Colour of transition metal compounds — why and how crystal field theory explains it

Question

Why are transition metal compounds coloured while s-block compounds are colourless? How does Crystal Field Theory (CFT) explain the origin of colour in coordination complexes?

(JEE Main, CBSE 12 — colour prediction from d-electron configuration is a high-value JEE question)

Solution — Step by Step

Colour in transition metal compounds arises from d-d electronic transitions. When white light falls on a complex, electrons in lower-energy d-orbitals absorb specific wavelengths and jump to higher-energy d-orbitals. The remaining (unabsorbed) wavelengths are transmitted or reflected — this is the colour we see.

The colour we observe is the complementary colour of the wavelength absorbed.

Absorbed Colour	Wavelength (nm)	Observed (Complementary) Colour
Violet	400-435	Yellow-green
Blue	435-480	Orange
Green	500-560	Red-purple
Yellow	560-595	Violet
Orange	595-650	Blue
Red	650-700	Green

In a free metal ion, all five d-orbitals have the same energy (degenerate). When ligands approach, their electron pairs repel the d-electrons, splitting the d-orbitals into two sets.

In octahedral complexes:

$t_{2g}$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — lower energy (pointing between ligands)
$e_g$ set ( $d_{z^2}$ , $d_{x^2-y^2}$ ) — higher energy (pointing directly at ligands)
Energy gap = $\Delta_o$ (crystal field splitting energy)

In tetrahedral complexes:

Splitting is reversed: $e$ set is lower, $t_2$ set is higher
$\Delta_t \approx \frac{4}{9}\Delta_o$ (smaller splitting)

The magnitude of $\Delta$ determines which wavelength of light is absorbed, and hence the colour.

Ligand strength (spectrochemical series):

I^- \lt Br^- \lt Cl^- \lt F^- \lt OH^- \lt H_2O \lt NH_3 \lt en \lt NO_2^- \lt CN^- \lt CO

Stronger field ligands cause larger $\Delta$ , absorbing higher energy (shorter wavelength) light.

Example: $[Ti(H_2O)_6]^{3+}$ is purple (absorbs green). If we replace $H_2O$ with stronger $NH_3$ , $\Delta$ increases, absorption shifts to higher energy, and the colour changes.

Oxidation state: Higher charge on the metal = stronger attraction for ligands = larger $\Delta$ = colour shift.

d-electron count: $d^0$ and $d^{10}$ complexes are colourless (no d-d transition possible — either no electrons to excite, or no empty d-orbital to move into).

flowchart TD
    A["Transition metal complex"] --> B{"d-electron count?"}
    B -->|"d⁰ or d¹⁰"| C["Colourless<br/>(no d-d transition possible)"]
    B -->|"d¹ to d⁹"| D["Coloured"]
    D --> E["Ligands split d-orbitals"]
    E --> F["Electron absorbs light = Δ"]
    F --> G["Observed colour = complementary<br/>of absorbed wavelength"]
    G --> H{"Δ depends on:"}
    H --> I["Ligand strength (spectrochemical series)"]
    H --> J["Metal oxidation state"]
    H --> K["Geometry (oct > tet)"]

Why This Works

The colour of transition metal compounds is a direct quantum mechanical phenomenon. The d-orbital splitting by the crystal field creates an energy gap that matches the energy of visible light photons ( $\Delta \approx 1.5 - 3.5$ eV, corresponding to 350-800 nm). S-block metal ions have no d-electrons (or completely empty d-orbitals), so no d-d transitions are possible, and they remain colourless.

The beauty of CFT is that it predicts colour changes with ligand substitution — replace weak-field $H_2O$ with strong-field $CN^-$ , and the colour must change because $\Delta$ changes.

Common Mistake

Students often predict that $[Zn(H_2O)_6]^{2+}$ should be coloured because Zn is a transition metal. But $Zn^{2+}$ is $d^{10}$ — all d-orbitals are fully occupied, and no d-d transition is possible. Similarly, $Sc^{3+}$ ( $d^0$ ) complexes are colourless. Always check the d-electron count of the metal ION (not the neutral atom) before predicting colour.

Quick colour prediction: $Cu^{2+}$ ( $d^9$ ) compounds are typically blue/green, $Fe^{3+}$ ( $d^5$ ) compounds are yellow/brown, $Co^{2+}$ ( $d^7$ ) compounds are pink/blue, $Ni^{2+}$ ( $d^8$ ) compounds are green, $Mn^{2+}$ ( $d^5$ high spin) compounds are very pale pink (spin-forbidden transitions). These colours shift with ligand changes.

Question

Solution — Step by Step

Why This Works

Common Mistake

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