Crystal field theory

Why the orbitals split at all. In an isolated metal ion, all five $d$ orbitals — $d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, d_{z^2}$ — are degenerate (equal energy) because nothing breaks their symmetry. Once six ligands approach along the $\pm x, \pm y, \pm z$ axes to form an octahedral complex, every $d$ orbital is pushed up in energy by the approaching negative charge, but not by the same amount. Two orbitals, $d_{x^2-y^2}$ and $d_{z^2}$, point their lobes directly along the axes, straight at the incoming ligands, so electrons in them feel stronger repulsion. The other three, $d_{xy}, d_{yz}, d_{xz}$, point between the axes, into the gaps, so they are repelled less. The five orbitals therefore separate into a higher-energy pair, called $e_g$, and a lower-energy triplet, called $t_{2g}$.

Naming the gap. The energy difference between the two sets is the crystal field splitting energy, written $\Delta_o$ for an octahedral field. Splitting must conserve the average (barycentre) energy of all five orbitals, which fixes the split precisely:

where: $\Delta_o$ is the octahedral crystal field splitting energy (measured in the same units as photon energy, since it is what a $d$–$d$ transition absorbs), $E(e_g)$ is the energy of the upper pair relative to the barycentre, and $E(t_{2g})$ is the energy of the lower triplet relative to the barycentre. In a tetrahedral field (four ligands at alternate corners of a cube, none pointing directly at any orbital) the pattern inverts — $t_2$ becomes the higher set and $e$ the lower — and the whole splitting shrinks, $\Delta_t \approx \tfrac{4}{9}\Delta_o$ for the same metal and ligands, since fewer ligands are present and their geometry is less confrontational.

How electrons fill the split levels. Whether electrons pair up inside the lower $t_{2g}$ set before entering $e_g$, or spread out singly across both sets first, is decided by comparing $\Delta_o$ with the pairing energy $P$ (the energy cost of forcing two electrons into one orbital). Strong-field ligands such as $CN^-, CO, NO_2^-$ give $\Delta_o > P$, so electrons pair in $t_{2g}$ first — a low-spin complex with fewer unpaired electrons. Weak-field ligands such as $F^-, H_2O, Cl^-$ give $\Delta_o < P$, so Hund's rule wins and electrons occupy all five orbitals singly before any pairing — a high-spin complex with more unpaired electrons.

Colour and magnetism from the same split. A photon of energy exactly $\Delta_o$ can promote an electron from $t_{2g}$ to $e_g$ — a $d$–$d$ transition — provided the ion is not $d^0$ or $d^{10}$ (both cases lack a partner orbital to jump to, or lack any gap to cross, so those complexes are colourless). The wavelength absorbed is fixed by $\Delta_o$; the complementary wavelength reaches the eye as the observed colour. The number of unpaired electrons left after high-spin or low-spin filling directly gives paramagnetic or diamagnetic behaviour.

Calculate the CFSE of $[CoF_6]^{3-}$, already established (Pass-1 result) to be high spin with the configuration $t_{2g}^4\,e_g^2$ and 4 unpaired electrons.

Step 1 — identify the filling. From the earlier analysis, $Co^{3+}$ is $3d^6$, and because $F^-$ is weak-field ($\Delta_o < P$), Hund's rule spreads electrons across all five orbitals before any pairing, giving $n_{t_{2g}} = 4$ and $n_{e_g} = 2$.

Step 2 — check for forced pairing. A fully spread-out $d^6$ arrangement can place only 5 electrons singly (one per orbital); the 6th electron must pair with one already in $t_{2g}$ (the lower, more accessible set). This is the same pairing that would occur in the free ion's own ground state once it holds 6 electrons across 5 orbitals, so no extra pairing is forced by the splitting itself: $x = 0$.

Step 3 — apply the CFSE formula. Substituting $n_{t_{2g}} = 4$, $n_{e_g} = 2$, $x = 0$:

Step 4 — interpret. The CFSE of $[CoF_6]^{3-}$ is $-0.4\,\Delta_o$, a modest but genuine stabilisation relative to the unsplit ion — consistent with $F^-$ being a weak-field ligand that produces only a small splitting to begin with, unlike a strong-field, low-spin case where forced pairing would add a destabilising $+xP$ term but the larger $\Delta_o$ term would usually still win out overall.

Competitive stretch

For a strong-field, low-spin $d^6$ ion such as $[Co(NH_3)_6]^{3+}$, all six electrons occupy $t_{2g}$ ($n_{t_{2g}} = 6, n_{e_g} = 0$), forcing 3 extra pairs beyond the free ion's single pair ($x = 3$), giving $\text{CFSE} = -0.4(6)\Delta_o + 3P = -2.4\,\Delta_o + 3P$ — a much larger orbital stabilisation, offset by the pairing cost, and it is this large negative CFSE that makes low-spin $d^6$ octahedral complexes exceptionally common and stable.

• Treating CFT as involving real covalent bonds — it is a purely electrostatic, point-charge model, unlike VBT's hybridisation picture.
• Forgetting that the $+0.6\,\Delta_o$ / $-0.4\,\Delta_o$ split is fixed by requiring the barycentre (weighted average) energy to stay unchanged — students sometimes invent arbitrary splits.
• Applying the octahedral $e_g$/$t_{2g}$ ordering to a tetrahedral complex — the tetrahedral pattern is inverted ($t_2$ higher, $e$ lower) and smaller in magnitude.
• Computing CFSE without checking whether the filling forces extra pairing beyond what the free ion already has — only the extra pairs contribute the $+xP$ penalty.

Ask the tutor: ask the AI tutor to compute the CFSE for any d-electron count and ligand field strength you're working through, and to check whether you've correctly counted the extra pairs.