Crystal Field Theory (CFT) — Δ_oct, Δ_tet, splitting diagrams: Visual walkthrough

Everything here supports its parent CFT topic note (index 3.4.8).

Step 1 — Five orbitals, one lonely ion, all equal

WHAT. Picture a metal ion floating alone in empty space. It has five $d$ orbitals. An "orbital" is just a region of space where an electron is likely to be found — think of it as a cloud with a particular shape. All five clouds sit at the same energy. We call same-energy states degenerate.

WHY start here. If the orbitals were already unequal, there'd be nothing to explain. We need the "before" picture so that later the "after" (the split) is visible as a change.

PICTURE. Look at Figure s01. On the left, all five orbitals are drawn as a single horizontal line — one shelf, five clouds, equal height. Height on this diagram means energy: higher line = more energy = less stable.

Figure — Crystal Field Theory (CFT) — Δ_oct, Δ_tet, splitting diagrams

Step 2 — Meet the two orbital families by their shape

WHAT. The five clouds are not all shaped alike. Two of them point their lobes straight along the $x$ , $y$ , $z$ axes; the other three point into the gaps between axes.

Family	Orbitals	Where the lobes point
"along-axis"	$d_{x^{2} - y^{2}}, d_{z^{2}}$	straight down the $\pm x, \pm y, \pm z$ directions
"between-axis"	$d_{x y}, d_{y z}, d_{x z}$	into the diagonal gaps

WHY this split matters. In Step 3 we place ligands on the axes. An orbital that points at a ligand will be affected very differently from one that points away. Shape decides energy — so we must sort by shape first.

PICTURE. Figure s02 shows one representative of each family: $d_{z^{2}}$ (a lobe straight up and down the $z$ -axis, cyan) versus $d_{x y}$ (four lobes tilted into the corners, amber). Trace with your eye where each cloud is densest — the cyan one is dense on the axis, the amber one is dense between axes.

Recall Which orbitals are "along axis"?

Which two $d$ orbitals point along the axes? ::: $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ . Which three point between the axes? ::: $d_{x y}$ , $d_{y z}$ , $d_{x z}$ .

Step 3 — Bring in six ligands on the axes (the octahedron)

WHAT. Now surround the ion with 6 ligands, one on each of the six axis-tips: $+ x, - x, + y, - y, + z, - z$ . Each ligand is modelled as a point negative charge — a tiny "keep away" magnet, because negative charges repel the negatively-charged electrons in the $d$ orbitals.

WHY point charges. CFT is deliberately simple: it ignores covalent bonding and asks only "how does electrostatic repulsion rearrange the orbital energies?" A point charge is the crudest, cleanest way to model "a ligand sits here and repels."

PICTURE. Figure s03 draws the six amber ligands at the axis-tips. Overlay the $d_{z^{2}}$ cloud (points at the $\pm z$ ligands — head-on collision) and the $d_{x y}$ cloud (points into the gaps — no ligand there). Notice: an along-axis orbital always meets a ligand face-on; a between-axis orbital always slides into an empty gap.

Step 4 — Repulsion pushes the two families apart

WHAT. Electrons in the along-axis orbitals ( $d_{x^{2} - y^{2}}, d_{z^{2}}$ ) sit right in front of the ligand charges → they feel strong repulsion → their energy goes UP. Electrons in the between-axis orbitals ( $d_{x y}, d_{y z}, d_{x z}$ ) dodge the ligands → weaker repulsion → energy goes DOWN (relative to the average).

We give the two new groups names:

$raised d_{x^{2} - y^{2}}, d_{z^{2}} \to == e_{g} == lowered d_{x y}, d_{y z}, d_{x z} \to == t_{2 g} ==$

Here $e$ means "a set of 2 orbitals", $t$ means "a set of 3 orbitals", and the subscript $g$ (gerade, German for "even") flags that the octahedron has a centre of symmetry.

WHY they move in opposite directions. All six ligands push every $d$ electron up somewhat (overall repulsion is up), but that overall rise is not the story. Relative to the average, the face-on orbitals rise and the gap-dodging orbitals fall. It's a see-saw about the average.

PICTURE. Figure s04: the single shelf from Step 1 splits into two shelves — a lower one labelled $t_{2 g}$ (3 lines) and an upper one labelled $e_{g}$ (2 lines). The vertical distance between them is the prize we're chasing: $Δ_{oc t}$ .

Step 5 — The barycentre rule: energy is only reshuffled, not created

WHAT. When we split the shelves, we cannot invent energy from nowhere. The rule: the weighted average energy stays exactly where the un-split line was (call that level zero). This is the barycentre (centre-of-gravity) rule.

WHY it must be true. Splitting is a redistribution. Turning ligands on rearranges which orbital is high and which is low, but the total electrostatic bookkeeping is fixed by the average interaction. So the ups must exactly cancel the downs when we weight each orbital by how many orbitals are in its group.

PICTURE. Figure s05 shows a literal see-saw: 3 orbitals ( $t_{2 g}$ ) sit below the pivot line, 2 orbitals ( $e_{g}$ ) sit above. For the plank to balance, the total downward pull (3 orbitals × their drop) must equal the total upward pull (2 orbitals × their rise).

Let each $t_{2 g}$ orbital drop by $x$ below zero, each $e_{g}$ orbital rise by $y$ above zero.

$orbitals in t_{2 g} 3 \cdot each drops x (- x) + orbitals in e_{g} 2 \cdot each rises y (+ y) = balance 0$

That's the balance condition. We also know the two shelves are $Δ_{oc t}$ apart:

$drop of t_{2 g} x + rise of e_{g} y = the gap Δ_{oc t}$

Step 6 — Solve, and out drop the famous fractions

WHAT. Solve the two equations for $x$ and $y$ .

From the balance equation: $3 x = 2 y \Rightarrow x = \frac{2}{3} y$ This says: because there are 3 lower orbitals but only 2 upper ones, each lower orbital only needs to drop $\frac{2}{3}$ as far as each upper one rises — fewer orbitals up top must each move further to keep balance.

Substitute into $x + y = Δ_{oc t}$ : $\frac{2}{3} y + y = Δ_{oc t} \Rightarrow \frac{5}{3} y = Δ_{oc t} \Rightarrow y = \frac{3}{5} Δ_{oc t}$

Then: $x = Δ_{oc t} - y = Δ_{oc t} - \frac{3}{5} Δ_{oc t} = \frac{2}{5} Δ_{oc t}$

WHY these exact numbers. The $\frac{3}{5}$ and $\frac{2}{5}$ are nothing but the other group's orbital count over the total: $e_{g}$ rises by $\frac{3}{5}$ (3 = number in $t_{2 g}$ ... no — it's the fraction that keeps the see-saw level). Concretely: $\frac{3}{5} = 0.6$ and $\frac{2}{5} = 0.4$ .

PICTURE. Figure s06 is the finished octahedral splitting diagram, fully labelled: dashed zero line (the barycentre), $t_{2 g}$ sitting $0.4 Δ_{oc t}$ below it, $e_{g}$ sitting $0.6 Δ_{oc t}$ above it, and a vertical amber arrow marking the full gap $Δ_{oc t}$ .

Recall Where do 0.4 and 0.6 come from?

Why does $t_{2 g}$ drop by $0.4Δ$ and $e_{g}$ rise by $0.6Δ$ ? ::: From the barycentre balance $3 x = 2 y$ with $x + y = Δ$ : solving gives $x = \frac{2}{5} Δ = 0.4Δ$ and $y = \frac{3}{5} Δ = 0.6Δ$ .

Step 7 — Edge case: what if a shelf is empty? (barycentre still holds)

WHAT. A common worry: "if I put electrons only in $t_{2 g}$ , doesn't the average energy drop below zero, breaking the balance?" No — the barycentre is a property of the orbital levels, not of the electrons. The zero line is fixed by the geometry (3 vs 2 orbitals), regardless of how many electrons you later pour in.

WHY this is the degenerate case. For $d^{0}$ (no electrons) and $d^{10}$ (all orbitals full) the electron average also sits exactly on zero, so those ions gain zero stabilisation from splitting. They are the "degenerate" endpoints where CFT predicts no crystal-field bonus.

PICTURE. Figure s07 shows three fillings side by side: $d^{0}$ (empty), $d^{3}$ ( $t_{2 g}^{3}$ , all three lower orbitals singly filled), and $d^{10}$ (all ten spots full). The dashed zero line sits in the same place in all three — the levels never moved; only the dots (electrons) changed.

The one-picture summary

Figure s08 compresses the whole derivation into one frame: on the left the five degenerate orbitals (Step 1); an arrow labelled "6 ligands on axes" (Step 3); on the right the split diagram with the see-saw balance annotation ( $3 \times 0.4Δ$ down $= 2 \times 0.6Δ$ up), the dashed barycentre, and the amber $Δ_{oc t}$ gap. Reading it left-to-right is the derivation.

Recall Feynman: retell the whole walkthrough in plain words

Five equal orbital-clouds float around a lonely metal ion (Step 1). Two of them (the "along-axis" ones) point straight along the $x, y, z$ arrows; three (the "between-axis" ones) hide in the gaps (Step 2). Now we drop six "keep-away" charges right on those arrows (Step 3). The clouds pointing at a charge get shoved uphill in energy — that's $e_{g}$ ; the clouds hiding in the gaps sink a little — that's $t_{2 g}$ (Step 4). But we can't create energy, so the see-saw must balance about the old average: three sinking orbitals versus two rising ones (Step 5). Working out the balance forces the risers up by $0.6Δ$ and the sinkers down by $0.4Δ$ — those fractions are just the orbital counts in disguise (Step 6). Finally, even if we leave shelves empty or fill them completely, the balance line never moves — it belongs to the geometry, not to the electrons; that's why $d^{0}$ and $d^{10}$ get zero crystal-field bonus (Step 7). And that gap $Δ_{oc t}$ is the single number that later decides colour, magnetism, and spin state.

Where this leads next: the size of $Δ_{oc t}$ is set by the ligand (Spectrochemical Series & Ligand Strength); comparing $Δ_{oc t}$ against pairing energy decides high-spin vs low-spin; a photon of energy $Δ_{oc t}$ gives colour; an uneven $e_{g}$ filling triggers Jahn–Teller Distortion; and summing electron energies gives CFSE. Compare the whole electrostatic picture against the orbital-overlap picture in Valence Bond Theory of Complexes, and name your complexes with Coordination Compounds — Nomenclature.