Crystal field theory

Why the degeneracy breaks. In an isolated gaseous metal ion, all five $d$-orbitals ($d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, $d_{z^2}$) have identical energy — there is no reason for any one to be favoured. Bring up six ligands, modelled as point negative charges, along the $\pm x$, $\pm y$, $\pm z$ axes (an octahedral field) and the picture changes: an electron sitting in an orbital that points directly along one of those axes feels more electrostatic repulsion than an electron in an orbital that points diagonally between the axes. Orbital energy rises or falls depending purely on geometry — how much lobe overlaps the ligand direction.

Octahedral splitting. The two orbitals that point straight at the ligands, $d_{x^2-y^2}$ and $d_{z^2}$, are pushed up together as the doubly degenerate $e_g$ set; the three orbitals that point between the ligands, $d_{xy}$, $d_{yz}$, $d_{xz}$, are pushed down together as the triply degenerate $t_{2g}$ set. The gap between them is the octahedral crystal field splitting energy:

where: $\Delta_o$ (also written $10Dq$) is the total octahedral splitting energy, and the $+0.6/-0.4$ weights follow from the "barycentre rule" — the weighted-average energy of all five orbitals must stay at the pre-splitting (spherical-field) value, so $2(0.6\Delta_o) + 3(-0.4\Delta_o) = 0$.

Tetrahedral splitting inverts and shrinks. With only four ligands at alternate corners of a cube, no $d$-orbital points directly at any ligand, so the whole splitting is smaller, and the sets swap roles: three orbitals ($t_2$) rise, two ($e$) fall. For the same metal, same ligands, same bond length,

so tetrahedral complexes almost never force electrons to pair — $\Delta_t$ is rarely large enough to beat the pairing energy $P$.

Spectrochemical series. Different ligands produce different $\Delta_o$ for the same metal, and decades of spectroscopic measurement rank them into a fixed order, weak-field to strong-field:

This order is empirical — CFT's pure point-charge model cannot fully explain why $\text{CN}^-$ and $\text{CO}$ (neutral or weakly charged) out-split $\text{F}^-$ (small and highly charged); that requires the covalency corrections of ligand field theory, which is beyond this course.

High-spin vs low-spin. This choice only arises for octahedral $d^4$–$d^7$ ions, because those are the only counts where an electron has a real choice between pairing in $t_{2g}$ or occupying the empty $e_g$. Compare $\Delta_o$ against the pairing energy $P$ (the extra repulsion cost of forcing two electrons into one orbital): if $\Delta_o < P$ (weak-field ligand), it is cheaper to spread electrons across all five orbitals first — high spin, maximum unpaired count. If $\Delta_o > P$ (strong-field ligand), it is cheaper to fill $t_{2g}$ completely before touching $e_g$ — low spin, fewer unpaired electrons.

Crystal field stabilization energy (CFSE). Once the $t_{2g}^x e_g^y$ configuration is fixed, the net electrostatic stabilization relative to an undivided spherical field is a straightforward count:

where: $n_{t_{2g}}$ and $n_{e_g}$ are the electron counts in each set. A pairing-energy penalty is added separately whenever the configuration forces more pairing than the free ion already had.

Jahn–Teller distortion. Whenever the ground-state configuration leaves the $e_g$ set unevenly occupied — the classic cases are $d^4$ high-spin and $d^9$ octahedral — the octahedron distorts, usually elongating along one axis, because an asymmetric arrangement can lower its total energy by removing that leftover degeneracy. This course needs only the qualitative recognition, not the vibronic-coupling derivation.

Problem. For the octahedral complex $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$ (high-spin $d^6$, since $\text{H}_2\text{O}$ sits on the weak/intermediate side of the spectrochemical series), state the configuration, the unpaired electron count, and the CFSE. Then repeat for $[\text{Fe}(\text{CN})_6]^{4-}$, the same $d^6$ ion with the strong-field ligand $\text{CN}^-$, and compare.

Step 1. Identify the ion and electron count. $\text{Fe}^{2+}$ is $d^6$ in both complexes — only the ligand differs, isolating the effect of field strength.

Step 2. For $\text{H}_2\text{O}$ (weak field, $\Delta_o < P$): electrons spread out first, so the configuration is $t_{2g}^4 e_g^2$ (high-spin). Unpaired count: $t_{2g}$ contributes 2 unpaired (one orbital paired, two single) and $e_g$ contributes 2 unpaired (both single) — total 4 unpaired electrons, matching the observed paramagnetism of the aqua complex.

Step 3. CFSE for the aqua complex, using the derivation above: $\text{CFSE} = (-0.4)(4) + (0.6)(2)\,\Delta_o = -0.4\,\Delta_o$ — a modest stabilization typical of high-spin $d^6$.

Step 4. For $\text{CN}^-$ (strong field, $\Delta_o > P$): all six electrons pack into $t_{2g}$ before $e_g$ is occupied, giving $t_{2g}^6 e_g^0$ (low-spin). Unpaired count: zero — $[\text{Fe}(\text{CN})_6]^{4-}$ is diamagnetic.

Step 5. CFSE for the cyanide complex: $\text{CFSE} = (-0.4)(6) + (0.6)(0)\,\Delta_o = -2.4\,\Delta_o$, six times larger in magnitude than the aqua case (ignoring the extra pairing-energy cost, which would trim but not erase this gap). The same $d^6$ metal ion swings from 4 unpaired electrons and $-0.4\Delta_o$ stabilization with a weak-field ligand to 0 unpaired electrons and $-2.4\Delta_o$ stabilization with a strong-field ligand — a prediction VBT alone cannot make but CFT explains directly from the spectrochemical series.

Competitive stretch

For $d^7$ ($\text{Co}^{2+}$): high-spin $t_{2g}^5 e_g^2$ gives $\text{CFSE} = -0.8\,\Delta_o$; low-spin $t_{2g}^6 e_g^1$ gives $\text{CFSE} = -1.8\,\Delta_o$ — the same weighting applies to any $d^n$ octahedral case once the filling order is fixed.

• Applying the octahedral $t_{2g}/e_g$ ordering to tetrahedral complexes. The pattern inverts for tetrahedral fields ($e$ lower, $t_2$ higher) — redraw the diagram, do not just relabel the octahedral one.
• Assuming every $d^n$ configuration can be high- or low-spin. Only octahedral $d^4$–$d^7$ show the distinction; $d^1$–$d^3$ and $d^8$–$d^{10}$ have one possible filling regardless of ligand field strength.
• Comparing $\Delta_o$ to zero instead of to $P$. A ligand only forces pairing if its $\Delta_o$ exceeds the pairing energy $P$ for that specific ion, not just because $\Delta_o$ looks "large" in absolute terms.
• Missing Jahn–Teller candidates. Check specifically for asymmetric $e_g$ occupancy; $d^4$ high-spin and $d^9$ are the classic strong distortion cases, not every configuration.

Ask the tutor: want to work through the $d^7$ high-spin versus low-spin CFSE comparison in full, or see why Jahn–Teller distortion specifically targets $d^4$ high-spin and $d^9$ ions?