Crystal Field Theory Explained: High Spin, Low Spin and Colour in Complexes (NEET & JEE)

Why is copper sulphate solution blue, while zinc sulphate is clear as water? Both are just salts of neighbouring d-block metals, yet one absorbs visible light and the other does not. Crystal Field Theory answers this in a single picture, and that picture also decides magnetism, spin state and several NEET and JEE questions every year.

Quick Summary

• In an octahedral complex, the five d orbitals split into a lower t2g set (3 orbitals) and a higher eg set (2 orbitals), separated by Δo.

• Weak-field ligands give high-spin complexes; strong-field ligands give low-spin ones, but only for d4 to d7 ions.

• Unpaired electrons give magnetism: μ = √(n(n+2)) BM.

• Colour comes from d-d transitions; no partially filled d set (d0 or d10) means no colour.

What Is Crystal Field Theory?

Crystal Field Theory (CFT) treats ligands as point negative charges that repel the d electrons of the central metal ion. Because the five d orbitals point in different directions, they are repelled by different amounts, so their energies split.

In an octahedral complex the ligands approach along the x, y and z axes. The eg orbitals (dx²−y², dz²) point straight at the ligands and rise in energy, while the t2g orbitals (dxy, dyz, dzx) point between them and drop. The t2g set sits at −0.4Δo and the eg set at +0.6Δo relative to the average energy.

High Spin or Low Spin: The Pairing Decision

Once the orbitals split, each incoming electron faces a choice: jump up to an eg orbital, or pair up in a t2g orbital and pay the pairing energy (P). If Δo < P, electrons spread out and the complex is high spin. If Δo > P, they pair first and the complex is low spin.

This choice exists only for d4, d5, d6 and d7 in octahedral fields. For d1 to d3 and d8 to d10, there is only one sensible way to fill the orbitals.

The Spectrochemical Series

Ligands are ranked by how large a Δ they create. A commonly memorised order, from weak to strong field, is:

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H2O < NH3 < en < NO2⁻ < CN⁻ < CO

Halides and water are usually weak field; CN⁻ and CO are strong field. A higher oxidation state of the metal and a heavier (4d/5d) metal also raise Δo.

Pro Tip: Always do the same three steps: find the metal's oxidation state, write its dn count, then look at the ligand. Students who drill this routine on a dozen complexes, such as the ones taught at Sukriti NeetJee, stop treating "high spin vs low spin" as a memory question and start treating it as arithmetic. The ligand tells you the field; the d-count tells you whether the field even matters.

Worked Examples: Magnetism

Take [Fe(H2O)6]2+: Fe2+ is d6, water is weak field, so the configuration is t2g4eg2 with 4 unpaired electrons (paramagnetic, μ ≈ 4.9 BM).

Now [Fe(CN)6]4−: same Fe2+ d6, but CN⁻ is strong field, so the configuration is t2g6 with zero unpaired electrons (diamagnetic). Likewise [CoF6]3− is high spin while [Co(NH3)6]3+ is low spin and diamagnetic.

Spin-only magnetic moments for n = 1 to 5 unpaired electrons are about 1.73, 2.83, 3.87, 4.90 and 5.92 BM. Memorise these and you save time in every exam.

CFSE: Counting the Stabilisation

Crystal field stabilisation energy for an octahedral complex is CFSE = (−0.4 × nt2g + 0.6 × neg)Δo (plus pairing energy terms where relevant). For low-spin d6, that is 6 × (−0.4) = −2.4Δo; for high-spin d5 it is exactly zero.

Why Complexes Are Coloured

When white light falls on a complex, an electron in a t2g orbital can absorb light matching the gap Δo and jump to eg. The solution then shows the complementary colour of the light absorbed. This is a d-d transition, so no unpaired room in the d set means no colour: Zn2+ (d10), Sc3+ (d0) and Cu+ (d10) compounds are colourless, while [Ti(H2O)6]3+ (d1) is purple.

Changing the ligand changes Δ and therefore the colour. That is why [Cu(H2O)4]2+ is light blue but turns deep blue when ammonia replaces water.

Tetrahedral Complexes

With four ligands, the splitting inverts: e orbitals go lower, t2 orbitals go higher, and Δt ≈ (4/9)Δo. Because this gap is small, tetrahedral complexes are almost always high spin.

Frequently Asked Questions

What is the difference between Δo and Δt?

Δo is the crystal field splitting in octahedral complexes and Δt is the splitting in tetrahedral ones. Since fewer ligands are involved and their orientation is less direct, Δt is only about 4/9 of Δo for the same metal and ligand.

Why does CN⁻ form low-spin complexes while F⁻ forms high-spin ones?

CN⁻ sits high in the spectrochemical series and creates a large Δo, larger than the pairing energy, so electrons pair in t2g. F⁻ is a weak-field ligand with a small Δo, so electrons occupy eg orbitals before pairing.

How do you calculate the number of unpaired electrons in a complex?

Find the oxidation state of the metal, write the dn count, and fill t2g and eg orbitals according to whether the ligand is strong or weak field. Count the singly occupied orbitals, then use μ = √(n(n+2)) BM for the magnetic moment.

Why are some transition metal complexes colourless?

Colour needs a d-d transition, which requires a partially filled d subshell with a vacant place for the excited electron. Ions with d0 or d10 configurations, such as Sc3+ and Zn2+, have no such transition in visible light.

Is Crystal Field Theory in the NEET and JEE syllabus?

Yes. Coordination compounds, including crystal field theory, colour and magnetic properties, are part of the Class 12 chemistry syllabus for both exams. Confirm the exact topic list in the latest official information bulletin at neet.nta.nic.in or jeemain.nta.nic.in.

Got a coordination-compound question that still trips you up? Leave it in the comments, and share this post with a fellow aspirant who keeps confusing high spin with low spin.