IIT-JAM Coordination Chemistry — VBT, CFT and MOT Compared

By Aniket Bhardwaj · 9 September 2026 · IIT-JAM Chemistry

Coordination chemistry is one of the most dependable parts of the inorganic section for an IIT-JAM aspirant, because the questions repeat in shape even when the complex changes. Almost every one of them asks the same three things: how many d-electrons does the metal have, how are they arranged, and what does that arrangement predict? Three bonding theories answer that — valence bond theory (VBT), crystal field theory (CFT) and molecular orbital theory (MOT). This article shows what each theory can and cannot do, and works the standard calculations in full.

Step 0 — get the d-electron count right

Every question begins here, and most wrong answers begin here too. Find the oxidation state of the metal from the overall charge, then remove electrons from the neutral atom — 4s before 3d for the first transition series.

Oxidation state of M = (overall charge on complex) − (sum of ligand charges)
dⁿ count = (group number for the neutral atom's valence electrons) − (oxidation state)

For [Fe(CN)₆]⁴⁻: six CN⁻ contribute −6, so the iron is −4 − (−6) = +2. Iron is [Ar]3d⁶4s²; removing two electrons (both from 4s) leaves d⁶. For [CoF₆]³⁻: 3 − 6 = −3 checks out with Co(III), and cobalt [Ar]3d⁷4s² minus three electrons gives d⁶ as well. Same count, completely different magnetic behaviour — that contrast is exactly what JAM likes to test.

Theory 1 — Valence bond theory (VBT)

VBT treats the metal–ligand bond as a coordinate bond: each ligand donates a lone pair into an empty hybrid orbital on the metal. You decide the hybridisation from the geometry, then check whether the d-electrons must pair up to make those orbitals empty.

Coordination number	Hybridisation	Geometry	Typical example
2	sp	Linear	[Ag(NH₃)₂]⁺
4	sp³	Tetrahedral	[NiCl₄]²⁻
4	dsp²	Square planar	[Ni(CN)₄]²⁻
6	d²sp³ (inner orbital)	Octahedral	[Fe(CN)₆]⁴⁻
6	sp³d² (outer orbital)	Octahedral	[FeF₆]³⁻

The useful VBT vocabulary for JAM is inner orbital (uses inner 3d orbitals, so the electrons had to pair, giving a low-spin complex) versus outer orbital (uses the outer 4d orbitals, electrons stay unpaired, high spin). VBT's weakness is that it never explains why CN⁻ forces pairing while F⁻ does not, and it says nothing at all about colour. That is the gap CFT fills.

Theory 2 — Crystal field theory (CFT) and CFSE

CFT models the ligands as point negative charges. In an octahedral field the five degenerate d orbitals split into a lower t_2g set (d_xy, d_yz, d_xz) and an upper e_g set (d_z², d_x²−y²). The gap is Δ_o. Measured from the average (barycentre), t_2g sits at −0.4Δ_o and e_g at +0.6Δ_o, so the weighted total stays zero: 3(−0.4) + 2(+0.6) = 0.

CFSE (octahedral) = [ −0.4 n(t_2g) + 0.6 n(e_g) ] Δ_o + (extra pairs) × P
Δ_t = (4/9) Δ_o · μ_spin-only = √[n(n+2)] BM

Here n(t_2g) and n(e_g) are electron counts, P is the pairing energy, and n in the magnetic formula is the number of unpaired electrons. Whether a d⁴–d⁷ complex is high or low spin is decided by a simple comparison: if Δ_o > P the electrons pair (low spin); if Δ_o < P they spread out (high spin).

Worked example 1 — the d⁶ pair

[CoF₆]³⁻, high spin. F⁻ is a weak-field ligand, so Δ_o < P. Filling gives t_2g⁴ e_g².
CFSE = 4(−0.4Δ_o) + 2(+0.6Δ_o) = −1.6Δ_o + 1.2Δ_o = −0.4Δ_o. Free-ion d⁶ already has one pair, and this arrangement also has one pair, so no extra pairing term.
Unpaired electrons n = 4 → μ = √[4(4+2)] = √24 = 4.90 BM. Paramagnetic.

[Fe(CN)₆]⁴⁻, low spin. CN⁻ is a strong-field ligand, so Δ_o > P. Filling gives t_2g⁶ e_g⁰.
CFSE = 6(−0.4Δ_o) + 0 = −2.4Δ_o. Three pairs are now present against one in the free ion, so two extra pairs must be paid for: total −2.4Δ_o + 2P.
Unpaired electrons n = 0 → μ = √[0(0+2)] = 0 BM. Diamagnetic.

Worked example 2 — converting CFSE into energy

A question gives Δ_o = 20 000 cm⁻¹ for [Cr(H₂O)₆]³⁺ and asks for the CFSE in kJ/mol.

Cr(III) is d³: t_2g³ e_g⁰ (no high/low spin question arises for d¹–d³).
CFSE = 3(−0.4Δ_o) = −1.2Δ_o = −1.2 × 20 000 = −24 000 cm⁻¹.
Conversion: 1 cm⁻¹ = 11.9627 J/mol.
Energy = −24 000 × 11.9627 = −287 104.8 J/mol = −287.1 kJ/mol.

Magnetic check: n = 3 → μ = √[3 × 5] = √15 = 3.87 BM.

The reference table you should be able to rebuild from memory

dⁿ	Field	Configuration	CFSE (Δ_o)	n unpaired	μ (BM)
d¹	either	t₂g¹	−0.4	1	1.73
d³	either	t₂g³	−1.2	3	3.87
d⁵	high spin	t₂g³ e_g²	0	5	5.92
d⁵	low spin	t₂g⁵	−2.0 (+2P)	1	1.73
d⁶	high spin	t₂g⁴ e_g²	−0.4	4	4.90
d⁶	low spin	t₂g⁶	−2.4 (+2P)	0	0
d⁸	either (oct.)	t₂g⁶ e_g²	−1.2	2	2.83
d⁹	either (oct.)	t₂g⁶ e_g³	−0.6	1	1.73

Two consequences worth remembering. A high-spin d⁵ complex such as [Mn(H₂O)₆]²⁺ has zero CFSE, which is part of why such complexes are pale and labile. And d⁹ (as in [Cu(H₂O)₆]²⁺) has an unevenly filled e_g set, so the octahedron distorts — the Jahn–Teller effect.

The spectrochemical series

Ligands are ordered by the size of Δ they produce:

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

Pure CFT cannot explain this order at all. A point-charge model predicts that the small, highly charged F⁻ should split the field more strongly than neutral CO — the opposite of what is observed. That failure is the reason MOT exists.

Theory 3 — Molecular orbital theory (MOT)

MOT combines the six ligand donor orbitals into symmetry-adapted combinations and mixes them with the metal's nine valence orbitals (one s, three p, five d). The result for an octahedral, σ-only complex is set out below — read the table as an energy ladder, lowest first.

Level (low → high)	Symmetry label	Character
1 (lowest)	a₁g, t₁u, e_g (bonding)	Six bonding MOs, mostly ligand in character; hold the 12 donated electrons
2	t₂g	Non-bonding in a σ-only complex; pure metal d
3	e_g*	Antibonding, mostly metal d
4 (highest)	a₁g, t₁u	Antibonding, empty

The gap between level 2 and level 3 is Δ_o — so MOT reproduces CFT's splitting rather than replacing it. Its extra power comes from π bonding:

π-donor ligands (F⁻, Cl⁻, OH⁻) have filled p orbitals that push the metal t_2g level up, shrinking Δ_o → weak field.
π-acceptor ligands (CO, CN⁻, phen) have empty π* orbitals that pull t_2g down, enlarging Δ_o → strong field.

That single idea explains the whole spectrochemical series in one sentence, and it is the standard "why" question in a JAM inorganic paper.

Common mistakes that cost marks

Removing 3d electrons before 4s. Fe²⁺ is d⁶, not d⁴ with a filled 4s. Fill 4s first, but empty it first as well.
Applying high/low spin to d¹, d², d³, d⁸, d⁹, d¹⁰. Those counts have only one possible octahedral arrangement — the question is meaningless there.
Forgetting that tetrahedral complexes are essentially always high spin, because Δ_t = (4/9)Δ_o is too small to beat P.
Adding the pairing term twice. Count only the pairs beyond those the free ion already had. Low-spin d⁶ has 3 pairs, free-ion d⁶ has 1, so the extra term is 2P, not 3P.
Using the spin-only formula where it fails. It is a good approximation for first-row complexes; for heavier metals and for ions with strong orbital contribution the experimental moment departs from it. Say so if the question invites a comment.
Giving CFSE without a sign. Stabilisation is negative. A bare "2.4Δ_o" can be marked down.

Where each theory is examined

Question type	Theory you need	What to write
Predict geometry and hybridisation	VBT	Orbital box diagram, inner vs outer orbital
High spin or low spin; magnetic moment	CFT	Compare Δ_o with P, then μ = √[n(n+2)]
Numerical CFSE in Δ_o or kJ/mol	CFT	−0.4/+0.6 weighting, then unit conversion
Colour and d–d transition energy	CFT / MOT	Δ_o corresponds to the absorbed photon
Why CO > CN⁻ > NH₃ > H₂O > F⁻	MOT	π-acceptor vs π-donor argument
Distorted geometry of Cu(II)	CFT	Jahn–Teller, uneven e_g occupancy

For the exact syllabus wording, the number of questions and the marking scheme, always read the current official IIT-JAM notification rather than any coaching summary — including this one.

Get the d-electron count right in seconds. Every CFSE and magnetic-moment question starts with the metal's configuration, and that is the step students rush. The free Electron Configuration Calculator writes out the full and noble-gas-core configuration for any element or ion, so you can check your dⁿ count before you start filling t_2g and e_g.

Open the Electron Configuration Calculator →

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