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The d- and f-block chapter carries many facts, and students often try to memorise every one: every configuration, every colour, every oxidation state. A better approach is to learn one skill well, which is working out the d-electron count of an ion. Magnetism, colour and much of the chemistry follow from it. This post shows how, with worked examples, then covers the lanthanoids and the mistakes that cost the most marks.
What do the syllabi include?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list the same d- and f-block unit: electronic configuration, trends in the first transition series (ionisation enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloys), the preparation, properties and uses of K₂Cr₂O₇ and KMnO₄, lanthanoid configurations, oxidation states and contraction, and actinoid configurations and oxidation states. The JEE (Advanced) 2026 brochure adds the oxoanions of chromium and manganese and lanthanoid and actinoid contractions. The NCERT Class 12 chapter on the d- and f-block elements is the base text. Check the current year's syllabus before you plan.
How do you write configurations for transition metal ions?
Start from the neutral atom. The NCERT chapter's table of ground-state configurations shows that most first-row metals are 3dⁿ4s², with two exceptions: chromium is 3d⁵4s¹ and copper is 3d¹⁰4s¹, because a half-filled or full d subshell is extra stable.
When the atom forms an ion, the 4s electrons leave first, even though 4s filled first.
Fe is 3d⁶4s². Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵.
Cu is 3d¹⁰4s¹. Cu⁺ is 3d¹⁰ and Cu²⁺ is 3d⁹.
Mn is 3d⁵4s². Mn²⁺ is 3d⁵ and Mn in MnO₄⁻ (+7) is 3d⁰.
The common mistake: removing 3d electrons first and writing Fe²⁺ as 3d⁴4s². That gives the wrong number of unpaired electrons and every later answer goes wrong.
Worked example 1: spin-only magnetic moment
For first-row transition metal ions, NCERT explains that the orbital contribution is largely quenched, so the magnetic moment is given by the spin-only formula μ = √[n(n + 2)] Bohr magnetons (BM), where n is the number of unpaired electrons.
Question: find the spin-only moment of the divalent ion of cobalt (Z = 27) in water.
Co is 3d⁷4s². Co²⁺ is 3d⁷.
Fill five d orbitals singly first, then pair: seven electrons give three unpaired.
μ = √[3 × 5] = √15 = 3.87 BM.
Working backwards: an ion has a measured moment of about 2.84 BM. Then n(n + 2) = 2.84² ≈ 8.07, so n = 2. A d² or d⁸ ion fits, for example V³⁺ or Ni²⁺.
Useful values to know: one unpaired electron gives 1.73 BM, two give 2.83, three 3.87, four 4.90 and five 5.92. The NCERT table lists observed values next to calculated ones; for Fe²⁺ the observed range is 5.3 to 5.5 BM against 4.90 calculated, a reminder that "spin-only" is an approximation.
Why are some transition metal ions coloured?
In a coloured ion, an electron absorbs visible light and jumps from a lower to a higher d orbital. The colour you see is the complementary colour of the light absorbed. That needs a partly filled d subshell. So:
d⁰ ions (Sc³⁺, Ti⁴⁺) and d¹⁰ ions (Zn²⁺, Cu⁺) are colourless.
Ions with 1 to 9 d electrons are usually coloured. NCERT's table of hydrated ions lists Ti³⁺ purple, V³⁺ green, Mn²⁺ pink, Fe³⁺ yellow, Fe²⁺ green, Ni²⁺ green and Cu²⁺ blue.
Worked example 2: which of Sc³⁺, Ti³⁺, Cu⁺, Ni²⁺ and Zn²⁺ are coloured in water? Their counts are d⁰, d¹, d¹⁰, d⁸ and d¹⁰. Only Ti³⁺ and Ni²⁺ have partly filled d subshells, so only they are coloured.
The same idea applies to complexes, where the ligands decide how far apart the d levels are. Our guide to naming coordination compounds covers the next chapter.
What is lanthanoid contraction and why does it matter?
Across the lanthanoids, electrons enter 4f orbitals, which shield the outer electrons poorly. Each step adds a proton, so the atoms and ions shrink steadily. The consequence is in the next period: the 5d metals are about the same size as the 4d metals above them. NCERT gives radii of 160 pm for zirconium and 159 pm for hafnium, which is why their chemistry is so alike.
Which lanthanoid oxidation states break the +3 rule?
The common state is +3, but NCERT notes exceptions that follow from stable empty, half-filled or full f subshells:
Ce⁴⁺ has a noble gas configuration and is a strong oxidant; E° for Ce⁴⁺/Ce³⁺ is +1.74 V.
Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) are reducing agents that change to +3.
Tb⁴⁺ (4f⁷) is an oxidant.
Actinoids show a wider range: the maximum oxidation state rises from +4 in thorium to +5, +6 and +7 in the next elements, according to the same chapter.
A practice routine for d- and f-block
Write configurations of every first-row ion in its common oxidation states, with unpaired electrons and spin-only moment.
For a mixed list of fifteen ions, predict coloured or colourless.
Balance the reactions of KMnO₄ and K₂Cr₂O₇ in acid with Fe²⁺, oxalate and iodide.
Explain lanthanoid contraction and its consequences in three sentences from memory.
Repeat steps 1 and 2 a week later without notes.
If a step such as orbital filling or d-orbital splitting is unclear, the Fermi Concept Library covers every concept in chemistry, each with its own lesson, so you can fix that idea before returning to the chapter. For the quantum numbers behind orbital filling, see quantum numbers and atomic structure.
Start now: write the d-electron count and spin-only moment for Cr³⁺, Mn²⁺ and Cu²⁺, then check them against the values above.
