d-Block and f-Block Elements (Class 12): Trends, Colour and Magnetic Behaviour
Why transition metals are coloured, where variable oxidation states come from, and what lanthanoid contraction actually explains.
Class 12 · Inorganic Chemistry · Concept · Updated 25 August 2026
Why the characteristic properties appear
- Variable oxidation states — ns and (n−1)d electrons are close in energy, so differing numbers can be lost.
- Colour — d–d transitions absorb visible light; the colour seen is the complement of the light absorbed.
- Paramagnetism — unpaired d electrons are attracted into a magnetic field.
- Catalytic activity — variable oxidation states let the metal accept and donate electrons through a cycle.
- Complex formation — small, highly charged ions with vacant d orbitals accept lone pairs from ligands.
Calculating magnetic moment
The spin-only formula is asked almost every year:
| Unpaired electrons (n) | μ (BM) |
|---|---|
| 1 | 1.73 |
| 2 | 2.83 |
| 3 | 3.87 |
| 4 | 4.90 |
| 5 | 5.92 |
Note that Zn2+, Cd2+ and Sc3+ have no unpaired d electrons, so they are diamagnetic and colourless — the standard exception used to test whether you understand the cause rather than the pattern.
Lanthanoid contraction and what it explains
Across the lanthanoid series, electrons enter inner 4f orbitals which shield poorly, so the effective nuclear charge felt by the outer electrons rises steadily and the ionic radius shrinks. That steady decrease is the lanthanoid contraction.
Its most examined consequence is that second and third transition series elements have nearly identical sizes — zirconium and hafnium, for instance, are so alike that they are notoriously difficult to separate chemically.
FAQs
Why are Zn²⁺ compounds colourless?
Zn²⁺ has a full d¹⁰ configuration, so no d–d transition is possible and no visible light is absorbed.
What causes lanthanoid contraction?
Poor shielding by inner 4f electrons lets effective nuclear charge rise across the series, steadily pulling the outer electrons inward.
Why do transition metals act as good catalysts?
Their variable oxidation states allow them to accept and release electrons during a reaction cycle, and their surfaces can adsorb reactants.
How is spin-only magnetic moment calculated?
Using μ = √(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.
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