Crystal Field Theory — Why Transition Metal Compounds Are Colourful

Why Do Transition Metal Compounds Have Colour?

When white light hits a solution of copper sulphate, the solution looks blue. When you dissolve potassium permanganate, it turns an intense purple. Iron(III) compounds are yellow or brown. What’s happening at the atomic level?

The answer lies in the d orbitals of transition metals — and how surrounding ligands (charged ions or neutral molecules) split those orbitals into different energy levels. Crystal Field Theory (CFT) is the model that explains this splitting, and with it, the magnetic properties, colours, and stability of coordination compounds.

Key Terms and Definitions

Transition Metal: An element whose atom, or at least one of its common ions, has an incompletely filled d subshell. Fe, Co, Ni, Cu, Cr, Mn, Ti are the classic examples.

Ligand: An ion or molecule that donates a lone pair of electrons to a central metal ion, forming a coordinate bond. Examples: H₂O, NH₃, Cl⁻, CN⁻, CO.

Coordination Compound: A compound in which a central metal atom or ion is bonded to ligands. Example: [Cu(H₂O)₄]²⁺, [Fe(CN)₆]³⁻.

Crystal Field Splitting (Δ): The energy difference between the two groups of d orbitals after the ligand field splits their degeneracy.

Strong Field Ligands: Ligands that cause large splitting (large Δ). Examples: CN⁻, CO, NO⁺, en, NH₃. These tend to cause low-spin complexes and are often colourless because Δ is so large that d-d transitions fall outside the visible range — or actually, they do absorb visible light but the absorbed frequency shifts.

Weak Field Ligands: Ligands that cause small splitting (small Δ). Examples: I⁻, Br⁻, Cl⁻, F⁻, H₂O. These tend to cause high-spin complexes.

The Spectrochemical Series (from weakest to strongest field):

I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < CN^- < CO

The spectrochemical series is a mandatory memorisation item for JEE Main, NEET, and CBSE Class 12. Questions like “which ligand causes greater splitting: NH₃ or Cl⁻?” are direct one-liners from this series. NH₃ > Cl⁻ always.

The Core Idea: How Ligands Split d Orbitals

An isolated transition metal ion has five d orbitals that are all degenerate (same energy):

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ , $d_{x^2-y^2}$ , $d_{z^2}$

When ligands approach the metal from specific directions, they do so with their lone pairs. These lone pairs create an electrostatic field that repels the d electrons of the metal. The repulsion is not equal for all d orbitals — it depends on the orbital’s orientation relative to the ligand positions.

Octahedral Crystal Field Splitting

In an octahedral complex (6 ligands along ±x, ±y, ±z axes):

$d_{x^2-y^2}$ and $d_{z^2}$ point directly at the approaching ligands → maximum repulsion → higher energy. These form the $e_g$ set.
$d_{xy}$ , $d_{xz}$ , $d_{yz}$ point between the ligand axes → less repulsion → lower energy. These form the $t_{2g}$ set.

The energy gap between $t_{2g}$ and $e_g$ is called Δ₀ (octahedral crystal field splitting energy).

e_g\ (\text{higher energy}) \quad \longleftrightarrow \quad \Delta_0 \quad \longleftrightarrow \quad t_{2g}\ (\text{lower energy})

$t_{2g}$ set: $d_{xy}, d_{xz}, d_{yz}$ — energy lowered by $\dfrac{2}{5}\Delta_0 = 0.4\Delta_0$ each

$e_g$ set: $d_{x^2-y^2}, d_{z^2}$ — energy raised by $\dfrac{3}{5}\Delta_0 = 0.6\Delta_0$ each

(These fractions maintain the “barycentre” — the average energy stays the same)

Tetrahedral Crystal Field Splitting

In a tetrahedral complex (4 ligands approach along alternate corners of a cube):

The $t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) is now closer to the ligands → higher energy
The $e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) is now lower energy

This is the reverse of octahedral! Also, Δ_t = $\dfrac{4}{9}$ Δ₀ (tetrahedral splitting is smaller because: fewer ligands (4 vs 6) and they don’t point directly at any d orbital).

High Spin vs Low Spin

Whether electrons fill the $t_{2g}$ and $e_g$ orbitals according to Hund’s rule (high spin) or pair up first in $t_{2g}$ (low spin) depends on comparing:

Pairing energy (P): Energy cost of forcing two electrons into the same orbital
Crystal field splitting energy (Δ₀)

If Δ₀ > P: Strong field ligand → electrons pair up in $t_{2g}$ first → low-spin complex (fewer unpaired electrons)

If Δ₀ < P: Weak field ligand → electrons fill $t_{2g}$ and $e_g$ before pairing → high-spin complex (more unpaired electrons)

Example — [Fe(CN)₆]³⁻ vs [FeF₆]³⁻:

Fe³⁺ has 5 d electrons.

CN⁻ is a strong field ligand: Δ₀ > P → low spin → d electrons: $t_{2g}^5\ e_g^0$ → 1 unpaired electron → paramagnetic (weakly)
F⁻ is a weak field ligand: Δ₀ < P → high spin → d electrons: $t_{2g}^3\ e_g^2$ → 5 unpaired electrons → strongly paramagnetic

Why Colour Occurs

When white light passes through a coordination compound, electrons in the lower $t_{2g}$ orbitals absorb photons of a specific energy and jump to the higher $e_g$ orbitals. The absorbed wavelength (colour) is removed from white light. The complementary colour is transmitted — what we see.

\Delta E = h\nu = \frac{hc}{\lambda}

The energy absorbed (Δ₀) determines the wavelength absorbed, and therefore the colour we perceive:

Colour Absorbed	Colour Observed (Complementary)
Violet	Yellow-green
Blue	Orange
Green	Red/Purple
Yellow	Violet
Orange	Blue
Red	Green

[Ti(H₂O)₆]³⁺ — Ti³⁺ has 1 d electron. The $t_{2g}^1$ electron absorbs green light (around 490–510 nm) and jumps to $e_g$ . The complementary colour of green is red-violet — which is what we observe. This is the simplest example of d-d transition explaining colour.

Why is [Fe(CN)₆]⁴⁻ yellow but [Fe(H₂O)₆]²⁺ pale green? CN⁻ causes larger Δ₀ than H₂O. Larger Δ₀ → higher frequency light absorbed → different position in the visible spectrum → different colour perceived.

Crystal Field Stabilisation Energy (CFSE)

Electrons in the $t_{2g}$ set are stabilised (lower energy by 0.4Δ₀ each); electrons in $e_g$ are destabilised (+0.6Δ₀ each). The net stabilisation due to crystal field is the CFSE.

\text{CFSE} = n_{t_{2g}} \times (-0.4\Delta_0) + n_{e_g} \times (+0.6\Delta_0)

For [Ti(H₂O)₆]³⁺: $t_{2g}^1 e_g^0$

\text{CFSE} = 1 \times (-0.4\Delta_0) + 0 = -0.4\Delta_0

For d³ complex (e.g., Cr³⁺ octahedral): $t_{2g}^3 e_g^0$

\text{CFSE} = 3 \times (-0.4\Delta_0) = -1.2\Delta_0

Complexes with large CFSE are more stable. d³ (Cr³⁺) and d⁶ low-spin (Fe²⁺ with strong ligands) have the highest CFSE in octahedral fields — this is why these configurations are unusually stable.

Solved Examples

Example 1 (CBSE): Why is [Ti(H₂O)₆]³⁺ coloured?

Ti³⁺ has 1 d electron ( $t_{2g}^1$ ). Visible light can excite this electron from $t_{2g}$ to $e_g$ (d-d transition). The absorbed energy corresponds to a specific wavelength in the visible region. Since some wavelengths are absorbed, the transmitted light is coloured. This is why the complex appears coloured.

Example 2 (JEE Main): Predict the number of unpaired electrons in [CoF₆]³⁻ and [Co(CN)₆]³⁻

Co³⁺ has 6 d electrons (d⁶).

[CoF₆]³⁻: F⁻ is a weak field ligand → high spin → $t_{2g}^4 e_g^2$ → 4 unpaired electrons

[Co(CN)₆]³⁻: CN⁻ is a strong field ligand → low spin → $t_{2g}^6 e_g^0$ → 0 unpaired electrons (diamagnetic)

Example 3 (JEE Advanced): Calculate CFSE for [Cr(H₂O)₆]³⁺

Cr³⁺ has d³ configuration. H₂O is a moderate field ligand; for d³, both high-spin and low-spin give the same filling: $t_{2g}^3 e_g^0$ .

\text{CFSE} = 3 \times (-0.4\Delta_0) + 0 = -1.2\Delta_0

Exam-Specific Tips

JEE Main 2024 had a question: “Arrange [CrCl₆]³⁻, [Cr(NH₃)₆]³⁺, [Cr(CN)₆]³⁻ in increasing order of Δ₀.” Use the spectrochemical series: Cl⁻ < NH₃ < CN⁻. So Δ₀ increases in the order: [CrCl₆]³⁻ < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]³⁻.

NEET regularly asks: “Which of the following is a strong field ligand?” and “Which complex is diamagnetic?” For diamagnetic: need 0 unpaired electrons, usually requires strong field ligand (CN⁻, CO) with d⁶ configuration.

Common Mistakes to Avoid

Mistake 1: Writing that d⁰ and d¹⁰ complexes are coloured. These cannot undergo d-d transitions (no electron to promote in d⁰; no vacancy to receive in d¹⁰). Such complexes are generally colourless. [Cu(H₂O)₄]²⁺ is Cu²⁺ (d⁹) — coloured. [Zn(H₂O)₆]²⁺ is Zn²⁺ (d¹⁰) — colourless.

Mistake 2: Mixing up the $t_{2g}$ and $e_g$ energy levels. In octahedral: $e_g$ is HIGHER ( $d_{x^2-y^2}$ , $d_{z^2}$ pointing at ligands). In tetrahedral: $t_2$ is higher. Students sometimes reverse these.

Mistake 3: Concluding that strong field ligands always give colourless complexes. Strong field ligands cause large Δ₀ — the absorbed wavelength shifts, but d-d transitions can still be in the visible range. [Fe(CN)₆]⁴⁻ is yellow; [Co(CN)₆]³⁻ is pale yellow — both are strong-field but coloured.

Practice Questions

Q1. How many unpaired electrons does [MnCl₆]³⁻ have?

Mn³⁺ has d⁴ configuration. Cl⁻ is a weak field ligand → high spin → $t_{2g}^3 e_g^1$ → 4 unpaired electrons

Q2. Why is [Cu(NH₃)₄]²⁺ deep blue while [Cu(H₂O)₄]²⁺ is pale blue?

NH₃ is a stronger field ligand than H₂O. Larger Δ₀ with NH₃ → absorbs different wavelength of visible light → deeper blue colour. The d-d transition energy is higher with NH₃ (absorbs at shorter wavelength in the orange/red region vs green for the water complex).

Q3. Calculate CFSE for a d⁶ high-spin octahedral complex.

d⁶ high-spin: $t_{2g}^4 e_g^2$
CFSE = 4(-0.4Δ₀) + 2(+0.6Δ₀) = -1.6Δ₀ + 1.2Δ₀ = -0.4Δ₀
Compare with low-spin d⁶: $t_{2g}^6$ → CFSE = 6(-0.4Δ₀) = -2.4Δ₀ (much more stable).

Jahn-Teller Distortion

Certain octahedral complexes show a Jahn-Teller distortion — the octahedron elongates along the z-axis (or compresses). This happens when the $e_g$ orbitals have unequal occupation (like $d^9$ Cu²⁺ with $t_{2g}^6 e_g^3$ ).

The distortion removes the degeneracy of the $e_g$ set, lowering the energy of the system. This is why Cu²⁺ complexes often show slightly distorted octahedral geometry and broad absorption bands.

JEE Advanced has asked about the Jahn-Teller effect in the context of why [Cu(H₂O)₆]²⁺ has a distorted octahedral structure. The answer: Cu²⁺ is $d^9$ with configuration $t_{2g}^6 e_g^3$ — the unequal filling of $e_g$ ( $d_{z^2}^1 d_{x^2-y^2}^2$ or vice versa) creates a geometric asymmetry that drives elongation.

Additional Solved Examples

Example 4 (JEE Advanced): CFSE Comparison

Compare the CFSE of $[Cr(H_2O)_6]^{2+}$ (d⁴) in high-spin and low-spin configurations.

High-spin d⁴: $t_{2g}^3 e_g^1$ CFSE $= 3(-0.4\Delta_0) + 1(+0.6\Delta_0) = -1.2 + 0.6 = -0.6\Delta_0$

Low-spin d⁴: $t_{2g}^4 e_g^0$ CFSE $= 4(-0.4\Delta_0) + 0 = -1.6\Delta_0$ (plus pairing energy penalty for one paired electron)

Net CFSE for low-spin $= -1.6\Delta_0 + P$ (where $P$ = pairing energy). Low-spin is favoured only when $\Delta_0 > P$ , i.e., with strong field ligands like CN⁻.

With H₂O (moderate field), Cr²⁺ is typically high-spin: CFSE $= -0.6\Delta_0$ .

Q4. Why is $[Fe(H_2O)_6]^{3+}$ light yellow while $[Fe(H_2O)_6]^{2+}$ is green?

Fe³⁺ is d⁵ (high-spin with H₂O: $t_{2g}^3 e_g^2$ ) and Fe²⁺ is d⁶ (high-spin: $t_{2g}^4 e_g^2$ ). The d-d transitions in Fe³⁺ are spin-forbidden (all electrons are unpaired in the ground state — any transition requires a spin flip), making the absorption very weak and the colour pale yellow. In Fe²⁺, d-d transitions are only Laporte-forbidden but spin-allowed, giving a more visible green colour. The different number of d electrons leads to different Δ₀ values and different absorbed wavelengths.

Q5. Arrange the following in order of increasing number of unpaired electrons: $[Co(NH_3)_6]^{3+}$ , $[CoF_6]^{3-}$ , $[Co(en)_3]^{3+}$ .

All contain Co³⁺ (d⁶).

$[Co(NH_3)_6]^{3+}$ : NH₃ is a strong field ligand → low-spin → $t_{2g}^6 e_g^0$ → 0 unpaired

$[Co(en)_3]^{3+}$ : en is a stronger field ligand than NH₃ → low-spin → $t_{2g}^6 e_g^0$ → 0 unpaired

$[CoF_6]^{3-}$ : F⁻ is a weak field ligand → high-spin → $t_{2g}^4 e_g^2$ → 4 unpaired

Order: $[Co(NH_3)_6]^{3+} = [Co(en)_3]^{3+}$ (0) $<$ $[CoF_6]^{3-}$ (4).

FAQs

Why doesn’t VBT explain colour? Valence Bond Theory describes hybridisation and geometry but doesn’t account for d orbital energy splitting. It has no mechanism to explain why compounds absorb visible light or why colour depends on the ligand. CFT explains both.

What is the charge transfer transition (vs d-d transition)? Some complexes get their intense colour from charge transfer transitions where an electron moves from ligand to metal (LMCT) or metal to ligand (MLCT). These are typically much more intense than d-d transitions (which are Laporte-forbidden). MnO₄⁻ (permanganate) is coloured by LMCT, not d-d (Mn is d⁰).

Why are tetrahedral complexes always high-spin? Δ_t = 4/9 Δ₀ — tetrahedral splitting is only about 4/9 of octahedral. This is usually less than the pairing energy, so electrons prefer to remain unpaired. It is therefore very rare to find a low-spin tetrahedral complex.