Crystal Field Theory - d-Orbital Splitting

Crystal Field Theory — A Better Model

Valence Bond Theory explains geometry and magnetism, but it struggles with colour and the detailed energetics. Crystal Field Theory (CFT) fills the gap. It is an electrostatic model: the metal-ligand bond is treated as a purely ionic/electrostatic attraction between the positive metal ion and the negative ligands (or the negative ends of dipolar ligands).

The key idea: in a free metal ion the five d orbitals are degenerate (same energy). When ligands approach, their negative fields repel the d electrons — but not equally, because the d orbitals point in different directions. This lifts the degeneracy and splits the d orbitals into groups of different energy. This splitting is what gives complexes their colour and magnetism.

Octahedral and Tetrahedral Splitting

Octahedral field: the six ligands approach along the x, y, z axes. The two d orbitals pointing along the axes ($d_{x^2-y^2}$ and $d_{z^2}$, the $e_g$ set) are repelled more (higher energy); the three pointing between the axes ($d_{xy}, d_{yz}, d_{zx}$, the $t_{2g}$ set) are repelled less (lower energy).

$$\text{Octahedral: } t_{2g}\ (\text{lower, } -0.4\Delta_o) \quad\text{and}\quad e_g\ (\text{higher, } +0.6\Delta_o)$$

The energy gap is the crystal field splitting energy, $\Delta_o$ (also called 10 Dq).

Tetrahedral field: the splitting is reversed and smaller. The $e$ set is lower and the $t_2$ set is higher, with

$$\Delta_t = \frac{4}{9}\,\Delta_o$$

Because $\Delta_t$ is so small, tetrahedral complexes are almost always high spin (no pairing).

The Spectrochemical Series and Spin State

How big is $\Delta_o$? It depends mainly on the ligand. Arranging ligands by the splitting they cause gives the spectrochemical series (weak field → strong field):

$$\text{I}^- < \text{Br}^- < \text{SCN}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{C}_2\text{O}_4^{2-} < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$$

• Weak-field ligands (small $\Delta_o$): electrons spread out to avoid pairing → high spin.
• Strong-field ligands (large $\Delta_o$): electrons pair in the lower $t_{2g}$ set → low spin.

The deciding comparison is $\Delta_o$ versus the pairing energy (P):

• If $\Delta_o < P$ (weak field) → high spin (electrons go to $e_g$ rather than pair).
• If $\Delta_o > P$ (strong field) → low spin (electrons pair in $t_{2g}$).

The net stabilisation from preferentially filling lower orbitals is the Crystal Field Stabilisation Energy (CFSE):

$$\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\,\Delta_o$$

[JEE Tip] For a $d^4$-$d^7$ octahedral ion you must decide high vs low spin from the ligand. CN$^-$, CO, NH$_3$ → low spin; H$_2$O, F$^-$, Cl$^-$ → high spin. This decision drives the unpaired-electron count, magnetism and colour.

Solved Examples

Example 1: Octahedral d-orbital sets

Name the two sets of d orbitals in an octahedral crystal field and their relative energies.

Solution: The lower-energy set is $t_{2g}$ ($d_{xy}, d_{yz}, d_{zx}$) and the higher-energy set is $e_g$ ($d_{x^2-y^2}, d_{z^2}$). The gap between them is $\Delta_o$.

Example 2: Tetrahedral splitting size

How does the tetrahedral splitting $\Delta_t$ compare with the octahedral $\Delta_o$?

Solution: $\Delta_t = \dfrac{4}{9}\Delta_o$ — much smaller. Because of this small gap, tetrahedral complexes are almost always high spin.

Example 3: High spin vs low spin

For a $d^6$ octahedral ion, give the $t_{2g}/e_g$ configuration in (a) a strong field and (b) a weak field.

Solution: (a) Strong field (low spin): all six pair in the lower set → $t_{2g}^6 e_g^0$ → 0 unpaired. (b) Weak field (high spin): $t_{2g}^4 e_g^2$ → 4 unpaired.

Example 4: Predict spin state from ligand

Is $[\text{Fe(CN)}_6]^{4-}$ high spin or low spin?

Solution: Fe$^{2+}$ is $d^6$. CN$^-$ is a strong-field ligand (large $\Delta_o > P$), so it is low spin: $t_{2g}^6 e_g^0$, 0 unpaired electrons (diamagnetic).

Example 5: Unpaired electrons in a weak-field complex

How many unpaired electrons does $[\text{Mn(H}_2\text{O)}_6]^{2+}$ have?

Solution: Mn$^{2+}$ is $d^5$. H$_2$O is weak field → high spin → $t_{2g}^3 e_g^2$ → 5 unpaired electrons ($\mu = 5.92$ BM).

Example 6: CFSE of a d3 ion

Calculate the CFSE (in units of $\Delta_o$) for an octahedral $d^3$ ion.

Solution: $d^3$ fills $t_{2g}^3 e_g^0$. CFSE $= (-0.4\times3 + 0.6\times0)\Delta_o = -1.2\,\Delta_o$.

Example 7: Spectrochemical order

Arrange Cl$^-$, H$_2$O, CN$^-$ and NH$_3$ in increasing field strength.

Solution: From the spectrochemical series: Cl$^-$ < H$_2$O < NH$_3$ < CN$^-$ (weak to strong).

Example 8: Why tetrahedral is high spin

Why are tetrahedral complexes nearly always high spin?

Solution: Because $\Delta_t$ is only $\tfrac{4}{9}\Delta_o$ — so small that it is almost always less than the pairing energy, so electrons prefer to occupy the upper orbitals singly rather than pair. Hence high spin.

Example 9: Strong vs weak field outcome

State the condition (in terms of $\Delta_o$ and pairing energy P) for a complex to be low spin.

Solution: Low spin occurs when $\Delta_o > P$ (strong field): the energy cost of pairing is less than the cost of promoting an electron to the $e_g$ set, so electrons pair in $t_{2g}$.

Example 10: CFSE of a low-spin d6 ion

Calculate the CFSE (in $\Delta_o$) for a low-spin octahedral $d^6$ ion (ignore pairing energy).

Solution: Low-spin $d^6$ is $t_{2g}^6 e_g^0$. CFSE $= (-0.4\times6 + 0.6\times0)\Delta_o = -2.4\,\Delta_o$.