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🧪 Chemistry · Class 12 · NEET & JEE

d- and f-Block Elements

Transition metals (d-block) and lanthanides/actinides (f-block). Known for coloured compounds, variable oxidation states, complex formation, and catalytic properties. Frequently tested in JEE and NEET.

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Reading time~10 min

Revision time~4 min

Last updated2026-07-19

1 Read the chapter ~10 min

🎯 Key Points

d-block: (n−1)d¹⁻¹⁰ ns⁰⁻² — close energy of (n−1)d and ns causes variable oxidation states
Colour comes from d-d electron transitions; d⁰ and d¹⁰ ions are colourless (no transition possible)
Lanthanide contraction: steady radius decrease across lanthanides (poor 4f shielding) → makes Zr/Hf, Nb/Ta nearly identical and hard to separate
Magnetic moment μ = √(n(n+2)) Bohr magnetons, where n = number of unpaired electrons
Key catalysts: Fe (Haber), V₂O₅ (Contact process), Ni/Pt (hydrogenation)

Aqueous solutions of transition metal ions show characteristic colours caused by electrons absorbing visible light to jump between split d-orbitals; the exact colour depends on the metal, its oxidation state, and surrounding ligands.

Transition Metals (d-Block)

Elements of Groups 3-12 with partially filled d orbitals. Include Fe, Cu, Zn, Cr, Mn, Ni, Co, Ti, V, and others.

General Properties

Hard, dense metals with high melting and boiling points
Good conductors of heat and electricity
Variable oxidation states (due to similar energies of 3d and 4s)
Form coloured ions and complex compounds
Many act as catalysts in industrial processes

Variable Oxidation States

Fe: +2 (FeSO₄, green), +3 (Fe₂O₃, rust)
Cr: +3 (Cr₂O₃, green), +6 (K₂Cr₂O₇, orange)
Mn: +2 (pale pink), +4 (MnO₂, black), +7 (KMnO₄, purple)
Cu: +1 (Cu₂O), +2 (CuSO₄, blue)

Chart of oxidation states 1 to 8 against atomic number for the three transition series Sc-Zn, Y-Cd and Lu-Hg; filled dots mark common states and open dots less common ones, peaking near the middle of each series.

Transition metals show variable oxidation states because their (n−1)d and ns electrons are close in energy and can all take part in bonding. The number of states rises to a maximum near the middle of each series (e.g. Mn reaches +7) and falls towards the ends. Image: Felix Wan / Andel, CC0, via Wikimedia Commons.

Coloured Compounds

Colour arises because d-electrons absorb visible light and jump to higher d-orbitals. The complementary colour is what we observe.

CuSO₄.5H₂O: Blue; K₂Cr₂O₇: Orange; KMnO₄: Purple
Zn, Sc, and Ti⁴⁺ are colourless (empty or fully filled d-orbitals)

Important Catalysts

Fe: Haber process (NH₃ synthesis)
V₂O₅: Contact process (H₂SO₄ manufacturing)
Ni or Pt: Hydrogenation of oils

f-Block Elements

Lanthanides (Ce to Lu): 4f filling; used in permanent magnets and lasers
Actinides (Th to Lr): 5f filling; mostly radioactive; include U and Pu (nuclear fuel)
Lanthanide contraction: steady decrease in size across lanthanides; affects properties of 5th and 6th period transition metals

General Electronic Configuration

d-block: (n-1)d¹⁻¹⁰ ns⁰⁻² ; the (n-1)d and ns orbitals are close in energy, which is the root cause of variable oxidation states
Cr has configuration 3d⁵4s¹ and Cu has 3d¹⁰4s¹ (not the expected 3d⁴4s² and 3d⁹4s²) because half-filled and fully-filled d-subshells offer extra stability
f-block: (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns² ; lanthanides fill 4f, actinides fill 5f

Why Transition Metal Ions are Coloured

In a free ion the five d-orbitals are degenerate (equal energy), but ligands split them into two sets (d-d splitting, explained by Crystal Field Theory)
An electron absorbs a specific wavelength of visible light to jump from a lower to a higher d-orbital (d-d transition); the colour seen is complementary to the colour absorbed
Ions with completely empty (d⁰, e.g., Sc³⁺, Ti⁴⁺) or completely filled (d¹⁰, e.g., Zn²⁺, Cu⁺) d-orbitals are colourless since no d-d transition is possible

Lanthanide Contraction

Steady decrease in atomic and ionic radii of lanthanides from La to Lu, caused by the poor shielding effect of 4f electrons (imperfect shielding allows effective nuclear charge to increase steadily).

Consequence: radii of second and third-row transition elements in the same group become very close (e.g., Zr and Hf have nearly identical radii), making their chemical separation difficult
This is why elements like Zr/Hf and Nb/Ta show very similar chemical properties

Magnetic Properties and Interstitial Compounds

Transition metal ions with unpaired d-electrons are paramagnetic; the magnetic moment increases with the number of unpaired electrons (mu = root of n(n+2) Bohr magnetons)
Transition metals form interstitial compounds with small atoms like H, C, N trapped in lattice gaps; these are hard, retain metallic conductivity, and have higher melting points than the pure metal
Alloy formation is common among transition metals because of similar atomic sizes, allowing one metal atom to be readily replaced by another in the lattice

Periodic Trends in the Transition Series

Atomic radii: Decrease across a series at first, then stay almost constant (added d-electrons shield poorly but electron–electron repulsion offsets the rising nuclear charge)
Ionisation enthalpy: Rises gradually across a series, but irregularly, owing to the stability of half-filled/filled d-subshells
Enthalpy of atomisation and melting point: High, and maximum near the middle of the series (Cr, Mo, W) where the number of unpaired d-electrons available for metallic bonding peaks; Zn, Cd, Hg have low values (no unpaired d-electrons)
Density: Increases across a series as radius falls and atomic mass rises

Potassium Dichromate (K₂Cr₂O₇)

Preparation: Chromite ore (FeCr₂O₄) fused with Na₂CO₃ in air → Na₂CrO₄; acidified to Na₂Cr₂O₇, then treated with KCl to crystallise K₂Cr₂O₇ (orange crystals)
Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ equilibrium: Orange dichromate in acid converts to yellow chromate in base and back (pH-dependent)
Oxidising action (in acidic medium, Cr⁶⁺ → Cr³⁺, gain of 6 e⁻): oxidises Fe²⁺ → Fe³⁺, I⁻ → I₂, H₂S → S, SO₃²⁻ → SO₄²⁻
Half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = +1.33 V); used in volumetric analysis as a primary standard

Potassium Permanganate (KMnO₄)

Preparation: Pyrolusite (MnO₂) fused with KOH and an oxidiser (KNO₃/air) → green K₂MnO₄, which is then oxidised (electrolytically or by Cl₂) to purple KMnO₄
Oxidising action depends on medium:
- Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.51 V); strongest, used in titrations (self-indicating)
- Neutral/faintly alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
- Strongly alkaline: MnO₄⁻ + e⁻ → MnO₄²⁻
In acidic medium it oxidises Fe²⁺ → Fe³⁺, oxalate (C₂O₄²⁻) → CO₂, I⁻ → I₂, and decolourises on reduction — the basis of permanganometry

Lanthanoids vs Actinoids

Feature	Lanthanoids (4f)	Actinoids (5f)
Common oxidation state	+3 (dominant)	+3, but many others (+4 to +7) common
Radioactivity	Only Pm	All are radioactive
Shielding by f-electrons	Better (4f)	Poorer (5f), so actinoid contraction is larger
Complex formation	Limited	Greater (higher charge, larger size)

Standard Electrode Potentials and Reactivity

Most first-series transition metals have negative E°(M²⁺/M), so they can displace H₂ from dilute acids, but the values are irregular due to differing ionisation and hydration/atomisation enthalpies
Cu has a POSITIVE E°(Cu²⁺/Cu) = +0.34 V (high atomisation + ionisation enthalpy not offset by hydration), so it does NOT liberate H₂ from dilute acids
E°(Mn²⁺/Mn) is unusually negative and E°(Zn²⁺/Zn) too, linked to the extra stability of the half-filled d⁵ (Mn²⁺) and filled d¹⁰ (Zn²⁺) configurations of the ions formed

Catalytic Behaviour of Transition Metals

Transition metals and their compounds are excellent catalysts largely because they show variable oxidation states and can form intermediate complexes with reactants, providing a low-activation-energy pathway
Their partially filled d-orbitals let them adsorb reactant molecules onto the metal surface (heterogeneous catalysis), weakening bonds and bringing reactants close together
Key examples: Fe in the Haber process (NH₃), V₂O₅ in the Contact process (H₂SO₄), finely divided Ni in hydrogenation of oils, and MnO₂ in the decomposition of KClO₃
The catalytic action of Fe²⁺/Fe³⁺ in the reaction between iodide and persulphate ions illustrates catalysis by oxidation-state cycling in solution

Complex Formation and Alloy Formation

Complex formation: Transition metal ions readily form coordination complexes because they have small size, high effective nuclear charge (high charge/size ratio) and available vacant d-orbitals to accept lone pairs from ligands, e.g. [Fe(CN)₆]^4-, [Cu(NH₃)₄]²⁺
Alloy formation: Because transition metals have very similar atomic radii, atoms of one metal can replace atoms of another in the crystal lattice, giving substitutional alloys such as brass (Cu-Zn), bronze (Cu-Sn) and steels
Alloys are generally harder, have higher melting points and are more resistant to corrosion than the constituent pure metals

Actinoids and Actinoid Contraction

Actinoids (Th to Lr) involve filling of the 5f orbitals; all are radioactive and elements beyond uranium (transuranics) are man-made
They show a wider range of oxidation states than lanthanoids (e.g. U shows +3, +4, +5, +6) because the 5f, 6d and 7s orbitals lie close in energy
Actinoid contraction: a steady decrease in atomic and ionic radii across the series; it is greater than lanthanoid contraction because 5f electrons shield the nuclear charge even more poorly than 4f electrons
Actinoids have a greater tendency to form complexes than lanthanoids, owing to their higher charges and larger sizes

🚀 JEE Advanced Edge

Why Mn shows the most oxidation states: Mn (3d⁵4s²) is roughly in the middle of the first transition series, giving it the maximum number of accessible oxidation states from +2 to +7 — every electron from 3d and 4s can potentially be involved in bonding, unlike elements at either extreme of the series.

Why Zn, Cd, Hg are NOT considered true transition metals: Despite being in the d-block, they have a completely filled d¹⁰ configuration in both the metal AND their common ions (Zn²⁺, Cd²⁺, Hg²⁺) — no partially-filled d-orbital ever exists, so they show none of the hallmark transition behaviours (no colour, no variable oxidation states, weak catalytic activity).

Calculating magnetic moment from oxidation state: For Fe³⁺ ([Ar]3d⁵), all 5 d-electrons are unpaired (high-spin, since d⁵ is exactly half-filled) → n=5 → μ = √(5×7) = √35 ≈ 5.92 BM, a frequently tested numerical.

2 Revise ~4 min before the exam

📐 Formula Sheet

Transition metals: partly filled d orbitals; variable oxidation states, coloured ions, catalytic activity
Variable oxidation states: arise because (n−1)d and ns energies are close, so different numbers of electrons participate
Colour: from d–d electronic transitions; ions with d⁰ or d¹⁰ (Sc³⁺, Zn²⁺) are colourless
Magnetic moment: μ = √(n(n + 2)) BM, n = unpaired electrons
Catalysis: variable states and surface adsorption make them good catalysts (Fe in Haber, V₂O₅ in Contact)
Alloys and interstitial compounds: small atoms (C, N, H) fit interstitially, giving hard materials like steel
Lanthanoid contraction: steady size decrease across the 4f series; makes 4d and 5d elements similar in size
KMnO₄ and K₂Cr₂O₇: important oxidisers; Mn goes +7→+2, Cr goes +6→+3

3 Practice apply it

✍️ Worked Examples

Example 1 — Why transition-metal ions are coloured
Q: Explain why Cu²⁺ is blue but Zn²⁺ is colourless.
Step 1 — Colour arises when d electrons absorb visible light to jump between split d orbitals (d–d transition).
Step 2 — Cu²⁺ is d⁹ — it has a partly filled d set, so such transitions occur and it absorbs red, appearing blue.
Step 3 — Zn²⁺ is d¹⁰ — the d orbitals are full, so no d–d transition is possible.
Answer: Cu²⁺ has a vacancy for a d–d transition; Zn²⁺'s full d¹⁰ does not, so it is colourless. Note: Sc³⁺ (d⁰) is colourless for the opposite reason — no d electrons at all.

Example 2 — Magnetic moment
Q: Calculate the spin-only magnetic moment of Fe²⁺ (d⁶) in a high-spin octahedral field.
Step 1 — High-spin d⁶ fills the five orbitals singly (5 electrons) then pairs one, leaving 4 unpaired.
Step 2 — Apply μ = √(n(n + 2)) with n = 4: = √(4 × 6) = √24.
Step 3 — Compute: ≈ 4.9 BM.
Answer: ≈ 4.9 BM. Note: a low-spin d⁶ complex would pair all electrons (n = 0) and be diamagnetic.

Example 3 — Lanthanoid contraction
Q: What is the lanthanoid contraction and one of its consequences?
Step 1 — Across the 4f series, each added electron enters an inner 4f orbital that shields poorly.
Step 2 — Effective nuclear charge rises steadily, so atomic and ionic radii shrink gradually across the series.
Step 3 — This offsets the expected size increase from period 5 to 6, making Zr and Hf (and Nb/Ta) almost identical in size and hard to separate.
Answer: the steady 4f-series size decrease makes second- and third-row transition metals similar in size. Note: it is why zirconium and hafnium occur together and resist separation.