Chapter 13: The d- and f- Block Elements

Introduction

Overview of d- and f-block Elements

• Four series of d-block elements correspond to filling of 3d, 4d, 5d, and 6d orbitals.
• d-block elements are transition metals with valence electrons entering d-orbitals.
• f-block elements involve filling of 4f and 5f orbitals, forming the lanthanide and actinide series.

Solved Examples

• Ex 1: Identify the block of element with electronic configuration [Ar] 3d⁵ 4s².
  Solution: Last electron enters 3d orbital → d-block element.

Practice Set

• Q1: Which orbital is being filled in the first series of d-block elements?
  A. 3p
  B. 3d
  C. 4s
  D. 4p
• Q2: The f-block elements are characterized by filling which orbitals?
  A. 3d and 4d
  B. 4f and 5f
  C. 5d and 6d
  D. 4p and 5p
• Q3: Transition elements are located between which blocks in the periodic table?
  A. s- and p-block
  B. p- and f-block
  C. s- and f-block
  D. p- and d-block

Answer Key

• Q1: B. 3d — First d-block series fills 3d orbitals.
• Q2: B. 4f and 5f — f-block elements fill 4f and 5f orbitals.
• Q3: A. s- and p-block — Transition elements lie between s- and p-block.

Memory Aids

• d-block = last electron in d-orbital.
• f-block = filling of f-orbitals (4f, 5f).
• Transition metals lie between s- and p-block in periodic table.

Transition Elements Position

Location in Periodic Table

• Transition elements are located in groups 3 to 12 of the periodic table.
• They lie between s-block (groups 1-2) and p-block (groups 13-18) elements.
• Called d-block elements because their last electron enters the d-orbital.

Solved Examples

• Ex 1: Determine the group of element with configuration [Ar] 3d⁷ 4s².
  Solution: Last electron in 3d, total d-electrons = 7 → Group 9 (Co group).

Practice Set

• Q1: Which group contains transition elements?
  A. 1 and 2
  B. 3 to 12
  C. 13 to 18
  D. Lanthanides
• Q2: Why are transition elements called d-block elements?
  A. Last electron enters s-orbital
  B. Last electron enters p-orbital
  C. Last electron enters d-orbital
  D. Last electron enters f-orbital
• Q3: Which of the following is NOT a transition element?
  A. Scandium
  B. Zinc
  C. Iron
  D. Copper

Answer Key

• Q1: B. 3 to 12 — Transition elements occupy groups 3 to 12.
• Q2: C. Last electron enters d-orbital — Defines d-block elements.
• Q3: B. Zinc — Zn has full d¹⁰ shell and is not a transition element.

Memory Aids

• Transition elements = groups 3–12.
• Last electron in d-orbital defines transition metals.
• Zinc, cadmium, mercury are exceptions (not transition metals).

Electronic Configurations of the d-Block Elements

General Electronic Configuration

• d-Block elements have outer electronic configuration: (n-1)d^1-10 ns^0-2
• Zn, Cd, Hg, and Cn have completely filled d-orbitals: (n-1)d¹⁰ ns²
• These fully filled elements are not considered transition elements

Transition Series

• Four transition series in d-block:
• 1^st series (3d): Sc (21) to Zn (30)
• 2^nd series (4d): Y (39) to Cd (48)
• 3^rd series (5d): La (57), Hf (72) to Hg (80)
• 4^th series (6d): Ac (89), Rf (104) to Cn (112)

Solved Examples

• Ex 1: Write the electronic configuration of Sc (Z=21).
• Solution: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s² or simplified as 3d¹ 4s².
• Ex 2: Identify if Zn is a transition element.
• Solution: Zn has configuration 3d¹⁰ 4s², fully filled d-orbitals; hence, not a transition element.

Practice Set

• Q1: What is the general electronic configuration of d-block elements?
• A. (n-1)d^1-10 ns^0-2
• B. ns² np^1-6
• C. (n-2)f^1-14 (n-1)d^0-1 ns²
• D. ns¹ np⁵
• Q2: Which of the following is NOT a transition element?
• A. Fe
• B. Zn
• C. Cu
• D. Ni
• Q3: The 3rd transition series starts from which element?
• A. La (57)
• B. Hf (72)
• C. Hg (80)
• D. Ac (89)

Answer Key

• Q1: A. (n-1)d^1-10 ns^0-2 - Defines d-block electron filling.
• Q2: B. Zn - Has fully filled d-orbitals, not a transition element.
• Q3: A. La (57) - Start of 3rd transition series.

Memory Aids

• Transition elements: d-orbitals partially filled (except Zn, Cd, Hg, Cn).
• Series order: 3d (Sc-Zn), 4d (Y-Cd), 5d (La-Hg), 6d (Ac-Cn).
• Zn, Cd, Hg, Cn are d-block but NOT transition elements.

General Properties of Transition Elements

Metallic Characteristics

• Exhibit high tensile strength, ductility, malleability, and metallic lustre.
• High thermal and electrical conductivity due to metallic bonding.
• Most have typical metallic crystal structures except Zn, Cd, Hg, and Mn.

Physical and Chemical Trends

• High melting and boiling points due to strong metallic bonds involving d-electrons.
• High enthalpies of atomisation reflect strong interatomic forces.
• Atomic and ionic sizes decrease across a period due to increasing nuclear charge; increase down a group.
• Lanthanoid contraction causes smaller atomic sizes in 4d and 5d series.
• Ionisation enthalpy generally increases across the series; exceptions due to stable electron configurations.
• Multiple oxidation states arise from similar energy of 4s and 3d electrons.
• Standard electrode potentials vary with electron configuration and oxidation state stability.
• Magnetic properties due to unpaired d-electrons; paramagnetism increases with unpaired electrons.
• Formation of coloured ions due to d-d electronic transitions absorbing visible light.
• Form complex compounds due to small size and ability to accept electron pairs.
• Act as catalysts by changing oxidation states and forming complexes.
• Form interstitial compounds by small atoms fitting into metal lattices.
• Form alloys easily due to similar atomic sizes allowing substitution in crystal lattices.

Solved Examples

• Ex 1: Explain why transition metals have high melting points.
  - Due to strong metallic bonds involving unpaired d-electrons, requiring more energy to break.
• Ex 2: Why do transition metals form coloured compounds?
  - d-d transitions absorb visible light, causing colour.

Practice Set

• Q1: Which property is NOT typical of transition metals?
  A. High melting point
  B. Multiple oxidation states
  C. Colourless ions in all oxidation states
  D. Formation of complex compounds
• Q2: Why do atomic sizes decrease across a transition series?
  A. Increase in nuclear charge
  B. Increase in electron shielding
  C. Lanthanoid contraction
  D. Decrease in nuclear charge
• Q3: Which factor contributes to catalytic activity of transition metals?
  A. Fixed oxidation state
  B. Ability to form coloured ions
  C. Variable oxidation states
  D. Large atomic size

Answer Key

• Q1: C - Some ions are coloured due to d-d transitions, not all are colourless.
• Q2: A - Increasing nuclear charge pulls electrons closer, reducing size.
• Q3: C - Variable oxidation states allow redox reactions essential for catalysis.

Memory Aids

• Transition metals = Multiple oxidation states + coloured ions + catalytic activity.
• Lanthanoid contraction = Smaller atomic size in 4d and 5d series.
• High melting point = Strong metallic bonds from unpaired d-electrons.

Colour and Electronic Transitions

Colour Origin

• Sc³⁺, Ti⁴⁺, V⁵⁺, Cr⁶⁺, Mn⁷⁺ have empty d-orbitals; no d-d transitions; colourless.
• VO₂⁺ (pale yellow), CrO₄²⁻ (deep yellow), MnO₄⁻ (intense purple) show colour due to charge transfer transitions.

Solved Examples

• Ex 1: Why is CrO₄²⁻ yellow?
  - Charge transfer from O to Cr causes absorption in visible region.

Practice Set

• Q1: Which ion is colourless?
  A. MnO₄⁻
  B. Ti⁴⁺
  C. VO₂⁺
  D. CrO₄²⁻
• Q2: Colour in transition metal ions arises mainly due to:
  A. d-d transitions
  B. s-p transitions
  C. Charge transfer
  D. Nuclear spin
• Q3: Which ion shows charge transfer colour?
  A. Sc³⁺
  B. MnO₄⁻
  C. Ti⁴⁺
  D. V⁵⁺

Answer Key

• Q1: B - Ti⁴⁺ has empty d-orbitals, no d-d transitions.
• Q2: A - Colour mainly from d-d electronic transitions.
• Q3: B - MnO₄⁻ shows charge transfer colour.

Memory Aids

• Empty d-orbitals = Colourless ions.
• Charge transfer = Intense colours in oxoanions.

Stability of Higher Oxidation States

Halides

• Ti forms TiX₄ with +4 oxidation state.
• Vanadium forms VF₅ (+5), chromium forms CrF₆ (+6).
• Mn does not form simple +7 halides; MnO₃F is known.
• Fluorine stabilises higher oxidation states due to high lattice energy and bond enthalpy.
• CoF₃ stable due to high lattice enthalpy.
• VF₅ and CrF₆ stable due to strong bonds.
• Other halides hydrolyse to oxohalides (VOX₃).
• Cu(II) halides known except iodide; Cu²⁺ oxidises I⁻ to I₂.
• Copper(I) compounds disproportionate: 2Cu⁺ → Cu²⁺ + Cu.
• Cu²⁺ more stable than Cu⁺ in aqueous solution due to hydration enthalpy.

Oxides

• Highest oxidation number equals group number: Sc₂O₃, TiO₂, V₂O₅, CrO₃, Mn₂O₇.
• No higher oxides beyond group 7 except Fe₂O₃ and Co₃O₄.
• Oxocations like TiO²⁺ and VO₂⁺ exist.
• Oxides more stable than fluorides at high oxidation states due to multiple bonding by oxygen.
• MnF₄ highest fluoride; Mn₂O₇ highest oxide.
• Mn₂O₇ structure: each Mn tetrahedrally surrounded by O; Mn-O-Mn bridge present.
• Tetrahedral ions: VO₄³⁻, CrO₄²⁻, MnO₄³⁻, MnO₄²⁻, MnO₄⁻.

Solved Examples

• Ex 1: Write the disproportionation reaction of Cu(I) in aqueous solution.
  - 2Cu⁺ → Cu²⁺ + Cu
• Ex 2: Why are fluorides more stable in higher oxidation states?
  - Due to high lattice energy and strong covalent bonds with fluorine.

Practice Set

• Q1: Which halide shows highest oxidation state for vanadium?
  A. VCl₃
  B. VF₅
  C. VBr₄
  D. VI₂
• Q2: Which oxide corresponds to chromium in highest oxidation state?
  A. CrO
  B. Cr₂O₃
  C. CrO₃
  D. CrO₂
• Q3: Why is Mn₂O₇ more stable than MnF₇ (hypothetical)?
  A. Oxygen forms multiple bonds
  B. Fluorine is less electronegative
  C. Mn prefers lower oxidation states
  D. Fluorides have lower lattice energy

Answer Key

• Q1: B - VF₅ shows +5 oxidation state for vanadium.
• Q2: C - CrO₃ corresponds to +6 oxidation state.
• Q3: A - Oxygen forms multiple bonds stabilizing higher oxidation states.

Memory Aids

• Fluorine stabilises high oxidation states by strong bonds and lattice energy.
• Highest oxide oxidation state = group number.
• Cu(I) disproportionates in aqueous solution.

Exceptions and Special Cases

Non-Typical Transition Elements

• Zn, Cd, Hg are d-block but not transition elements (pseudo transition metals).
• Anhydrous FeSO₄ and CuSO₄ are white due to absence of crystal field splitting despite unpaired electrons.
• Cu²⁺ (3d⁹, E° = 0.34 V) is more stable than Cu⁺ (3d¹⁰, E° = 0.54 V) except with large anions.
• Examples: CuI more stable than CuI₂, CuCN more stable than Cu(CN)₂.
• Among d-block, tungsten (W, Z=74) has highest melting point (3410°C), mercury (Hg, Z=80) lowest (-38.9°C).

Solved Examples

• Ex 1: Why is anhydrous CuSO₄ white?
  - No crystal field splitting without water ligands; no colour.
• Ex 2: Which d-block element has highest melting point?
  - Tungsten (W) at 3410°C.

Practice Set

• Q1: Which is NOT a typical transition metal?
  A. Fe
  B. Zn
  C. Cr
  D. Ni
• Q2: Why is Cu²⁺ more stable than Cu⁺ in aqueous solution?
  A. Higher ionisation energy
  B. More negative hydration enthalpy
  C. Larger ionic radius
  D. Lower electrode potential
• Q3: Which element has the lowest melting point among d-block?
  A. W
  B. Hg
  C. Fe
  D. Cu

Answer Key

• Q1: B - Zn is not a typical transition metal.
• Q2: B - Cu²⁺ has more negative hydration enthalpy stabilizing it.
• Q3: B - Mercury (Hg) has lowest melting point (-38.9°C).

Memory Aids

• Zn, Cd, Hg = pseudo transition metals.
• Cu²⁺ more stable than Cu⁺ in aqueous except with large anions.
• W highest melting point; Hg lowest among d-block.

Some Important Compounds of Transition Elements

Oxides and Oxoanions of Metals

• Transition metals form oxides by combining with oxygen at high temperatures.
• Except Sc, metals up to Mn form ionic monoxides (MO).
• Highest oxidation state in oxides equals group number.
• Examples: Sc₂O₃, Cr₂O₃, TiO₂, Mn₂O₇, V₂O₅.
• After Mn, only Fe₂O₃ and Co₃O₄ higher oxides are known.
• Increasing oxidation number increases covalent character and acidic nature of oxides.
• Examples: V₂O₃ (basic), V₂O₄ (less basic), V₂O₅ (amphoteric).
• V₂O₅ reacts with alkalis and acids to form VO₄³⁻ and VO₂⁺ respectively.
• CrO is basic; Cr₂O₃ amphoteric; CrO₃ and Mn₂O₇ acidic, forming acids in water.
• Reactions: CrO₃ + H₂O → H₂CrO₄; Mn₂O₇ + H₂O → 2HMnO₄.

Potassium Dichromate (K₂Cr₂O₇)

Preparation

• Obtained by fusion of chromite ore (FeCr₂O₄):
• 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + Fe₂O₃ + 8CO₂
• Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Structure

• Chromate ion (CrO₄²⁻) is tetrahedral.
• Dichromate ion (Cr₂O₇²⁻) consists of two tetrahedra sharing one corner; Cr–O–Cr bond angle is 126°.

Properties

• Orange crystals, moderately soluble in cold water, readily soluble in hot water.
• In aqueous solution, equilibrium exists between dichromate and chromate ions:

• In alkaline medium: Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O (yellow solution)
• In acidic medium: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O (orange solution)
• Strong oxidising agent in acidic medium:
• Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
• Oxidises I⁻ to I₂, H₂S to S, Sn²⁺ to Sn⁴⁺, Fe²⁺ to Fe³⁺.
• Chromyl chloride test: Metal chloride + K₂Cr₂O₇ + conc. H₂SO₄ → red vapours of CrO₂Cl₂.

Potassium Permanganate (KMnO₄)

Preparation

• From pyrolusite ore (MnO₂):
• 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
• K₂MnO₄ converted to KMnO₄ by electrochemical oxidation:
• Anode: MnO₄²⁻ → MnO₄⁻ + e⁻ (green to purple)
• Cathode: 2H⁺ + 2e⁻ → H₂
• In acidic/neutral medium, manganate disproportionates:
• 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Laboratory Preparation

• Mn²⁺ salt oxidised by peroxodisulphate:
• 2Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺

Structure

• Manganate and permanganate ions are tetrahedral.
• Manganate (green) is paramagnetic (1 unpaired electron); permanganate (purple) is diamagnetic.
• π-bonding via overlap of oxygen p-orbitals and manganese d-orbitals.

Properties

• Dark purple crystals, sparingly soluble in water, decompose on heating:
• 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
• Strong oxidising agent in acidic and alkaline media.
• In acidic medium:
• MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Oxidises I⁻ to I₂, Fe²⁺ to Fe³⁺, oxalate to CO₂, H₂S to S, sulphite to sulphate, nitrite to nitrate.
• In neutral/faintly alkaline medium:
• MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
• Used in volumetric titrations with dilute H₂SO₄ (not HCl or HNO₃).

Uses

• Strong oxidant in organic and analytical chemistry.
• Bleaching agent for wool, cotton, silk, and textiles.
• Decolourises oils due to strong oxidising power.

The Inner Transition Elements (f-Block)

Definition and Series

• Elements with last electron entering f-orbital are f-block elements (inner transition elements).
• Consist of two series:
• Lanthanoids: 14 elements after lanthanum.
• Actinoids: 14 elements after actinium.

Solved Examples

Ex 1

• Write the reaction for preparation of potassium dichromate from chromite ore.
• Answer: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + Fe₂O₃ + 8CO₂

Ex 2

• Show the equilibrium between chromate and dichromate ions in aqueous solution.
• Answer: Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O

Ex 3

• Write the half-reaction for reduction of permanganate ion in acidic medium.
• Answer: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Practice Set

Q1

• Which oxide of vanadium is amphoteric?
• A) V₂O₃
• B) V₂O₄
• C) V₂O₅
• D) VO

Q2

• In acidic medium, potassium dichromate acts as a strong oxidising agent. What is the product formed after reduction?
• A) Cr⁶⁺
• B) Cr³⁺
• C) Cr²⁺
• D) Cr metal

Q3

• Which of the following statements about potassium permanganate is FALSE?
• A) It is purple and sparingly soluble in water.
• B) It decomposes on heating to give K₂MnO₄, MnO₂, and O₂.
• C) It acts as a reducing agent in acidic medium.
• D) It oxidises Fe²⁺ to Fe³⁺ in acidic medium.

Answer Key

• Q1: C) V₂O₅ - It is amphoteric; V₂O₃ is basic, V₂O₄ less basic.
• Q2: B) Cr³⁺ - Dichromate reduces to Cr³⁺ in acidic medium.
• Q3: C) It acts as a reducing agent in acidic medium - FALSE; it is a strong oxidising agent.

Memory Aids

• Dichromate-chromate equilibrium shifts yellow (alkaline) ↔ orange (acidic).
• Potassium permanganate is purple; manganate is green.
• KMnO₄ oxidises Fe²⁺ to Fe³⁺ in acidic medium.
• Chromyl chloride test: red vapours indicate presence of chloride ion.
• f-Block = inner transition elements = lanthanoids + actinoids.

Lanthanoids Overview

Electronic Configurations

• General symbol: Ln
• Common configuration: 6s² with variable 4f occupancy (4fⁿ, n=1 to 14)
• Tripositive ions (Ln³⁺) have 4fⁿ configurations

Atomic and Ionic Sizes

• Atomic and ionic radii decrease from La to Lu (lanthanoid contraction)
• Cause: Poor shielding by 4f electrons increases effective nuclear charge
• Atomic radii decrease irregularly; ionic radii decrease regularly

Oxidation States

• Predominant oxidation state: +3
• Other states: +2 and +4 due to stable empty (4f⁰), half-filled (4f⁷), fully-filled (4f¹⁴) subshells
• Examples: Ce⁴⁺ (4f⁰), Eu²⁺ (4f⁷), Tb⁴⁺ (4f⁷), Yb²⁺ (4f¹⁴)
• Ce⁴⁺ is a strong oxidant (E° = +1.74 V), used as analytical reagent
• Pr, Nd, Tb, Dy show +4 in oxides (MO₂)
• Eu²⁺ and Yb²⁺ are strong reductants
• Samarium exhibits +2 and +3 states

General Characteristics

• Appearance: Silvery white, soft metals
• React with air: Tarnish in moist air, burn in oxygen forming Ln₂O₃
• Hardness increases with atomic number; samarium is very hard
• Melting points: 1000–1200 K; samarium melts at 1623 K
• Good conductors of heat and electricity
• Density varies smoothly except for Eu, Yb, Sm, Tm
• Colored ions due to f-f transitions; La³⁺ and Lu³⁺ are colorless
• Magnetic properties: La³⁺, Lu³⁺ diamagnetic; others paramagnetic
• Ionization enthalpy: 1st ~600 kJ/mol, 2nd ~1200 kJ/mol; 3rd varies due to stability of f-orbitals
• Standard electrode potentials: E°(Ln³⁺/Ln) ~ -2.2 to -2.4 V; Eu exception at -2.0 V

Chemical Behaviour

• React with H₂ on gentle heating to form hydrides
• React with carbon on heating to form carbides (Ln₃C, LnC₂, Ln₂C₃)
• React with dilute acids to liberate H₂: 2Ln + 6H⁺ → 2Ln³⁺ + 3H₂
• Burn in halogens to form halides: 2Ln + 3X₂ → 2LnX₃
• Burn in oxygen to form oxides: 4Ln + 3O₂ → 2Ln₂O₃
• Form basic oxides and hydroxides: M₂O₃ and M(OH)₃

Solved Examples

Ex 1

• Explain why lanthanoid contraction occurs.
• Answer: Poor shielding by 4f electrons increases effective nuclear charge, pulling electrons closer and decreasing radii.

Ex 2

• Write the reaction of lanthanoids with dilute acid.
• Answer: 2Ln + 6H⁺ → 2Ln³⁺ + 3H₂

Practice Set

Q1

Which lanthanoid ion is colorless due to fully filled 4f orbitals?

• A. Eu³⁺
• B. La³⁺
• C. Tb³⁺
• D. Gd³⁺

Q2

What is the predominant oxidation state of lanthanoids?

• A. +2
• B. +3
• C. +4
• D. +1

Q3

Which lanthanoid exhibits both +2 and +3 oxidation states?

• A. Samarium
• B. Cerium
• C. Ytterbium
• D. Terbium

Answer Key

• Q1: B. La³⁺ - has empty 4f orbitals (4f⁰), making it colorless.
• Q2: B. +3 - most stable and common oxidation state for lanthanoids.
• Q3: A. Samarium - exhibits both +2 and +3 states due to stable half-filled configuration.

Memory Aids

• Lanthanoid contraction: Poor 4f shielding → radius decreases across series.
• Stable oxidation states: +3 common; +2 and +4 due to empty, half, full 4f subshells.
• Colorless ions: La³⁺ (4f⁰), Lu³⁺ (4f¹⁴).

Actinoids Overview

Electronic Configuration

• Actinoids include 14 elements from Thorium (Th) to Lawrencium (Lr).
• General configuration: 5f^0–14 6d^0–2 7s².
• Electrons fill 5f orbitals except Thorium (Z=90); filling continues till Lawrencium (Z=103).
• Irregularities due to extra stability at f⁰, f⁷, and f¹⁴ configurations.
• All actinoids are radioactive.

Ionic Radii

• Ionic radii decrease regularly across the series (actinoid contraction).
• Contraction is greater than lanthanoids due to poor shielding by 5f electrons.
• 5f orbitals extend beyond 6s and 6p orbitals; 4f orbitals are buried deeper.

Oxidation States

• Most common oxidation state is +3.
• Higher states +4, +5, +6, +7 occur in Th, Pa, U, Np respectively.
• Multiple oxidation states due to small energy gap between 5f, 6d, and 7s subshells.

Solved Examples

• Ex 1: Write the electronic configuration of Uranium (Z=92).
  • U: [Rn] 5f³ 6d¹ 7s²
• Ex 2: Identify the common oxidation states of Neptunium (Np).
  • +3, +4, +5, +6, +7 (varies with compounds)

Practice Set

• Q1: Which element marks the start of the actinoid series?
  • A. Thorium
  • B. Actinium
  • C. Uranium
  • D. Lawrencium
• Q2: What is the most common oxidation state of actinoids?
  • A. +2
  • B. +3
  • C. +4
  • D. +5
• Q3: Why do actinoids show multiple oxidation states?
  • A. Large energy gap between 5f and 6d orbitals
  • B. Small energy gap between 5f, 6d, and 7s subshells
  • C. Only 5f electrons involved
  • D. Due to radioactive decay

Answer Key

• Q1: B - Actinium starts the actinoid series (Z=89).
• Q2: B - +3 is the most common oxidation state in actinoids.
• Q3: B - Small energy gap between 5f, 6d, and 7s allows multiple oxidation states.

Memory Aids

• Actinoids = 14 elements from Ac (89) to Lr (103).
• Common oxidation state +3; higher states in early actinoids.
• Actinoid contraction > lanthanoid contraction due to 5f poor shielding.

General Characteristics

Reactivity and Appearance

• Silvery white metals, tarnish rapidly in air forming oxide coating.
• Not attacked by alkalies; less reactive towards acids.

Colour and Magnetism

• Colored ions due to f–f transitions except Ac³⁺ (5f⁰), Cm³⁺ (5f⁷), Th⁴⁺ (5f⁰) which are colorless.
• Strongly paramagnetic.

Physical Properties

• High densities except Thorium and Americium.
• High melting and boiling points; no regular trend with atomic number.
• Low ionization energies.

Chemical Properties

• Strong reducing agents.
• Highly electropositive metals.

Solved Examples

• Ex 1: Explain why actinoids tarnish rapidly in air.
  • They form oxide coating due to high reactivity with oxygen.
• Ex 2: Why are most actinoid ions colored?
  • Due to f–f electronic transitions in 5f orbitals.

Practice Set

• Q1: Which actinoid ion is colorless?
  • A. Cm³⁺
  • B. Ac³⁺
  • C. U³⁺
  • D. Np³⁺
• Q2: Actinoids are strong reducing agents because they have:
  • A. High ionization energy
  • B. Low ionization energy
  • C. High electronegativity
  • D. Low atomic size
• Q3: Which property does NOT show a regular trend across actinoids?
  • A. Ionic radii
  • B. Melting point
  • C. Oxidation state
  • D. Paramagnetism

Answer Key

• Q1: B - Ac³⁺ (5f⁰) is colorless.
• Q2: B - Low ionization energy makes actinoids strong reducing agents.
• Q3: B - Melting and boiling points show no regular trend.

Memory Aids

• Colorless ions: Ac³⁺, Th⁴⁺, Cm³⁺.
• Actinoids tarnish fast due to oxide layer formation.
• Strong reducing agents due to low ionization energy.

Comparison: Lanthanoids vs Actinoids

Oxidation States

• Lanthanoids mostly +3; some +2 and +4.
• Actinoids show +3 and higher states (+4 to +7).

Radioactivity

• Lanthanoids mostly non-radioactive except Promethium.
• All actinoids are radioactive.

Cation Formation and Basicity

• Lanthanoids do not form oxo-cations.
• Actinoids form oxo-cations (e.g., UO₂²⁺, PuO₂²⁺).
• Actinoid compounds are more basic than lanthanoids.

Complex Formation

• Lanthanoids have less tendency to form complexes.
• Actinoids readily form complex compounds.

Solved Examples

• Ex 1: Identify which series forms oxo-cations.
  • Actinoids form oxo-cations; lanthanoids do not.
• Ex 2: Which series is more radioactive?
  • Actinoids are all radioactive; lanthanoids mostly not.

Practice Set

• Q1: Which series shows higher oxidation states?
  • A. Lanthanoids
  • B. Actinoids
  • C. Both equal
  • D. Neither
• Q2: Which series forms oxo-cations?
  • A. Lanthanoids
  • B. Actinoids
  • C. Both
  • D. None
• Q3: Which statement is true?
  • A. Lanthanoids are mostly radioactive.
  • B. Actinoids rarely form complexes.
  • C. Actinoids are more basic than lanthanoids.
  • D. Lanthanoids have higher oxidation states than actinoids.

Answer Key

• Q1: B - Actinoids show higher oxidation states (+4 to +7).
• Q2: B - Actinoids form oxo-cations like UO₂²⁺.
• Q3: C - Actinoids are more basic than lanthanoids.

Memory Aids

• Actinoids: radioactive, multiple oxidation states, oxo-cations.
• Lanthanoids: mostly +3, less reactive, no oxo-cations.
• Actinoids form complexes easily; lanthanoids less so.

Some Applications of d- and f-Block Elements

Key Industrial Uses

• Iron and steel: Primary materials for construction and infrastructure.
• Titanium dioxide (TiO₂): Widely used as a white pigment in paints and coatings.
• Manganese dioxide (MnO₂), Zinc (Zn), Nickel (Ni), Cadmium (Cd): Essential components in battery manufacturing.
• Copper (Cu), Silver (Ag), Gold (Au): Used in coins, jewelry, and ornaments due to their malleability and luster.
• Vanadium pentoxide (V₂O₅): Catalyst in the Contact process for sulphuric acid production.
• Titanium tetrachloride (TiCl₄) with trimethylaluminium (Al(CH₃)₃): Catalysts in the polymerization of ethylene to polythene.
• Iron catalyst: Used in Haber’s process for ammonia synthesis from N₂ and H₂.
• Nickel catalyst: Facilitates hydrogenation of unsaturated fats to saturated fats.
• Nickel complexes: Catalyze polymerization of alkynes and other organic compounds.
• Palladium chloride (PdCl₂): Catalyst in Wacker’s process converting ethyne to ethanal.
• Silver bromide (AgBr): Light-sensitive compound used in photographic films.
• Neodymium and praseodymium oxides: Used in colored glasses for goggles; absorb yellow sodium light, protecting glass blowers' eyes.

Solved Examples

• Ex 1: Identify the catalyst used in the Haber process.
  Solution: Iron catalyst is used to synthesize ammonia (NH₃) from nitrogen and hydrogen gases.
• Ex 2: Write the catalyst combination used in polythene manufacture.
  Solution: TiCl₄ with Al(CH₃)₃ acts as catalyst in ethylene polymerization to polythene.
• Ex 3: Name the catalyst for oxidation of ethyne to ethanal.
  Solution: PdCl₂ is used as catalyst in Wacker’s process.

Practice Set

• Q1: Which element is primarily used as a catalyst in the Haber process?
  A. Nickel
  B. Iron
  C. Palladium
  D. Vanadium
  
• Q2: TiO₂ is mainly used in which industry?
  A. Battery
  B. Pigment
  C. Photographic
  D. Polymerization
  
• Q3: Which of the following is used as a catalyst in the manufacture of sulphuric acid?
  A. V₂O₅
  B. MnO₂
  C. PdCl₂
  D. Fe
  

Answer Key

• Q1: B. Iron
  Iron acts as the catalyst in Haber’s process for ammonia synthesis.
• Q2: B. Pigment
  TiO₂ is widely used as a white pigment in paints and coatings.
• Q3: A. V₂O₅
  Vanadium pentoxide catalyzes the Contact process for sulphuric acid production.

Memory Aids

• Haber’s process catalyst: Fe (Iron) for Ammonia.
• Polythene catalyst: TiCl₄ + Al(CH₃)₃.
• Wacker’s process catalyst: PdCl₂ for ethyne oxidation.
• Battery metals: MnO₂, Zn, Ni, Cd.
• Photographic industry: AgBr light sensitivity.
• Colored goggles: Nd and Pr oxides absorb yellow sodium light.