Class 12 d and f Block Elements Guide 2027 — Transition Metals, Lanthanide Contraction, KMnO₄

In short: d and f Block Elements (Chapter 8) carries 5 marks in CBSE Class 12 Chemistry 2027 and covers the properties that make transition metals unique — variable oxidation states, colour, magnetic behaviour, catalytic activity, and complex formation. The chapter also covers lanthanide contraction and the preparation and properties of two important compounds: KMnO₄ (potassium permanganate) and K₂Cr₂O₇ (potassium dichromate). Board exams consistently test electronic configurations, reasons for properties, and the chemistry of Mn and Cr compounds.

d and f Block Elements is Chapter 8 in the NCERT textbook and forms the core of inorganic chemistry for Class 12. It carries approximately 5 marks on the board paper, typically split between one 3-mark question on general properties of transition metals or lanthanide contraction and one 2-mark question on KMnO₄/K₂Cr₂O₇ reactions or electronic configuration. This chapter requires a mix of conceptual understanding and factual memorisation — you need to know WHY transition metals show these properties, not just list them.

For the complete marks distribution, see the Class 12 Chemistry chapter-wise weightage. For all formulas and reactions, refer to the complete formulas and reactions list.

Electronic Configuration of d-Block Elements

The d-block elements are those in which the last electron enters the d-orbital. They span Groups 3 to 12 in the periodic table across four series:

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

Important exceptions in 3d series:

• Chromium (Cr, Z=24): Expected [Ar] 3d⁴ 4s² → Actual [Ar] 3d⁵ 4s¹ (half-filled d-orbital stability)
• Copper (Cu, Z=29): Expected [Ar] 3d⁹ 4s² → Actual [Ar] 3d¹⁰ 4s¹ (fully-filled d-orbital stability)

Memory trick — "Cr and Cu are rebels — they steal one electron from 4s to get d⁵ and d¹⁰ because half-filled and fully-filled d-orbitals are extra stable."

Transition Metals vs d-Block Elements

Not all d-block elements are transition metals. A transition metal is defined as an element that has a partially filled d-orbital in its ground state or in any of its common oxidation states.

Zinc (Zn) is a d-block element but NOT a transition metal because it has a completely filled d-orbital (3d¹⁰) in both its ground state and its common +2 oxidation state (Zn²⁺ = [Ar] 3d¹⁰).

Similarly, Scandium (Sc) shows only +3 oxidation state (Sc³⁺ = [Ar] 3d⁰), so some textbooks debate its classification, though NCERT includes it.

General Properties of Transition Metals

1. Variable Oxidation States

Transition metals show multiple oxidation states because both (n-1)d and ns electrons can participate in bonding. The number of oxidation states increases up to the middle of the series (Mn shows the maximum: +2 to +7) and then decreases.

Key oxidation state patterns:

• Minimum oxidation state = number of ns electrons (usually +2)
• Maximum oxidation state = total number of ns + (n-1)d electrons (up to Mn)
• After Mn, the maximum oxidation state decreases because of increasing nuclear charge holding d-electrons tightly
• The +2 oxidation state is the most common (loss of two 4s electrons)

Memory trick — "Mn is the king of oxidation states (+2 to +7). After Mn, it's downhill because d-electrons become more core-like."

2. Colour of Transition Metal Ions

Transition metal ions are coloured because they have partially filled d-orbitals. When visible light falls on the ion, a d-electron absorbs a specific wavelength and jumps from a lower-energy d-orbital to a higher-energy d-orbital (d-d transition). The colour we see is the complementary colour of the absorbed wavelength.

Ions with d⁰ or d¹⁰ configuration are COLOURLESS because d-d transitions are not possible.

• Sc³⁺ (d⁰) — colourless
• Ti⁴⁺ (d⁰) — colourless
• Zn²⁺ (d¹⁰) — colourless
• Cu⁺ (d¹⁰) — colourless

Coloured ions: Cu²⁺ (d⁹, blue), Fe³⁺ (d⁵, yellow), Cr³⁺ (d³, green), Mn²⁺ (d⁵, pale pink), Co²⁺ (d⁷, pink), Ni²⁺ (d⁸, green)

3. Magnetic Properties

Transition metal ions with unpaired d-electrons are paramagnetic (attracted by a magnetic field). The magnetic moment depends on the number of unpaired electrons.

Spin-only magnetic moment formula:

μ = √(n(n+2)) BM

where n = number of unpaired electrons, BM = Bohr Magneton

Memory trick — "More unpaired electrons = higher magnetic moment = more paramagnetic. d⁰ and d¹⁰ = diamagnetic (zero unpaired)."

4. Catalytic Activity

Transition metals and their compounds are excellent catalysts because:

• They have variable oxidation states — can provide an alternative reaction pathway with lower activation energy
• They have the ability to form temporary bonds with reactants (adsorption on surface)
• They can form interstitial compounds with small molecules

Common examples:

• Fe in Haber's process (N₂ + 3H₂ → 2NH₃)
• V₂O₅ in Contact process (2SO₂ + O₂ → 2SO₃)
• Ni in hydrogenation of oils
• MnO₂ in decomposition of KClO₃
• Pt/Pd in catalytic converters

5. Interstitial Compounds

Small atoms (H, C, N, B) get trapped in the voids (interstices) of the transition metal crystal lattice to form interstitial compounds.

Properties: Higher melting points than pure metals, very hard, retain metallic conductivity, chemically inert.

Examples: TiC (extremely hard, used in cutting tools), steel (iron with interstitial carbon), TiH₂

6. Alloy Formation

Transition metals form alloys readily because they have similar atomic sizes. Atoms of one metal can easily replace atoms of another in the crystal lattice (substitutional alloy).

Examples: Brass (Cu + Zn), Bronze (Cu + Sn), Stainless steel (Fe + Cr + Ni + C)

Lanthanide Contraction

The f-block elements include:

• Lanthanides (4f series): Ce (Z=58) to Lu (Z=71)
• Actinides (5f series): Th (Z=90) to Lr (Z=103)

General electronic configuration: [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s² (for lanthanides)

What Is Lanthanide Contraction?

As we move across the lanthanide series (Ce to Lu), each element has one more proton and one more 4f electron. The 4f orbitals have very poor shielding ability — they do not effectively screen the outer electrons from the increasing nuclear charge. The result is a steady decrease in atomic and ionic radii across the series.

Consequence: The size decrease accumulates across 14 elements. By the time we reach the 5d transition metals (Hf, Ta, W...), their radii are almost the SAME as the 4d metals above them (Zr, Nb, Mo...).

Effects of lanthanide contraction:

1. Hf ≈ Zr, Ta ≈ Nb, W ≈ Mo in size — this makes 4d and 5d elements hard to separate
2. Lanthanides have similar chemical properties — difficult to separate from each other
3. Basic strength of Ln(OH)₃ decreases from La(OH)₃ (most basic) to Lu(OH)₃ (least basic) — smaller ion holds OH⁻ less tightly

Memory trick — "Lanthanide contraction = f-orbitals are lousy shields. 14 elements of poor shielding add up to a significant size shrinkage."

Common Oxidation State of Lanthanides

The most stable oxidation state of all lanthanides is +3. Some show +2 or +4 when it leads to empty (f⁰), half-filled (f⁷), or fully filled (f¹⁴) configurations.

• Ce⁴⁺ (f⁰ — empty, extra stable)
• Eu²⁺ (f⁷ — half-filled, extra stable)
• Yb²⁺ (f¹⁴ — fully filled, extra stable)

Important Compounds: KMnO₄ and K₂Cr₂O₇

Potassium Permanganate (KMnO₄)

Preparation: 2MnO₂ + 4KOH + O₂ →(fuse)→ 2K₂MnO₄ + 2H₂O (green manganate) 3K₂MnO₄ + 2CO₂ → 2KMnO₄ + MnO₂ + 2K₂CO₃

Or by electrolytic oxidation: K₂MnO₄ → KMnO₄ (at anode, oxidation)

Properties:

• Deep purple/violet crystalline solid
• Mn is in +7 oxidation state
• Strong oxidising agent in both acidic and alkaline media
• Self-indicator in titrations — the pink/purple colour disappears as long as the reducing agent is present

Reactions in acidic medium (important): 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂ (Mn goes from +7 to +2 — five-electron reduction)

Memory trick — "KMnO₄ in acid: Mn goes from +7 to +2 (gains 5 electrons). In neutral/alkaline: Mn goes to +4 (MnO₂, brown precipitate)."

Potassium Dichromate (K₂Cr₂O₇)

Preparation: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ →(roast)→ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Chromate-Dichromate equilibrium: 2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O (In acidic solution: orange dichromate. In alkaline solution: yellow chromate)

Properties:

• Orange crystalline solid
• Cr is in +6 oxidation state
• Strong oxidising agent in acidic medium

Reactions in acidic medium: K₂Cr₂O₇ + 4H₂SO₄ + 3C₂H₅OH → K₂SO₄ + Cr₂(SO₄)₃ + 3CH₃CHO + 7H₂O (Cr goes from +6 to +3 — three-electron reduction per Cr atom, total 6 electrons)

Breathalyser test: K₂Cr₂O₇ in acidic medium turns from orange to green when it oxidises ethanol — used in drunk driving tests.

Board Exam Question Types

Type 1: Electronic Configuration and Properties (3 marks)

"Write the electronic configuration of Cr and Cu. Explain why they are exceptions."

Approach: Write the expected configuration first, then the actual one. Explain that half-filled (d⁵) and fully-filled (d¹⁰) d-orbitals have extra stability due to symmetrical distribution of electrons and maximum exchange energy.

Type 2: Reason-Based Questions on Properties (3 marks)

"Why do transition metals show variable oxidation states? Why are their ions coloured? Why do they form coloured compounds?"

Approach: For variable oxidation states — explain that both ns and (n-1)d electrons have similar energies and can participate in bonding. For colour — explain d-d transitions (absorption of visible light by partially filled d-orbitals). State the condition: d⁰ and d¹⁰ ions are colourless.

Type 3: Lanthanide Contraction (2-3 marks)

"What is lanthanide contraction? What are its consequences?"

Approach: Define it (steady decrease in ionic radii across lanthanide series due to poor shielding by 4f electrons). Give two consequences: (1) Similar radii of 4d and 5d elements, (2) Difficulty in separating lanthanides.

Type 4: KMnO₄ / K₂Cr₂O₇ Reactions (2 marks)

"Write the balanced equation for the reaction of KMnO₄ with oxalic acid in acidic medium."

Approach: Balance the equation, showing Mn going from +7 to +2 (acidic medium). Mention the colour change (purple to colourless). For K₂Cr₂O₇, mention Cr going from +6 to +3 (orange to green).

Type 5: Magnetic Moment Calculation (2 marks)

"Calculate the magnetic moment of Fe²⁺ ion."

Approach: Write the electronic configuration of Fe²⁺ ([Ar] 3d⁶). Count unpaired electrons (4 in high-spin). Apply μ = √(n(n+2)) = √(4×6) = √24 = 4.9 BM.

Common Mistakes That Cost Marks

1. Confusing d-block with transition metals. Zn, Cd, Hg are d-block elements but NOT transition metals (their d-orbitals are completely filled in both ground state and common oxidation states). Board exams specifically ask this distinction.

2. Writing wrong electronic configuration for Cr and Cu. Cr is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²). Cu is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²). These exceptions are asked almost every year. Always explain the reason (half-filled/fully-filled stability).

3. Claiming all transition metal compounds are coloured. Compounds of Sc³⁺ (d⁰), Ti⁴⁺ (d⁰), Zn²⁺ (d¹⁰), and Cu⁺ (d¹⁰) are colourless because d-d transitions require partially filled d-orbitals. Stating this exception shows understanding.

4. Not balancing KMnO₄ equations properly. In acidic medium, Mn goes from +7 to +2 (gain of 5 electrons). The coefficient of KMnO₄ and the reducing agent must balance the electron transfer. Practice balancing by the half-reaction method.

5. Defining lanthanide contraction incorrectly. Students often say "decrease in size across the periodic table" — this is too vague. Specifically state: "steady decrease in atomic and ionic radii across the lanthanide series (Ce to Lu) due to the poor shielding effect of 4f electrons."

6. Forgetting that 4s electrons are lost first in ionisation. When forming ions, transition metals lose 4s electrons BEFORE 3d electrons (even though 3d is filled after 4s). Fe²⁺ is [Ar] 3d⁶ (NOT [Ar] 3d⁴ 4s²). This is because in the ion, 3d is lower in energy than 4s.

4-Week Practice Plan

Week 1: Electronic Configuration and General Properties

• Read NCERT Chapter 8, Sections 8.1-8.4
• Write electronic configurations of all 3d series elements and their +2, +3 ions
• Study the five general properties with reasons (variable oxidation states, colour, magnetic properties, catalytic activity, interstitial compounds)
• Reference: NCERT Exercise Q1-Q6

Week 2: Deeper Properties and Calculations

• Practice magnetic moment calculations for 10 different ions
• Study catalytic activity with specific examples (Haber, Contact, hydrogenation)
• Learn alloy formation and interstitial compounds
• Solve comparison-type questions (why Cu²⁺ is coloured but Cu⁺ is not)
• Reference: NCERT Exercise Q7-Q12

Week 3: f-Block, Lanthanide Contraction, KMnO₄, K₂Cr₂O₇

• Read Sections 8.5-8.7 (f-block elements)
• Study lanthanide contraction — causes, consequences, effect on 5d elements
• Learn KMnO₄ and K₂Cr₂O₇ preparation, properties, and reactions
• Practice balancing redox equations in acidic medium
• Reference: NCERT Exercise Q13-Q20

Week 4: Board Paper Practice

• Solve d and f Block questions from the last 5 years of CBSE board papers
• Time yourself: 3-mark property questions in 7 minutes
• Focus on reason-based and assertion-reason type questions
• Final drill: write all electronic configurations of 3d elements and their ions from memory

Quick Revision Card for Exam Day

• d-block config: (n-1)d¹⁻¹⁰ ns¹⁻² | Cr: [Ar] 3d⁵ 4s¹ | Cu: [Ar] 3d¹⁰ 4s¹
• Transition metal: Partially filled d-orbital (Zn is NOT a transition metal)
• Variable oxidation states: Both ns and (n-1)d electrons participate
• Colour: d-d transitions (d⁰ and d¹⁰ = colourless)
• Magnetic moment: μ = √(n(n+2)) BM (n = unpaired electrons)
• Catalysts: Fe (Haber), V₂O₅ (Contact), Ni (hydrogenation)
• Lanthanide contraction: Poor 4f shielding → size decreases Ce to Lu
• Lanthanides: +3 is most stable. Ce⁴⁺ (f⁰), Eu²⁺ (f⁷) exceptions
• KMnO₄ (acidic): Mn⁷⁺ → Mn²⁺ (5e⁻ gain), purple → colourless
• K₂Cr₂O₇ (acidic): Cr⁶⁺ → Cr³⁺ (3e⁻ gain per Cr), orange → green

For guided practice on d and f block elements and inorganic reactions, try Padhaao's AI tutor.

Frequently Asked Questions

How many marks does d and f Block Elements carry in CBSE 2027?

d and f Block Elements carries approximately 5 marks in CBSE Class 12 Chemistry 2027. The questions are usually reason-based (why transition metals show variable oxidation states, why their compounds are coloured) or involve KMnO₄/K₂Cr₂O₇ reactions. Electronic configuration questions, especially for Cr and Cu, appear very frequently.

Why is zinc not considered a transition metal?

Zinc has the electronic configuration [Ar] 3d¹⁰ 4s². Its 3d orbital is completely filled in the ground state. Even in its most common oxidation state (+2), the configuration is [Ar] 3d¹⁰ — still completely filled. Since a transition metal must have a partially filled d-orbital in its ground state or common oxidation states, zinc does not qualify. It is a d-block element but not a transition metal.

How do you calculate the magnetic moment of a transition metal ion?

Use the spin-only formula μ = √(n(n+2)) Bohr Magneton, where n is the number of unpaired electrons. First write the electronic configuration of the ion (remember: 4s electrons are lost before 3d). Count the unpaired electrons in the d-orbitals. For example, Fe²⁺ is [Ar] 3d⁶ with 4 unpaired electrons, giving μ = √(4 × 6) = √24 = 4.9 BM.

What is lanthanide contraction and why does it happen?

Lanthanide contraction is the steady decrease in the atomic and ionic radii of lanthanide elements from Cerium (Z=58) to Lutetium (Z=71). It happens because the 4f electrons that are added across the series have very poor shielding ability. They do not effectively screen the outer electrons from the increasing nuclear charge, so the effective nuclear charge increases steadily, pulling the electron cloud inward and reducing the size.

Why do transition metals act as good catalysts?

Transition metals act as catalysts for two main reasons. First, they have variable oxidation states, which allows them to provide an alternative reaction pathway with lower activation energy by temporarily changing their oxidation state during the reaction. Second, they have the ability to form bonds with reactants by adsorbing them on their surface (or through forming intermediate compounds), bringing the reactant molecules closer together and in proper orientation for reaction.

What is the chromate-dichromate equilibrium?

In aqueous solution, chromate ions (CrO₄²⁻, yellow) and dichromate ions (Cr₂O₇²⁻, orange) exist in equilibrium: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. In acidic conditions, the equilibrium shifts right (orange dichromate is stable). In alkaline conditions, it shifts left (yellow chromate is stable). This colour change is a classic board exam question and is used analytically.

d and f Block Elements is the backbone of inorganic chemistry in Class 12. The electronic configurations tell you the oxidation states, the oxidation states explain the colour and magnetism, and the chemistry of KMnO₄ and K₂Cr₂O₇ shows these principles in action. Study the WHY behind each property, practise the calculations, and those 5 marks will be well within reach.