and others. The resulting interstitial compounds are chemically inert and have higher melting points than pure metals. These compounds are strong and durable, and they maintain metallic conductivity.

eg : TiH1.3,VH0.54

14)Alloy formation

Alloys are homogenous mixes of more than one metal. Transition metals are generally smaller in size. So they can displace another metal from the crystal lattices and there by forming alloys. Alloys are hard and have high melting points. Ferrous alloys include chromium, vanadium, tungsten, manganese, and molybdenum, for example. Brass (alloy of copper + zinc), stainless steel, bronze (alloy of copper + tin), and so on are some additional examples.

Some Important Compounds of Transition Metals

Oxides and Oxoanions of Metals

  1. Potassium dichromate (K2Cr2O7)

Preparation

Initially, chromite ore was fused with sodium or potassium carbonate in the presence of air.

4FeCr2O4+8Na2CO3+7O28Na2CrO4+2FeO3+8CO2

A sodium chromate solution is filtered and then acidified with a sulfuric acid solution, yielding an orange sodium dichromate solution that can be crystallized.

2Na2CrO4+2H+>Na2Cr2O7+2Na++H2O

Because sodium dichromate is more soluble than potassium dichromate, it is fused with KCl, resulting in the formation of orange potassium dichromate crystals.

Na2Cr2O7+2KClK2Cr2O7+2NaCl

The dichromates and chromates exist in equilibrium at pH 4 and can be interconnected.

2CrO42+2H2+Cr2O72+H2O

Cr2O72+2OH2CrO42+H2O

In the presence of an acidic medium, the yellow colour of chromate changes to an orange-colored dichromate, whereas in the presence of a basic medium, the dichromate transforms back to chromate.

2CrO42+2H+2HCrO4(Hydrogen chromate)

2HCrO4Cr2O72+H2O Dichromate (orange)

Properties

  • In an acidic media, potassium dichromate is a powerful oxidizing agent.

Cr2O72+14H++6 electron 2Cr3++7H2O

  • Heat causes potassium dichromate to decompose, resulting in the formation of potassium chromate, chromic oxide, and oxygen

4 K2Cr2O7-Heat 4 K2CrO4+2CrO3+3O2

  1. Potassium permanganate (KMnO4)

  • Potassium Permanganate is a dark purple solid made up of two ions: potassium and permanganate. Because it is a strong oxidizing agent with medicinal qualities, it is frequently used to clean wounds and treat dermatitis.

  • In the presence of air or an oxidizing agent, fusion of powdered Pyrolusite ore with an alkali metal hydroxide like KOH results in the formation of dark green potassium Manganate, which disproportionates in a neutral or acidic solution and results in the formation of potassium permanganate.

2MnO2+4KOH+O22 K2MnO4+2H2O3MnO42+4H+2MnO4+MnO2+2H2O

  • Commercially, potassium permanganate is made by alkaline oxidative fusion of Pyrolusite ore followed by electrolytic oxidation of manganate (4) ion.

2MnO2+4KOH+O22 K2MnO4+2H2OMnO42++( electrolytic oxidation )MnO4+e

f-block Elements

f-block elements are those that have gradually filled f orbitals. The elements of the inner transition metals' 4f series are known as lanthanoids, while the elements of the 5f series are known as actinoids.

Lanthanoid

  • Lanthanoid has an electronic configuration of [Xe]4fn+15d1 6s2 or [Xe] 4fn5d16 s2, with a valence shell electronic configuration of 4 f1-146s2
  • They have oxidation states of +3, +2, and +4
Atomic NumberNameSymbolElectronic ConfigurationRadii/pm
LnLn²⁺Ln³⁺Ln⁴⁺LnLn³⁺
57LanthanumLa5d¹6s²4f⁰187106
58CeriumCe4f¹5d¹6s²4f¹4f⁰183103
59PraseodymiumPr4f³6s²4f²4f¹182101
60NeodymiumNd4f⁴6s²4f³4f²18199
61PromethiumPm4f⁵6s²4f⁴4f³18198
62SamariumSm4f⁶6s²4f⁵4f⁴18096
63EuropiumEu4f⁷6s²4f⁷4f⁶19995
64GadoliniumGd4f⁷5d¹6s²4f⁷4f⁷5d¹18094
65TerbiumTb4f⁹6s²4f⁸4f⁷17892
66DysprosiumDy4f¹⁰6s²4f⁹4f⁸17791
67HolmiumHo4f¹¹6s²4f¹⁰4f⁹17689
68ErbiumEr4f¹²6s²4f¹¹4f¹⁰17588
69ThuliumTm4f¹³6s²4f¹²4f¹¹17487
70YtterbiumYb4f¹⁴6s²4f¹⁴4f¹³17386
71LutetiumLu4f¹⁴5d¹6s²4f¹⁴86
  • The first members of this series are mildly reactive to calcium, and their behaviour gradually resembles that of aluminium as the atomic number increases.
  • Lanthanides mix with hydrogen when exposed to mild heat.
  • Carbides and halides are formed when they are heated with carbon (in the presence of halogens while burning).
  • Lanthanides react with dilute acids to produce hydrogen gas.

General Characteristics:

Lanthanoids are soft, silvery-white metals that tarnish in air and show increasing hardness with atomic number (samarium is steel-hard). Their melting points range from 10001200 K, with Sm melting at 1623 K . They have metallic structures, conduct heat and electricity well, and show smooth trends in propertiesexcept Eu, Yb, Sm, and Tm. Most Ln3+ ions are coloured due to f-electron transitions, except La3+ and Lu3+. The absorption bands are narrow. All Ln3+ ions (except f0 and f14 ) are paramagnetic.

Their first and second ionisation enthalpies ( 600 and 1200 kJ/mol ) are similar to calcium. Variations in third ionisation enthalpy reflect stability of empty, half-filled, and full f-orbitals, seen in La, Gd, and Lu. Chemically, early lanthanoids resemble calcium, while heavier ones behave more like aluminium. The standard electrode potentials lie between -2.2 and -2.4 V , with Eu at -2.0 V . They form hydrides, oxides (Ln2O3), hydroxides (Ln(OH)3), halides, carbides, and liberate H2 from acids. Hydroxides are strongly basic, like those of alkaline earth metals.

Uses:

  • Mischmetall ( 95%Ln+Fe ) is used in Mg-alloys for bullets and lighters.
  • Mixed oxides act as petroleum cracking catalysts.
  • Some Ln oxides serve as phosphors in TVs and fluorescent displays.

Lanthanide Contraction

As we move through the lanthanoid series, the atomic number, or the number of electrons and protons, gradually increases by one. The effective nuclear charge increases as electrons are added to the same shell. As the atomic number rises, so does the number of electrons in the 4f orbital, which has a weak shielding effect and increases the effective nuclear charge on the outside electrons. As a result, the size of lanthanoids reduces with increasing atomic number, a phenomenon known as lanthanoid contraction. The features of the second and third transition series are comparable as a result of lanthanoid contraction.

Actinoids

The elements Th to Lr are members of the Actinoids series, which consists of 14 elements. They are radioactive substances. The former elements have extended half-lives, whereas the subsequent elements, such as lawrencium with atomic number 103, have half-lives ranging from one day to three minutes. The electronic configuration of actinoids is 7s2, with inconstant occupancy of the 5f and 6d subshells. Because of the poor screening effect of nuclear charge exerted by the f electrons, the ionic radii gradually decrease across the series. Actinoid contraction is the term for this.

Comparison Between Lanthanoid and Actinoid

  • Atomic and ionic sizes

Actinoids' ionic radii gradually decrease across the series, similar to lanthanoids, due to the poor screening effect of nuclear charge exerted by the f electrons.

  • Oxidation states

The lanthanide have oxidation states of +3. Due to the additional stability of fully-filled and half-filled orbitals, some elements may show + 2 and + 4 oxidation states.

Actinoids, on the other hand, have a + 3 oxidation state. Due to the similar energies of 5f, 6d, and 7s, they also have different oxidation states.

ElementSymbolAtomic NumberCommon Oxidation StatesMost Stable Oxidation State(s)
ActiniumAc89+3+3
ThoriumTh90+4+4
ProtactiniumPa91+3, +4, +5+5
UraniumU92+3, +4, +5, +6+6
NeptuniumNp93+3, +4, +5, +6, +7+5
PlutoniumPu94+3, +4, +5, +6, +7+4, +5
AmericiumAm95+2, +3, +4, +5, +6+3
CuriumCm96+3, +4+3
BerkeliumBk97+3, +4+3
CaliforniumCf98+3, +4+3
EinsteiniumEs99+3+3
FermiumFm100+2, +3+3
MendeleviumMd101+2, +3+3
NobeliumNo102+2, +3+2
LawrenciumLr103+3+3
  • Chemical reactivity

The lanthanide series' earlier members are more reactive. With rising atomic numbers, they resemble Al. Finely split actinoids are highly reactive metals that produce a mixture of oxide and hydride when introduced to boiling water. Actinoids mix with the majority of non-metallic elements at moderate temperatures. The action of alkalies has little effect on actinides, while nitric acid has a minor effect due to the creation of a protective oxide layer.

General Characteristics and Comparison with Lanthanoids

The actinoids are silvery, highly reactive metals with varied structures due to greater irregularities in metallic radii compared to lanthanoids. They readily react with boiling water, non-metals, and HCl, but are less affected by HNO₃ due to protective oxide layers. Alkalis have no effect. Their magnetic properties are complex, and though the variation follows the number of unpaired 5f electrons, lanthanoids show higher magnetic susceptibilities.The early actinoids have lower ionisation enthalpies than lanthanoids, as 5f electrons are more shielded and less tightly held, making them more available for bonding.

While later actinoids resemble lanthanoids more closely, even early actinoids show gradual property trends. Both lanthanoid and actinoid contractions affect the elements that follow, though the lanthanoid contraction has a greater chemical impact due to better-known properties of subsequent elements.

Some Applications ApplicationsApplications of d- and f-Block Element

Iron and steel are among the most vital construction materials, produced by reducing iron oxides, removing impurities, and adding carbon along with alloying elements like Cr,Mn, and Ni . Some metal compounds serve specialized roles-for example, TiO is used in pigments, and MnO2 powers dry cells. The battery industry depends on Zn and Ni/Cd systems. Group 11 elements (Cu, Ag, Au) are traditionally known as coinage metals, though today's UK coins use copper-coated steel or Cu/Ni alloys, while Ag and Au coins are mostly for collectors.

Many transition metals and their compounds are key catalysts in industry:

  • V2O5 for oxidizing SO2 in sulphuric acid production
  • TiCl4+Al(CH3)3 (Ziegler catalyst) for making polythene
  • Fe in the Haber process (NH3 production from N2/H2)
  • Ni for hydrogenating fats
  • PdCl2 in the Wacker process (oxidation of ethyne to ethanal)
  • Ni complexes in polymerizing alkynes and aromatics
  • The photographic industry relies on AgBr for its light sensitivity

The d- and f- block Elements Previous Year Question and Answer

Question: Consider the following reactions

A+NaCl+H2SO4CrO2Cl2+ Side Products  Little  amount CrO2Cl2( Vapour )+NaOH B+NaCl+H2O B+H+C+H2O

The number of terminal ' O ' present in the compound ' C ' is______ .

Answer:

Cr2O72+NaCl+H2SO4CrO2Cl2

CrO2Cl2( Vapour )+NaOHNa2CrO4+NaCl+H2O

Na2CrO4+HNa2Cr2O7+H2O

Na2Cr2O72Na+Cr2O72

No of terminal " O " =6

Hence, the answer is 6.

Question: The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. Cr2+
B. Fe2+
C. Fe3+
D. Co2+
E. Mn3+

Choose the correct answer from the options given below:

  1. A, C and E only
  2. A, D and E only
  3. B and E only
  4. A, B and E only

Answer: Given magnetic moment =4.9 B.M.
We know M.M =n(n+2) B.M.
Where, n No. of unpaired e

4.9=n(n+2)

Solving this equation, we find that n=4.

(A) 24Cr2+[Ar]3 d4 (4 unpaired e)
(B) 26Fe2+[Ar]3 d6 (4 unpaired e)
(C) 26Fe3+[Ar]3 d5 (5 unpaired e)
(D) 27Co2+[Ar]3 d7(3 unpaired e)
(E) 25Mn3+[Ar]3 d4(4 unpaired e )

The metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are Cr2+,Fe2+, and Mn3+. Because, the value of n=4 for these ions.

Hence, the correct answer is option (4).

Question: The element having [Ar] 3 d104 s1 electronic configuration is

  1. Cu
  2. Zn
  3. Cr
  4. Mn

Answer: Given that the electronic configuration is: [Ar] 3d104s1

The number of electrons in Argon is 18, there are 10 electrons in the d orbital and 1 electron in the s orbital.

According to this configuration, the number of electrons is 29.

The given electronic configuration and the number of electrons correspond to the element Copper.

Copper has an atomic number of 29, and its electronic configuration is [Ar] 3d104s1.

Hence, the correct answer is option (1).

Approach to solve questions of Class 12 Chemistry The d-and f-block elements

The chapter The d- and f-Block Elements talks about the properties, trends and chemical behavior of transition and inner-transition elements which play a crucial role in industrial chemistry, catalysis and coordination compounds. Students should follow structured approach given below to grasp the concepts of the chapter.

1. Read NCERT Line-by-Line

This chapter is highly factual and many board questions come directly from NCERT.
Highlight trends, exceptions, colour of compounds, and oxidation states.
Pay close attention to the table of electronic configurations of d - and f-block elements.
2. Understand the electronic configuration

Try to memorize the general outer electronic configurations like
d-block- [noble gas] (n1)d110ns02
f-block- [noble gas] (n2)f114(n1)d01ns2
Also learn and remember exceptions (e.g., Cr, Cu, Mo, Ag, Au).

3. Focus on Variable Oxidation States

Learn the common oxidation states of each group of transition elements.
Practice identifying elements that show multiple oxidation states (e.g., Mn: +2 to +7).
Understand which oxidation states are more stable and why.
Use this knowledge to solve questions on redox reactions and coloured ions.
4. Colour and Magnetic Properties

Understand that colour in transition metal compounds is due to d-d electronic transitions.
Learn which ions are coloured (partially filled d-orbitals) and which are colourless (e.g., Zn2+ ).
Magnetic behavior depends on number of unpaired electrons; use this to calculate magnetic moment
μ=n(n+2)BM

where n= number of unpaired electrons.

5. Practice conceptual and reasoning-based questions

Practice question like- Why some metals are coloured? Why Zn,Cd,Hg are not transition elements? Why transition metals form complexes? Also, learn the differences between lanthanides and actinides.
Be sure to practice Assertion-Reason and MCQs based on these properties.

CBSE Class 12 Chemistry Chapter-wise Notes

Here's is a list of NCERT notes for Class 12 chemistry. Students can use the link to access the notes of other chapters as well.

Subject-Wise NCERT Exemplar Solutions

The hyperlinks of subject wise exempler solutions are given below:

Subject-Wise NCERT Solutions

The hyperlinks of subject-wise ncert solutions are given below:

NCERT Books and Syllabus

Frequently Asked Questions (FAQs)

1. 1. What are the discussed about lanthanoid contraction in NCERT Class 12 Chemistry chapter 4?

Ans- Lanthanoid contraction is defined as a decrease in atomic and ionic radii with an increase in atomic number. In a lanthanoid series, the size of the ion decreases as it progresses from La3+ to Lu3+. The lanthanoid contraction is a reduction in the size of the series. The imperfect shielding of a 4f electron by a separate election from the same subshell causes this event. These are discussed in Class 12 The d and f block elements notes. Students can further explore Ch 4 Chemistry Class 12 notes to gain a holistic understanding of the structure, properties, and trends exhibited by d and f block elements. 

2. 2. What is mean by disproportionation according to The d and f block elements Class 12 notes?

Ans-. The reaction is considered to have experienced a disproportionation of the oxidation state when the oxidation of any element in the molecule grows in one product and reduces in another. Disproportionation is the reaction that occurs when a particular element performs both self-oxidation and self-reduction at the same time.

3. 3. What are the consequences of lanthanoid contraction as per The d and f block elements Class 12 pdf download?

Ans- Because of the lanthanoid contraction, the size of the next elements (Hf – Hg) becomes very similar to the size of the previous row members (Zr – Cd), making them difficult to distinguish. The properties of lanthanoids are essentially same due to minor differences in atomic radii, making separation difficult.

4. 4. What are the characteristics of d-block elements?

Ans- Each horizontal row in the d-block has ten d-subshell elements and can hold a maximum of ten electrons. Transition metals are d-block elements having a partially filled d-subshell.They exhibit variable valency. They have the ability to form coloured compounds.

5. 5. Why do transition metals have different oxidation states?

Ans-Because the energy difference between the (n-1) d-orbital and the ns-orbital is so small, transition elements have a wide range of oxidation states. Because the orbitals have such a little energy difference, both energy levels can be utilized to establish a bond. As a result, transition elements can have a wide range of oxidation states.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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