Crystal field theory — splitting in octahedral and tetrahedral complexes

Question

How does Crystal Field Theory (CFT) explain the splitting of d-orbitals in octahedral and tetrahedral complexes? Why are their splitting patterns different?

Solution — Step by Step

In an isolated metal ion, all five d-orbitals have the same energy (degenerate). When ligands approach, their lone pairs repel the d-electrons. But the repulsion is not equal for all five d-orbitals — it depends on the orbital’s orientation relative to the ligands.

This unequal repulsion causes the d-orbitals to split into two energy groups. The energy gap is called crystal field splitting energy ( $\Delta$ ).

In an octahedral complex, 6 ligands approach along the $x$ , $y$ , and $z$ axes.

The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point directly at the ligands $\to$ maximum repulsion $\to$ higher energy. These form the $e_g$ set.

The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point between the axes (45 degrees away from ligands) $\to$ less repulsion $\to$ lower energy. These form the $t_{2g}$ set.

\text{Splitting: } t_{2g} \text{ (lower, 3 orbitals)} \quad e_g \text{ (higher, 2 orbitals)}

\Delta_{oct} = \text{energy gap between } t_{2g} \text{ and } e_g

In a tetrahedral complex, 4 ligands approach from alternate corners of a cube — they are oriented between the axes.

Now the situation reverses: $d_{xy}$ , $d_{xz}$ , $d_{yz}$ point closer to the ligands $\to$ higher energy ( $t_2$ set). The $d_{x^2-y^2}$ and $d_{z^2}$ are farther away $\to$ lower energy ( $e$ set).

\text{Splitting: } e \text{ (lower, 2 orbitals)} \quad t_2 \text{ (higher, 3 orbitals)}

The splitting is inverted compared to octahedral. Also, $\Delta_{tet} = \frac{4}{9}\Delta_{oct}$ (roughly half) because there are fewer ligands and they do not point directly at any orbital.

graph TD
    A[Free metal ion: 5 degenerate d-orbitals] --> B{Ligand geometry?}
    B -->|Octahedral: 6 ligands on axes| C["t2g (lower) + eg (higher)"]
    B -->|Tetrahedral: 4 ligands between axes| D["e (lower) + t2 (higher)"]
    C --> E["Delta_oct = large"]
    D --> F["Delta_tet = 4/9 Delta_oct"]
    E --> G{Strong field ligand?}
    G -->|Yes: CN-, CO| H[Low spin complex]
    G -->|No: Cl-, H2O| I[High spin complex]

Why This Works

The splitting pattern depends entirely on geometry — specifically, which d-orbitals feel more repulsion from the ligand arrangement.

Spectrochemical series ranks ligands by their splitting power:

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

Strong field ligands (right side) cause large $\Delta$ $\to$ electrons pair up in lower orbitals $\to$ low spin complex. Weak field ligands (left side) cause small $\Delta$ $\to$ electrons spread across all orbitals $\to$ high spin complex.

Property	Octahedral	Tetrahedral
Ligands	6	4
Higher energy set	$e_g$ (2 orbitals)	$t_2$ (3 orbitals)
Lower energy set	$t_{2g}$ (3 orbitals)	$e$ (2 orbitals)
Splitting energy	$\Delta_{oct}$	$\Delta_{tet} \approx \frac{4}{9}\Delta_{oct}$
Low spin possible?	Yes (strong field)	Rarely (splitting too small)

Alternative Method

To predict magnetic properties quickly:

Find the metal’s d-electron count
Determine if the ligand is strong or weak field (spectrochemical series)
Fill electrons in the split orbitals accordingly:
- Strong field: fill $t_{2g}$ completely before putting electrons in $e_g$
- Weak field: follow Hund’s rule across both sets
Count unpaired electrons $\to$ magnetic moment $\mu = \sqrt{n(n+2)}$ BM

Common Mistake

Students often assume tetrahedral complexes can be low spin. Because $\Delta_{tet}$ is only $\frac{4}{9}$ of $\Delta_{oct}$ , the splitting is almost always too small to force electron pairing. So tetrahedral complexes are almost always high spin. JEE Advanced has tested this — a question gave $[\text{CoCl}_4]^{2-}$ and asked whether it is high or low spin. The answer is high spin (tetrahedral with weak field Cl $^-$ ). Students who applied octahedral logic got it wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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