IIT-JAM d- and f-Block Chemistry — Configurations, Colour and Magnetism

By Aniket Bhardwaj · 21 September 2026 · IIT-JAM Chemistry

The transition elements look like a memory unit and are not one. Almost every question that gets asked about them comes from three quantitative ideas — how the d electrons are counted, how the five d orbitals split in a ligand field, and how many of those electrons stay unpaired. Get those three right and colour, magnetic moment, stability of oxidation states and even the shape of a complex all follow. The f block adds one extra idea, the lanthanide contraction, and one extra complication: spin-only magnetism stops working. This guide sets out all of that with the arithmetic done in full.

The two formulae that do most of the work

Spin-only magnetic moment: μ_s.o. = √[n(n + 2)] BM
where n = number of unpaired electrons, BM = Bohr magneton

Crystal field stabilisation energy (octahedral):
CFSE = (−0.4 × n_{t_2g} + 0.6 × n_{e_g}) Δ_o + (extra pairs) × P

In the octahedral field the three t_2g orbitals fall by 0.4Δ_o each and the two e_g orbitals rise by 0.6Δ_o each, so that the weighted average is unchanged — that is the "barycentre" rule, and it is the reason for the 0.4 and 0.6. P is the pairing energy, counted only for pairs that the ligand field forces which the free ion did not already have. In a tetrahedral field the splitting is inverted and much smaller, Δ_t ≈ (4/9)Δ_o, which is why tetrahedral complexes are essentially always high spin.

Configurations — the part students get wrong first

Two rules, applied in this order, settle every case in the 3d series.

Filling: 4s fills before 3d in the neutral atom, giving the general form (n−1)d^xns². The two anomalies are Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹, because a half-filled or completely filled d set is extra stable (exchange energy plus a symmetrical charge distribution).
Ionisation: the 4s electrons leave first, always. Once the 3d orbitals are occupied they lie below 4s in the ion, so Fe²⁺ is 3d⁶, not 3d⁴4s². Writing an ion's configuration by removing d electrons first is the single most common error in this unit.

Worked example 1 — configurations and spin-only moments

Write the d-electron count and the spin-only magnetic moment for Ti³⁺, Mn²⁺, Fe³⁺, Co²⁺ and Ni²⁺ in a weak (high-spin) field.

Ti (Z = 22) = [Ar]3d²4s². Remove 4s², then one 3d → Ti³⁺ = 3d¹, n = 1.
μ = √[1 × (1 + 2)] = √3 = 1.73 BM (check: 1.73² = 2.993)

Mn (Z = 25) = [Ar]3d⁵4s². Mn²⁺ = 3d⁵, all five unpaired, n = 5.
μ = √[5 × 7] = √35 = 5.92 BM (check: 5.92² = 35.05)

Fe (Z = 26) = [Ar]3d⁶4s². Remove 4s² and one 3d → Fe³⁺ = 3d⁵, n = 5, μ = 5.92 BM — the same as Mn²⁺, because magnetic moment depends on the d count, not on the element.

Co (Z = 27) = [Ar]3d⁷4s². Co²⁺ = 3d⁷. High spin in octahedral field: t_2g⁵e_g², so n = 3.
μ = √[3 × 5] = √15 = 3.87 BM (check: 3.87² = 14.98)

Ni (Z = 28) = [Ar]3d⁸4s². Ni²⁺ = 3d⁸, t_2g⁶e_g², n = 2.
μ = √[2 × 4] = √8 = 2.83 BM (check: 2.83² = 8.01)

Reading the formula backwards is just as common in the exam: given μ = 4.90 BM, solve n(n + 2) = 24.01, so n² + 2n − 24 = 0, giving n = (−2 + √(4 + 96))/2 = (−2 + 10)/2 = 4 unpaired electrons — a high-spin d⁴ or d⁶ ion.

High spin or low spin — the Δ_o versus P contest

Only d⁴ to d⁷ have a choice. If Δ_o > P the electrons pair up in t_2g (low spin, strong field); if Δ_o < P they spread out (high spin, weak field). Ligands are ranked by the Δ_o they produce — the spectrochemical series:

I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < bpy < phen < NO₂⁻ < CN⁻ < CO

The order is not explained by simple electrostatics — π-donor ligands such as the halides lower Δ_o, and π-acceptor ligands such as CN⁻ and CO raise it sharply. That is why cyanide and carbon monoxide sit at the strong-field end even though CO carries no charge.

Worked example 2 — CFSE and magnetism for the same metal ion

Compare [Fe(H₂O)₆]³⁺ and [Fe(CN)₆]³⁻.

Both are Fe(III), so both are d⁵. Only the ligand differs.

Aqua complex, weak field, high spin: t_2g³e_g².
CFSE = (−0.4 × 3) + (0.6 × 2) = −1.2 + 1.2 = 0.
No extra pairs are forced, so no P term. n = 5 → μ = √35 = 5.92 BM.

Cyanido complex, strong field, low spin: t_2g⁵e_g⁰.
CFSE = (−0.4 × 5) + (0.6 × 0) = −2.0 Δ_o.
The free ion had five unpaired electrons and no pairs; this arrangement has two pairs, so add +2P. Net CFSE = −2.0Δ_o + 2P.
n = 1 → μ = √3 = 1.73 BM.

The chemistry follows straight from the numbers: high-spin d⁵ has zero CFSE, which is why weak-field d⁵ complexes are kinetically labile and why Mn(II) salts are so faintly coloured, while the strongly stabilised low-spin d⁶ (t_2g⁶, CFSE = −2.4Δ_o) is the reason [Co(NH₃)₆]³⁺ and [Fe(CN)₆]⁴⁻ are so inert.

Worked example 3 — turning a colour into Δ_o

[Ti(H₂O)₆]³⁺ absorbs most strongly at about 500 nm. Find Δ_o in kJ mol⁻¹, and say what colour the solution looks.

Energy of one photon, E = hc/λ, with h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m s⁻¹, λ = 500 nm = 5.00 × 10⁻⁷ m:

hc = 6.626 × 10⁻³⁴ × 2.998 × 10⁸ = 1.9865 × 10⁻²⁵ J m
E = 1.9865 × 10⁻²⁵ ÷ 5.00 × 10⁻⁷ = 3.973 × 10⁻¹⁹ J per photon

Per mole, multiply by N_A = 6.022 × 10²³ mol⁻¹:
E = 3.973 × 10⁻¹⁹ × 6.022 × 10²³ = 2.393 × 10⁵ J mol⁻¹ = 239 kJ mol⁻¹

Cross-check by the wavenumber route: ν̄ = 1/λ = 1/(500 × 10⁻⁷ cm) = 20 000 cm⁻¹. Since 1 cm⁻¹ ≡ 11.96 J mol⁻¹, 20 000 × 11.96 = 239 200 J mol⁻¹ = 239 kJ mol⁻¹. The two routes agree, so the arithmetic is safe.

Colour: the solution transmits what it does not absorb. Absorption near 500 nm (green) leaves the complementary colour, and the ion is seen as violet. This is a d¹ ion, so exactly one d–d transition is possible and the band is a single broad hump — which is also why the absorption maximum can be read straight off as Δ_o for d¹ only. For d² and beyond, several transitions overlap and Δ_o must be pulled out of a Tanabe–Sugano analysis instead.

Why d–d bands are weak, and charge-transfer bands are not

A d–d transition is Laporte forbidden in a centrosymmetric complex (g → g) and is also spin forbidden if it requires a spin flip. The molar absorptivity is therefore small — typically only a few units to a few tens of L mol⁻¹ cm⁻¹ — so d–d colours are pale. Vibrations that momentarily destroy the centre of symmetry are what let the transition happen at all. Contrast that with MnO₄⁻ and CrO₄²⁻: these are d⁰ ions with no d electrons to promote, so their intense colour cannot be a d–d band at all. It is a ligand-to-metal charge transfer, which is fully allowed and therefore hundreds of times stronger. Being asked to explain the colour of permanganate is a standard trap; "d–d transition" is the wrong answer.

Oxidation states and the shape of the series

Observation	Explanation to write in the exam
Highest oxidation state peaks in the middle (Mn reaches +7)	Both 4s and 3d electrons are available; after Mn the 3d electrons are held too tightly to all be used
+2 becomes more common on the right	Increasing effective nuclear charge; the 3d electrons become core-like
Higher oxidation states appear as oxides and fluorides	Only the small, highly electronegative O and F can stabilise a high positive centre
Sc³⁺ and Zn²⁺ compounds are white	d⁰ and d¹⁰ respectively — no d–d transition is possible
Sc and Zn are often excluded from "transition metals"	Neither has a partly filled d subshell in its common oxidation state
4d and 5d metals favour low spin more than 3d	Larger, more diffuse orbitals give a bigger Δ and a smaller pairing energy

The f block — lanthanides

The 4f orbitals are buried beneath the filled 5s and 5p shells. They barely reach the ligands, and three consequences follow directly from that one fact:

+3 is the dominant oxidation state for essentially the whole series, because the 4f electrons are not accessible for variable bonding the way 3d electrons are. The exceptions are driven by empty, half-filled or filled f shells: Ce⁴⁺ (4f⁰, a useful oxidising agent in titrations), Eu²⁺ (4f⁷), Tb⁴⁺ (4f⁷) and Yb²⁺ (4f¹⁴).
Colours and spectra are sharp and almost ligand-independent. f–f transitions are shielded from the ligand field, so lanthanide absorption bands are line-like rather than the broad humps seen for d–d bands.
The lanthanide contraction. Across the series the nuclear charge rises while each added electron enters a poorly shielding 4f orbital, so the ionic radius shrinks steadily. By the end of the series the accumulated shrinkage is large enough that the 5d elements are almost the same size as their 4d congeners — Zr and Hf in particular are so alike that their separation is genuinely difficult, and the 5d metals are unusually dense and unreactive.

Worked example 4 — why spin-only fails for the lanthanides

Predict the magnetic moment of Nd³⁺ (4f³).

For 4f ions, spin–orbit coupling is large and orbital angular momentum is not quenched, so the spin-only formula does not apply. Use the Landé expression:

μ = g√[J(J + 1)] BM, with g = 1 + [J(J+1) + S(S+1) − L(L+1)] ÷ [2J(J+1)]

For f³: three unpaired electrons, so S = 3/2. Placing them in the m_ℓ = +3, +2, +1 orbitals by Hund's rule gives L = 3 + 2 + 1 = 6. The shell is less than half full, so J = L − S = 6 − 1.5 = 4.5.

J(J+1) = 4.5 × 5.5 = 24.75 · S(S+1) = 1.5 × 2.5 = 3.75 · L(L+1) = 6 × 7 = 42

g = 1 + (24.75 + 3.75 − 42) ÷ (2 × 24.75) = 1 + (−13.5 ÷ 49.5) = 1 − 0.2727 = 0.7273

√24.75 = 4.975 (check: 4.975² = 24.75)

μ = 0.7273 × 4.975 = 3.62 BM

The spin-only answer for three unpaired electrons would have been √15 = 3.87 BM, which is clearly different. Measured moments for Nd(III) salts sit close to the Landé value, not the spin-only one. Gd³⁺ is the honourable exception: f⁷ has L = 0, so J = S and the two formulae collapse into the same answer, μ = √(7 × 9) = √63 = 7.94 BM. Whenever a question gives a lanthanide moment that spin-only cannot reproduce, that is the point being tested.

Actinides in brief

The 5f orbitals are less buried than 4f, so the early actinides (Th to Am) show a much wider range of oxidation states — uranium reaches +6, neptunium +7 — before settling back towards +3 in the later members. All actinides are radioactive, the later ones are synthetic and available only in tiny amounts, and an actinide contraction runs parallel to the lanthanide one. For JAM purposes the safe summary is: compare the two series, do not try to memorise actinide chemistry in the same detail as lanthanide chemistry.

Common mistakes that cost marks

Removing 3d electrons before 4s when writing an ion. Fe²⁺ is 3d⁶. Every magnetic-moment answer collapses if this step is wrong.
Assuming a tetrahedral complex can be low spin. Δ_t is roughly 4/9 of Δ_o and almost never beats the pairing energy — treat tetrahedral as high spin unless the question says otherwise.
Forgetting the pairing-energy term in CFSE. Quoting −2.4Δ_o for low-spin d⁶ without the +2P is an incomplete answer.
Calling the colour of MnO₄⁻ a d–d transition. It is d⁰. The band is charge transfer, which is also why it is so intense.
Using the spin-only formula for lanthanides. It works for Gd³⁺ by accident of L = 0, and fails for nearly everything else.
Saying the absorbed colour is the observed colour. The observed colour is the complement of what is absorbed.
Treating the lanthanide contraction as a lanthanide-only fact. Its most examined consequence is about the 4d/5d transition metals, especially Zr and Hf.

How to prepare this unit

Theme	What you must be able to do without hesitation
Configurations	Write any 3d atom or ion, including Cr and Cu, and know which electrons leave first
Spin-only moment	Go both ways — n to μ, and μ back to n by solving the quadratic
CFSE	Compute for any dⁿ, octahedral and tetrahedral, with the P term where it applies
Spectrochemical series	Predict high or low spin from the ligand, and justify it with Δ_o versus P
Colour	Convert λ to Δ_o in kJ mol⁻¹ or cm⁻¹, and name the complementary colour
Charge transfer	Recognise d⁰ and d¹⁰ ions and explain intense colour without d–d bands
Lanthanides	State the contraction, its consequences, the +3 rule and the f⁰/f⁷/f¹⁴ exceptions
Landé formula	Derive S, L and J for a given fⁿ and compute μ

Treat that as a revision checklist, not as a prediction of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.

Check your d-electron counts quickly. Every magnetic and CFSE answer above starts from a configuration, and that is the step where marks are lost. The electron configuration calculator writes out the configuration for an element so you can confirm the neutral atom before you strip the 4s electrons off by hand.

Open the Electron Configuration Calculator →

Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches — classroom at the Gurugram coaching centre and live online classes for students anywhere in India. Details at abcchemistry.in.