Crystal field theory — splitting in octahedral and tetrahedral complexes

Question

Explain crystal field splitting in octahedral and tetrahedral complexes. How does the magnitude of splitting ( $\Delta$ ) determine whether a complex is high-spin or low-spin? Why is there no low-spin tetrahedral complex?

(JEE Advanced 2023 asked about high-spin/low-spin prediction; JEE Main regularly tests colour explanation)

Solution — Step by Step

In a free metal ion, all five d-orbitals have the same energy (degenerate). When ligands approach, their electron pairs repel the d-electrons. Orbitals pointing towards ligands are raised in energy, while those pointing between ligands are lowered.

This splitting of energy levels is called crystal field splitting.

In an octahedral complex (6 ligands along x, y, z axes), the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point directly at the ligands — they go up in energy (called the $e_g$ set).

The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point between the ligands — they go down (called the $t_{2g}$ set).

The energy gap between $t_{2g}$ and $e_g$ is $\Delta_o$ (octahedral splitting energy).

In a tetrahedral complex (4 ligands between axes), the splitting is inverted: $t_2$ set goes up, $e$ set goes down.

The key result: $\Delta_t = \frac{4}{9}\Delta_o$ . Tetrahedral splitting is always less than half of octahedral splitting for the same metal-ligand combination.

When filling electrons into split d-orbitals, there is a competition between:

$\Delta$ (splitting energy): favours putting electrons in lower orbitals (pairing)
P (pairing energy): opposes putting two electrons in the same orbital

If $\Delta > P$ : electrons pair up in $t_{2g}$ before filling $e_g$ → low-spin complex. If $\Delta < P$ : electrons spread out to avoid pairing → high-spin complex.

Since $\Delta_t$ is always small ( $\frac{4}{9}\Delta_o$ ), it is almost always less than P. That is why low-spin tetrahedral complexes are practically non-existent.

flowchart TD
    A["Free metal ion<br/>5 degenerate d-orbitals"] --> B{"Ligand geometry?"}
    B -->|Octahedral| C["Splitting: t₂g below, eg above<br/>Gap = Δo"]
    B -->|Tetrahedral| D["Splitting: e below, t₂ above<br/>Gap = Δt = 4/9 Δo"]
    C --> E{"Δo vs Pairing energy P?"}
    E -->|"Δo > P (strong field)"| F["Low-spin<br/>Fewer unpaired e⁻"]
    E -->|"Δo < P (weak field)"| G["High-spin<br/>More unpaired e⁻"]
    D --> H["Almost always high-spin<br/>Δt too small for pairing"]

Why This Works

The colour of transition metal complexes comes from d-d transitions — an electron absorbs a photon and jumps from $t_{2g}$ to $e_g$ (in octahedral complexes). The energy of the absorbed photon equals $\Delta_o$ , which determines what colour of light is absorbed. The complementary colour is what we see.

Strong field ligands (CN⁻, CO, NH₃) cause large $\Delta_o$ → low-spin complexes. Weak field ligands (I⁻, Br⁻, Cl⁻, F⁻) cause small $\Delta_o$ → high-spin complexes. The spectrochemical series ranks ligands by their splitting ability.

Alternative Method

Memorise the spectrochemical series for quick prediction: $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}$ . Ligands from CN⁻ onwards are almost always strong field (low-spin for d⁴ to d⁷). Water and below are usually weak field (high-spin).

Common Mistake

Students apply the high-spin/low-spin concept to d¹, d², d³, d⁸, d⁹, and d¹⁰ configurations. But for these configurations, the filling is the same regardless of $\Delta$ . The distinction only matters for d⁴ through d⁷ — these are the cases where the electron can either go into a higher orbital or pair up in a lower one. For d³ (like Cr³⁺), all three electrons go into $t_{2g}$ regardless. For d⁸ (like Ni²⁺ in octahedral), the configuration is always $t_{2g}^6 e_g^2$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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