Chapter 13: The d- and f- Block Elements

Introduction

Overview

• Four series of d-block elements formed by filling 3d, 4d, 5d, and 6d electron shells.
• These series together make up the d-block elements.
• f-block elements fill 4f and 5f orbitals in two long periods.

Solved Examples

• Ex 1: Identify the block of element with electronic configuration [Ar] 3d⁵ 4s².
  Answer: d-block (3d electrons filling).

Practice Set

• Q1: Which orbitals are filled in d-block elements?
  A. s-orbitals
  B. p-orbitals
  C. d-orbitals
  D. f-orbitals
• Q2: The f-block elements fill which orbitals?
  A. 3d
  B. 4f and 5f
  C. 4d
  D. 5d
• Q3: How many series of d-block elements are there?
  A. Two
  B. Three
  C. Four
  D. Five

Answer Key

• Q1: C - d-orbitals are filled in d-block elements.
• Q2: B - f-block elements fill 4f and 5f orbitals.
• Q3: C - There are four series of d-block elements.

Memory Aids

• D-block = last electron in d-orbital.
• f-block = filling 4f and 5f orbitals.
• Four d-block series: 3d, 4d, 5d, 6d.

Transition Elements

Position in Periodic Table

• Transition elements lie between s- and p-block elements in the long form of the periodic table.
• Called d-block elements because last electron enters d-orbital.

Solved Examples

• Ex 1: Identify the position of Fe in the periodic table.
  Answer: Between s- and p-block, in d-block.

Practice Set

• Q1: Transition elements are located between which blocks?
  A. s- and p-block
  B. p- and f-block
  C. s- and f-block
  D. p- and d-block
• 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 a transition element?
  A. Na
  B. Ca
  C. Fe
  D. Ne

Answer Key

• Q1: A - Transition elements lie between s- and p-block.
• Q2: C - Last electron enters d-orbital.
• Q3: C - Fe is a transition element.

Memory Aids

• Transition elements = d-block elements.
• Located between s- and p-block in periodic table.
• Last electron enters d-orbital.

Electronic Configurations of the d-Block Elements

General electronic configuration

• d-block elements have the general configuration (n-1)d^1-10 ns^0-2.
• Electrons fill the (n-1)d and ns orbitals progressively across the series.

Filled d-orbitals in Zn, Cd, Hg, and Cn

• Elements Zn, Cd, Hg, and Cn have configuration (n-1)d¹⁰ ns².
• Their d and s orbitals are completely filled in ground and common oxidation states.
• They are not considered transition elements due to fully filled d-orbitals.

Transition series

• d-block consists of four transition series based on the d-orbital being filled:
• 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)

Electron configuration pattern

• Transition elements fill d-orbitals progressively across each series.
• Example: Sc (21) has configuration 3d¹ 4s².
• 4th series includes filling 5f orbitals (actinides).
• Electron configurations explain chemical properties and trends of transition metals.

Solved Examples

• Ex 1: Write the electronic configuration of Fe (Z=26).
  • Fe is in 1st transition series (3d).
  • Configuration: [Ar] 3d⁶ 4s².
• Ex 2: Identify if Zn is a transition element.
  • Zn configuration: [Ar] 3d¹⁰ 4s².
  • Fully filled d-orbitals, so 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 np⁶
  • C. (n-1)f^1-14 ns²
  • D. ns¹ np⁵
• Q2: Which element is NOT considered a transition element?
  • A. Fe
  • B. Zn
  • C. Cu
  • D. Ni
• Q3: What is the electronic configuration of the 2nd transition series element Cd (Z=48)?
  • A. [Kr] 4d¹⁰ 5s²
  • B. [Kr] 4d⁹ 5s²
  • C. [Kr] 4d⁸ 5s²
  • D. [Kr] 4d¹⁰ 5s¹

Answer Key

• Q1: A. (n-1)d^1-10 ns^0-2 - Defines d-block filling pattern.
• Q2: B. Zn - Fully filled d-orbitals, not a transition element.
• Q3: A. [Kr] 4d¹⁰ 5s² - Cd has completely filled d and s orbitals.

Memory Aids

• Transition series order: 3d (Sc-Zn), 4d (Y-Cd), 5d (La/Hf-Hg), 6d (Ac/Rf-Cn).
• Zn, Cd, Hg, Cn: d¹⁰ s² configuration, not transition metals.
• Remember: Transition elements have incomplete d-orbitals in common oxidation states.

General Properties of Transition Elements

Metallic Characteristics

• High tensile strength, ductility, malleability, thermal & electrical conductivity.
• Metallic lustre common except Zn, Cd, Hg, Mn.
• Typical metallic structures at room temperature except Zn, Cd, Hg, Mn.

Physical Properties

• High melting and boiling points due to strong metallic bonding.
• High enthalpies of atomisation from strong atomic forces.
• Atomic and ionic sizes decrease across series; increase down groups.
• Lanthanoid contraction causes size anomalies.
• Ionisation enthalpy generally increases across series.

Chemical Properties

• Multiple oxidation states due to similar energy d-orbitals.
• Variable standard electrode potentials.
• Magnetic properties from unpaired electrons; paramagnetism increases with unpaired electrons.
• Formation of coloured ions via d-d transitions; exceptions with empty d-orbitals.
• Complex compound formation due to small size and electron acceptance.
• Catalytic activity from variable oxidation states and complex formation.
• Interstitial compounds form by small atoms fitting in metal lattices.
• Alloy formation facilitated by similar atomic sizes.

Solved Examples

• Ex 1: Explain why transition metals have high melting points.
  Strong metallic bonding due to many valence electrons involved in bonding.
• Ex 2: Why do transition metals form coloured ions?
  Electrons jump between split d-orbitals absorbing visible light.

Practice Set

• Q1: Which element is NOT a typical transition metal?
  A. Fe B. Zn C. Cr D. Ni
• Q2: What causes the magnetic properties in transition metals?
  A. Paired electrons B. Unpaired electrons C. Full d-orbitals D. Empty s-orbitals
• Q3: Why do transition metals show multiple oxidation states?
  A. Different s-orbitals B. Similar energy d-orbitals C. Large atomic size D. High ionisation energy

Answer Key

• Q1: B - Zn is a non-typical transition metal with full d-orbitals.
• Q2: B - Unpaired electrons cause magnetic properties.
• Q3: B - Similar energy d-orbitals allow multiple oxidation states.

Memory Aids

• Zn, Cd, Hg are pseudo transition metals (no typical d-block behavior).
• High melting points = strong metallic bonds + many valence electrons.
• Colour arises from d-d electron transitions.

Colour and d-d Transitions

Colour Origin

• Transition metal ions with unpaired d-electrons show colour due to d-d transitions.
• Empty d-orbitals (e.g., Sc³⁺, Ti⁴⁺, V⁵⁺, Cr⁶⁺, Mn⁷⁺) are colourless.
• Charge transfer causes colour in ions like VO₂⁺ (pale yellow), CrO₄^2- (deep yellow), MnO₄^- (intense purple).

Solved Examples

• Ex 1: Why is MnO₄^- intensely purple?
  Charge transfer transitions absorb visible light strongly.

Practice Set

• Q1: Which ion is colourless due to empty d-orbitals?
  A. Ti⁴⁺ B. Fe²⁺ C. Cu²⁺ D. Mn²⁺
• Q2: What causes colour in VO₂⁺ ions?
  A. d-d transitions B. Charge transfer C. No colour D. Full d-orbitals
• Q3: Which ion shows intense purple colour?
  A. CrO₄^2- B. MnO₄^- C. VO₂⁺ D. Sc³⁺

Answer Key

• Q1: A - Ti⁴⁺ has empty d-orbitals, no d-d transitions.
• Q2: B - Charge transfer causes colour in VO₂⁺.
• Q3: B - MnO₄^- is intensely purple due to charge transfer.

Memory Aids

• Empty d-orbitals = colourless ions.
• Charge transfer ions: VO₂⁺, CrO₄^2-, MnO₄^-.

Stability of Higher Oxidation State

Halides

• Ti forms TiX₄ with oxidation state +4.
• Vanadium forms VF₅ (+5), chromium forms CrF₆ (+6).
• Mn does not form simple +7 halides; MnO₃F is known.
• Fluorine stabilises higher oxidation states via high lattice energy and bond enthalpy.
• CoF₃ stable due to high lattice enthalpy.
• VF₅ and CrF₆ stable due to high bond enthalpy.
• Vanadium pentahalide: only VF₅ known; others hydrolyse to VOX₃.
• Cu(II) halides known except iodide; Cu²⁺ oxidises I^- to I₂.
• Reaction: 2Cu²⁺_aq + 4I^-_aq → Cu₂I_2s + I_2s
• Copper(I) compounds disproportionate: 2Cu⁺ → Cu²⁺ + Cu
• Cu²⁺ more stable than Cu⁺ in aqueous solution due to hydration enthalpy.

Oxides

• Highest oxidation number matches group number: Sc₂O₃, TiO₂, V₂O₅, CrO₃, Mn₂O₇.
• No higher oxides beyond group 7 except Fe₂O₃, Co₃O₄.
• Oxocations like TiO²⁺, VO₂⁺ exist.
• Oxides more stable than fluorides at high oxidation states.
• Example: MnF₄ vs Mn₂O₇.
• Oxygen forms multiple bonds, increasing oxide stability.
• Mn₂O₇ structure: each Mn tetrahedrally surrounded by O; Mn-O-Mn bridge present.
• Tetrahedral ions: VO₄^3-, CrO₄^2-, MnO₄^3-, MnO₄^2-, MnO₄^-.

Solved Examples

• Ex 1: Write the formula of the highest oxide of Mn.
  Answer: Mn₂O₇
• Ex 2: Why are oxides more stable than fluorides at high oxidation states?
  Oxygen forms multiple bonds with metals, increasing stability.

Practice Set

• Q1: Which halide shows highest oxidation state for vanadium?
  A. VF₃ B. VF₄ C. VF₅ D. VCl₅
• Q2: What happens to copper(I) compounds in aqueous solution?
  A. Stable B. Disproportionate C. Oxidised to Cu(II) D. Reduced to Cu metal
• Q3: Which oxide corresponds to Mn in +7 oxidation state?
  A. MnO₂ B. Mn₂O₇ C. MnO₄^- D. MnO

Answer Key

• Q1: C - VF₅ is the only known pentahalide of vanadium(V).
• Q2: B - Copper(I) compounds disproportionate in aqueous solution.
• Q3: B - Mn₂O₇ is the highest oxide of Mn (+7 state).

Memory Aids

• Fluorides stabilise high oxidation states due to lattice and bond enthalpy.
• Cu(I) compounds disproportionate: 2Cu⁺ → Cu²⁺ + Cu.
• Highest oxide formula matches group number for early transition metals.

Special Notes on Transition Elements

Key Points

• All transition elements are d-block elements.
• Zn, Cd, Hg are pseudo transition elements (not true transition metals).
• Anhydrous FeSO₄ and CuSO₄ are white due to no 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.
• W (atomic no. 74) has highest melting point (3410°C) among d-block; Hg (80) has lowest (-38.9°C).

Solved Examples

• Ex 1: Why are anhydrous FeSO₄ and CuSO₄ white?
  No crystal field splitting in absence of water.
• Ex 2: Which d-block element has the highest melting point?
  Answer: Tungsten (W), 3410°C.

Practice Set

• Q1: Which is NOT a true transition metal?
  A. Fe B. Zn C. Ni D. Co
• Q2: Why is Cu²⁺ more stable than Cu⁺ in aqueous solution?
  A. Lower ionisation energy B. Higher hydration enthalpy C. Larger size D. More unpaired electrons
• 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 a pseudo transition metal.
• Q2: B - Cu²⁺ has higher hydration enthalpy stabilising it.
• Q3: B - Mercury (Hg) has the lowest melting point (-38.9°C).

Memory Aids

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

Some Important Compounds of Transition Elements

Oxides and Oxoanions of Metals

• Transition metals form oxides with oxygen at high temperatures; formula MO up to Mn (except Sc).
• Highest oxidation number in oxides equals group number.
• Examples: Sc₂O₃, Cr₂O₃, TiO₂, Mn₂O₇, V₂O₅.
• After Mn, only Fe₂O₃ and Co₃O₄ higher oxides known.
• Covalent character and acidic nature increase with metal oxidation state.
• Examples: V₂O₅, Cr₂O₃, Mn₂O₇ have low melting points; Mn₂O₇ is covalent green oil.
• Oxide acidity trend: V₂O₃ (basic), V₂O₄ (less basic), V₂O₅ (amphoteric).
• V₂O₅ reacts with alkali and acid:
  • V₂O₅ + 6NaOH → 2Na₃VO₄ + 3H₂O
  • V₂O₅ + 2HCl → 2VO₂Cl + H₂O
• CrO is basic; Cr₂O₃ amphoteric; CrO₃ and Mn₂O₇ acidic, forming acids in water:
  • CrO₃ + H₂O → H₂CrO₄
  • Mn₂O₇ + H₂O → 2HMnO₄
  • 2CrO₃ + H₂O → H₂Cr₂O₇

Potassium Dichromate (K₂Cr₂O₇)

Preparation

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

Structure

• Chromate ion (CrO₄^2-) is tetrahedral.
• Dichromate ion (Cr₂O₇^2-) consists of two tetrahedra sharing one corner; Cr-O-Cr bond angle 126°.

Properties

• Orange crystals, moderately soluble in cold water, readily soluble in hot water.
• In aqueous solution, equilibrium between dichromate and chromate ions:
  • Cr₂O₇^2- + H₂O ⇌ 2CrO₄^2- + 2H⁺
• In alkaline medium, equilibrium shifts forward (yellow solution): Cr₂O₇^2- + 2OH^- → 2CrO₄^2- + H₂O
• In acidic medium, equilibrium shifts backward (orange solution): 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
• Strong oxidising agent in acidic medium:
  • Cr₂O₇^2- + 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.
• Reaction: K₂Cr₂O₇ + 4KCl + 6H₂SO₄ → 2CrO₂Cl₂ + 6KHSO₄ + 3H₂O

Potassium Permanganate (KMnO₄)

Preparation

• Prepared from pyrolusite ore (MnO₂): 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
• K₂MnO₄ converted to KMnO₄ by:
  • Electrochemical oxidation: MnO₄^2- → MnO₄^- + e^- (green to purple)
  • In acidic/neutral medium, manganate disproportionates: 3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂ + 2H₂O
• Commercially prepared by alkaline fusion of MnO₂ followed by electrolytic oxidation.

Laboratory Preparation

• Mn²⁺ salt oxidised by peroxodisulphate: 2Mn²⁺ + 5S₂O₈^2- + 8H₂O → 2MnO₄^- + 10SO₄^2- + 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 with manganese d-orbitals.

Properties

• Dark purple crystals, sparingly soluble in water, decomposes at 513 K: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
• Strong oxidising agent in acidic and alkaline medium.
• 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

• Analytical chemistry and organic synthesis oxidant.
• Bleaching of wool, cotton, silk, and textiles.
• Decolourisation of oils.

The Inner Transition Elements (f-Block)

Definition and Series

• Elements with last electron entering f-orbital are f-block or inner transition elements.
• Consist of lanthanoids (14 elements after lanthanum) and actinoids (14 elements after actinium).

Lanthanoids Overview

Electronic Configurations

• Common outer configuration: 6s²
• Variable 4f electron occupancy: 4fⁿ (n=1 to 14)
• Most stable oxidation state: +3 (tripositive ions)

Atomic and Ionic Sizes

• Atomic and ionic radii decrease from La to Lu (lanthanoid contraction)
• 4f electrons poorly shield nuclear charge, increasing effective nuclear charge
• Atomic radii decrease irregularly; ionic radii decrease regularly

Oxidation States

• Predominant oxidation state: +3
• +2 and +4 states occur due to stable empty (4f⁰), half-filled (4f⁷), and fully-filled (4f¹⁴) subshells
• Examples: Ce⁴⁺ (4f⁰), Eu²⁺ (4f⁷), Tb⁴⁺ (4f⁷), Yb²⁺ (4f¹⁴)
• Ce⁴⁺ is a strong oxidant, Eb0 = 1.74 V, slow water oxidation
• Pr, Nd, Tb, Dy show +4 only in oxides (MO₂)
• Eu²⁺ and Yb²⁺ are strong reductants
• Samarium shows both +2 and +3 states

General Characteristics

• Silvery white, soft metals; tarnish in moist air
• Burn in oxygen to form Ln₂O₃ type oxides
• Hardness increases with atomic number; samarium is very hard
• Melting points mostly 1000-1200 K; samarium melts at 1623 K
• Good conductors of heat and electricity
• Density changes smoothly except for Eu, Yb, Sm, Tm
• Colored ions due to f-f transitions; La³⁺ and Lu³⁺ are colorless
• La³⁺ and Lu³⁺ are diamagnetic; others paramagnetic
• Ionisation enthalpies: 1st ~600 kJ/mol, 2nd ~1200 kJ/mol, 3rd irregular
• Standard electrode potentials: Eb0 (Ln³⁺/Ln) ~ -2.2 to -2.4 V (Eu exception: -2.0 V)

Chemical Behaviour

• React with H₂ on heating to form hydrides
• Form carbides (Ln₃C, LnC₂, Ln₂C₃) with carbon on heating
• 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 Ce⁴⁺ is a strong oxidant but reacts slowly with water.

• Ce⁴⁺ has noble gas configuration (4f⁰), making it stable.
• Standard electrode potential Eb0 = 1.74 V indicates strong oxidizing power.
• Slow reaction with water due to kinetic factors despite thermodynamic favorability.

Practice Set

• Q1: Which lanthanoid ion is colorless due to fully filled 4f orbitals?
  A. Eu³⁺
  B. La³⁺
  C. Gd³⁺
  D. Tb³⁺
• Q2: What causes the lanthanoid contraction?
  A. Poor shielding by 4f electrons
  B. Increase in atomic mass
  C. Increase in 5d electron count
  D. Decrease in nuclear charge
• Q3: Which lanthanoid shows both +2 and +3 oxidation states?
  A. Samarium
  B. Cerium
  C. Ytterbium
  D. Terbium

Answer Key

• Q1: B. La³⁺ - Has empty 4f⁰ orbitals, colorless ion.
• Q2: A. Poor shielding by 4f electrons - Leads to increased effective nuclear charge and contraction.
• Q3: A. Samarium - Exhibits both +2 and +3 oxidation states.

Memory Aids

• Lanthanoid contraction: Poor 4f shielding → radius decreases
• Stable oxidation states: Empty (4f⁰), half-filled (4f⁷), full (4f¹⁴)
• Colorless ions: La³⁺ (4f⁰), Lu³⁺ (4f¹⁴)
• Samarium: Only lanthanoid with +2 and +3 states

Actinoids Overview

General Information

• Actinoids include 14 elements from Thorium (Th) to Lawrencium (Lr).
• They are radioactive elements.
• General electronic configuration: 5f^0-14 6d^0-2 7s².
• Electrons fill 5f orbitals except Thorium (Z=90), with irregularities due to extra stability at f⁰, f⁷, and f¹⁴ configurations.

Electronic Configuration and Properties

• 5f orbitals extend beyond 6s and 6p orbitals, unlike 4f orbitals which are buried deep.
• Actinoid contraction is greater than lanthanoids due to poor shielding by 5f electrons.
• Ionic radii decrease regularly across the series.
• Most common oxidation state is +3; others include +4, +5, +6, +7 (e.g., Th, Pa, U, Np).
• Small energy gap between 5f, 6d, and 7s subshells allows multiple oxidation states.

Solved Examples

• Ex 1: Write the general electronic configuration of actinoids.
  Answer: 5f^0-14 6d^0-2 7s².
• Ex 2: Why do actinoids show multiple oxidation states?
  Answer: Due to small energy differences between 5f, 6d, and 7s subshells.

Practice Set

• Q1: Which element starts the actinoid series?
  A. Actinium
  B. Thorium
  C. Uranium
  D. Lawrencium
• Q2: What is the most common oxidation state of actinoids?
  A. +2
  B. +3
  C. +4
  D. +5
• Q3: Why is actinoid contraction greater than lanthanoid contraction?
  A. Better shielding by 5f electrons
  B. Poor shielding by 5f electrons
  C. 4f orbitals extend beyond 6s
  D. 5f orbitals are buried deep

Answer Key

• Q1: A - Actinium starts the actinoid series (Z=89).
• Q2: B - +3 is the most common oxidation state in actinoids.
• Q3: B - Poor shielding effect of 5f electrons causes greater contraction.

Memory Aids

• Actinoids = 5f filling, start at Actinium (Z=89).
• Multiple oxidation states due to close energy levels of 5f, 6d, 7s.
• Actinoid contraction > lanthanoid contraction due to poor 5f shielding.

Oxidation States

Oxidation State Variations

• Most actinoids show +3 oxidation state.
• Higher oxidation states (+4, +5, +6, +7) occur in some elements: Th (+4), Pa (+5), U (+6), Np (+7).
• Small energy gap between 5f, 6d, and 7s subshells allows multiple oxidation states.

Solved Examples

• Ex 1: Identify the highest oxidation state of Neptunium.
  Answer: +7.
• Ex 2: Which oxidation state is common to all actinoids?
  Answer: +3.

Practice Set

• Q1: Which actinoid shows +7 oxidation state?
  A. Thorium
  B. Protactinium
  C. Uranium
  D. Neptunium
• Q2: What is the common oxidation state for all actinoids?
  A. +2
  B. +3
  C. +4
  D. +5
• Q3: Why do actinoids show multiple oxidation states?
  A. Large energy gap between subshells
  B. Small energy gap between 5f, 6d, 7s
  C. Only 5f electrons involved
  D. Due to radioactive nature

Answer Key

• Q1: D - Neptunium shows +7 oxidation state.
• Q2: B - +3 is common to all actinoids.
• Q3: B - Small energy gap between 5f, 6d, and 7s subshells allows multiple states.

Memory Aids

• +3 oxidation state is universal in actinoids.
• Higher states (+4 to +7) appear in early actinoids (Th, Pa, U, Np).
• Small subshell energy gaps enable variable oxidation states.

General Characteristics

Physical and Chemical Properties

• Silvery white metals, tarnish rapidly in air forming oxide coating.
• Not attacked by alkalies; less reactive towards acids.
• Colored ions due to f-f transitions except Ac³⁺ (5f⁰), Cm³⁺ (5f⁷), Th⁴⁺ (5f⁰) which are colorless.
• Strongly paramagnetic.
• High densities except Thorium and Americium.
• High melting and boiling points with no regular trend.
• Low ionization energies.
• Strong reducing agents.
• Highly electropositive metals.

Solved Examples

• Ex 1: Why are most actinoid ions colored?
  Answer: Due to f-f electronic transitions.
• Ex 2: Which actinoid ions are colorless?
  Answer: Ac³⁺, Cm³⁺, Th⁴⁺.

Practice Set

• Q1: What happens to actinoids in air?
  A. Remain shiny
  B. Tarnish forming oxide coating
  C. React violently
  D. Dissolve
• Q2: Which actinoid ions are colorless?
  A. Ac³⁺, Cm³⁺, Th⁴⁺
  B. All actinoid ions
  C. Only Ac³⁺
  D. None
• Q3: What is the magnetic property of actinoids?
  A. Diamagnetic
  B. Paramagnetic
  C. Ferromagnetic
  D. Non-magnetic

Answer Key

• Q1: B - Actinoids tarnish rapidly forming oxide coating.
• Q2: A - Ac³⁺, Cm³⁺, Th⁴⁺ are colorless due to electronic configuration.
• Q3: B - Actinoids are strongly paramagnetic.

Memory Aids

• Actinoids tarnish in air forming oxide layers.
• Colorless ions: Ac³⁺, Cm³⁺, Th⁴⁺.
• Strongly paramagnetic due to unpaired f electrons.

Applications of d- and f-Block Elements

Key Industrial Uses

• Iron and steel: Primary materials in construction.
• TiO₂: Used as a white pigment in paints and coatings.
• MnO₂, Zn, Ni, Cd: Essential components in battery manufacturing.
• Cu, Ag, Au: Used for coins and ornamental purposes due to their malleability and luster.
• V₂O₅: Catalyst in the production of sulphuric acid.
• TiCl₄ with Al(CH₃)₃: Catalyst system for polythene synthesis.
• Iron catalyst: Used in Haber’s process to synthesize ammonia (NH₃).
• Ni catalyst: Facilitates hydrogenation of fats to produce saturated fats.
• Ni complexes: Catalyze polymerization of alkynes and other organic compounds.
• PdCl₂: Catalyst in Wacker’s process converting ethyne (C₂H₂) to ethanal (CH₃CHO).
• AgBr: Light-sensitive compound used in photographic films.
• Neodymium and praseodymium oxides: Used in colored glasses for goggles, absorbing yellow sodium light, protecting glass blowers.

Solved Examples

• Ex 1: Write the catalyst used in the Haber process.
  Answer: Iron catalyst (Fe).
• Ex 2: Identify the catalyst for hydrogenation of fats.
  Answer: Nickel (Ni) catalyst.
• Ex 3: Name the catalyst used in Wacker’s process.
  Answer: Palladium chloride (PdCl₂).

Practice Set

• Q1: Which element is primarily used in the manufacture of steel?
  A. Copper
  B. Iron
  C. Nickel
  D. Zinc
• Q2: TiO₂ is mainly used in which industry?
  A. Battery
  B. Pigment
  C. Photographic
  D. Polymer
• Q3: Which catalyst is used in the oxidation of ethyne to ethanal?
  A. Iron
  B. Nickel
  C. Palladium chloride
  D. Vanadium pentoxide

Answer Key

• Q1: B. Iron - Iron is the main element in steel production.
• Q2: B. Pigment - TiO₂ is widely used as a white pigment.
• Q3: C. Palladium chloride - PdCl₂ catalyzes Wacker’s process.

Memory Aids

• Iron = Steel (Construction)
• TiO₂ = White pigment
• MnO₂, Zn, Ni, Cd = Batteries
• Cu, Ag, Au = Coins & Ornaments
• V₂O₅ = Sulphuric acid catalyst
• TiCl₄ + Al(CH₃)₃ = Polythene catalyst
• Fe = Haber’s process (NH₃)
• Ni = Hydrogenation of fats
• PdCl₂ = Wacker’s process
• AgBr = Photographic films
• Nd, Pr oxides = Colored goggles