Crystal Field Theory (CFT) — Δ_oct, Δ_tet, splitting diagrams: Worked examples

First, let's fix notation and lay out the full space of cases so nothing hides.

Definition Notation used on this page

$Δ \equiv Δ_{oc t}$ . Whenever you see a bare $Δ$ in a CFSE line for an octahedral complex, it means $Δ_{oc t}$ — the octahedral splitting energy, measured in $kJ mol^{- 1}$ (or, per photon, in J). We only write the subscript when both geometries appear together.
$P$ = pairing energy: the extra energy it costs to force a second electron into an already-occupied orbital, because two electrons in the same small region repel each other (and lose exchange stabilisation). Typical magnitude is a few hundred $kJ mol^{- 1}$ — the same order as $Δ_{oc t}$ , which is exactly why the two can compete. High-spin vs low-spin is literally the contest " $Δ$ vs $P$ ".
BM = Bohr magneton, the natural unit of magnetic moment for an electron. When we quote $μ = 3.87 BM$ we mean "3.87 Bohr magnetons" — you can treat BM simply as the unit that the spin-only formula spits out, the way metres come out of a distance formula.

The scenario matrix

Every problem in Crystal Field Theory is a point in this table. The two axes are electron count $d^{n}$ and geometry + field strength.

Cell	$d^{n}$ range	Geometry / field	What makes it special
A	$d^{1}, d^{2}, d^{3}$	octahedral, any field	Filling forced — no spin choice, same answer weak or strong
B	$d^{4}$ – $d^{7}$	octahedral, weak field	High-spin: singly-fill before pairing
C	$d^{4}$ – $d^{7}$	octahedral, strong field	Low-spin: pair in $t_{2 g}$ first
D	$d^{8}, d^{9}, d^{10}$	octahedral, any field	Filling forced again
E	any $d^{n}$	tetrahedral	Always high-spin, order flipped, no $g$
F	$d^{5}$ high-spin	octahedral	Degenerate/zero case: CFSE $= 0$
G	$d^{10}$	any	Fully-filled limiting case: CFSE $= 0$ , colourless
H	colour	octahedral	Photon energy $\leftrightarrow$ wavelength $\leftrightarrow$ complementary colour
I	exam twist	mixed	Same ion, two ligands → magnetism flip

The 10 worked examples below hit every cell A–I (with the $d^{4}$ contrast in cell B/C spelled out explicitly).

Two energy patterns we reuse constantly (all $Δ$ in $kJ mol^{- 1}$ ):

$E (t_{2 g}) = - 0.4 Δ_{oc t}$ , $E (e_{g}) = + 0.6 Δ_{oc t}$ (octahedral).
$E (e) = - 0.6 Δ_{t e t}$ , $E (t_{2}) = + 0.4 Δ_{t e t}$ (tetrahedral — order flipped).

CFSE = (electrons in lower set) $\times$ (lower energy) + (electrons in upper set) $\times$ (upper energy), plus one $P$ for every electron pair beyond the free-ion baseline above. Units of CFSE are therefore the units of $Δ$ (and $P$ ): $kJ mol^{- 1}$ .

And the magnetism check, from the spin-only formula: $μ = n (n + 2) BM, n = number of unpaired electrons.$

Example 1 — Cell A: $d^{3}$ , filling is forced

Step 1 — Find $d^{n}$ . Cr is $[Ar] 3 d^{5} 4 s^{1}$ ; Cr $^{3 +}$ removes $4 s^{1}$ and two $3 d$ → $3 d^{3}$ . Why this step? CFSE and unpaired-count depend only on how many $d$ electrons there are and where they sit, so we must nail $n$ first — everything downstream is built on it.

Step 2 — Place the electrons. Three electrons, three $t_{2 g}$ orbitals, one each (Hund's rule — spread out to avoid pairing cost). Config: $t_{2 g}^{3} e_{g}^{0}$ . Why this step? With exactly three orbitals and three electrons, there is no reason to climb to $e_{g}$ regardless of $Δ$ . The weak/strong distinction is irrelevant — that's the whole point of cell A.

Step 3 — Count unpaired. Each of the three $t_{2 g}$ orbitals holds one electron, so all three are unpaired → $n = 3$ . Why this step? Magnetism ( $μ$ ) is set entirely by the number of unpaired electrons, so we tally them before applying the spin-only formula.

Step 4 — CFSE. $3$ electrons in $t_{2 g}$ , none in $e_{g}$ ; free-ion baseline for $d^{3}$ is 0 pairs and the complex has 0 pairs → no $P$ : $CFSE = 3 (- 0.4 Δ_{oc t}) = - 1.2 Δ_{oc t} (kJ mol^{- 1}) .$ Why this step? Multiply each set's occupancy by its orbital energy and sum — this is the definition of CFSE, the net electrostatic stabilisation the splitting buys us.

Verify: $μ = 3 (3 + 2) = 15 \approx 3.87$ BM, the textbook value for Cr $^{3 +}$ . Weak-field $[CrF_{6}]^{3 -}$ gives the identical config — confirming cell A has no spin choice. ✓

Example 2 — Cell B: $d^{4}$ vs $d^{6}$ weak field (high-spin)

Step 1 — $d^{n}$ for both. Cr is $[Ar] 3 d^{5} 4 s^{1}$ ; Cr $^{2 +}$ removes $4 s^{1}$ and one $3 d$ → $3 d^{4}$ . Fe is $[Ar] 3 d^{6} 4 s^{2}$ ; Fe $^{2 +}$ removes $4 s^{2}$ → $3 d^{6}$ . Why this step? The whole filling pattern hinges on $n$ ; getting it wrong poisons every later number.

Step 2 — Field. Both ligands are weak (H $_{2}$ O is mild; F $^{-}$ is left of centre in the spectrochemical series) → $Δ_{oc t} < P$ → high-spin. Why this step? Weak field means the climb to $e_{g}$ ( $Δ$ ) is cheaper than pairing ( $P$ ), so electrons spread out singly before any pairing.

Step 3a — Fill $d^{4}$ . Fill each of the five orbitals singly in energy order until electrons run out: $t_{2 g}$ gets 3 singles, then the 4th electron goes to an empty $e_{g}$ orbital (cheaper than pairing). Config $t_{2 g}^{3} e_{g}^{1}$ ; all four unpaired → $n = 4$ . Why this step? The $d^{4}$ choice — $t_{2 g}^{4} e_{g}^{0}$ (low-spin) versus $t_{2 g}^{3} e_{g}^{1}$ (high-spin) — is the very fork that splits cells B and C. Weak field takes the high-spin branch.

Step 3b — Fill $d^{6}$ . All five orbitals get one electron first (5 singles), then the 6th must pair, landing in the lowest available orbital, a $t_{2 g}$ . Config $t_{2 g}^{4} e_{g}^{2}$ . Now count unpaired: $t_{2 g}^{4}$ means one orbital is doubly filled (a pair, 0 unpaired) and the other two hold one each (2 unpaired); $e_{g}^{2}$ means two singles (2 unpaired). Total $n = 2 + 2 = 4$ . Why this step? With more electrons than orbitals, one pair is forced; we track exactly which orbital pairs so the unpaired count is unambiguous.

Step 4 — CFSE. $d^{4}$ : free-ion baseline $max (0, 4 - 5) = 0$ pairs; complex has 0 pairs → no $P$ . $CFSE = 3 (- 0.4Δ) + 1 (+ 0.6Δ) = - 1.2Δ + 0.6Δ = - 0.6 Δ_{oc t}$ . $d^{6}$ : free-ion baseline $max (0, 6 - 5) = 1$ pair; the high-spin complex also has just 1 pair → no extra $P$ . $CFSE = 4 (- 0.4Δ) + 2 (+ 0.6Δ) = - 1.6Δ + 1.2Δ = - 0.4 Δ_{oc t}$ . Why this step? CFSE compares each complex to its own barycentre, and we only charge $P$ for pairs created beyond the free-ion baseline — otherwise we'd double-count intrinsic pairing.

Verify: $d^{4}$ : $μ = 4 (4 + 2) = 24 \approx 4.90$ BM (Cr $^{2 +}$ high-spin). $d^{6}$ : $μ = 4 (4 + 2) = 24 \approx 4.90$ BM — the measured high-spin Fe $^{2 +}$ value. ✓

Example 3 — Cell C: $d^{6}$ strong field (low-spin)

Step 1 — $d^{n}$ . Same Fe $^{2 +}$ → $3 d^{6}$ . Why this step? Fixing $n$ lets us compare directly with the high-spin twin — only the arrangement, not the electron count, changes.

Step 2 — Field. CN $^{-}$ is at the strong end of the series → $Δ_{oc t} > P$ → low-spin. Why this step? A big gap makes pairing in $t_{2 g}$ ( $P$ ) cheaper than the climb to $e_{g}$ ( $Δ$ ), so electrons pair down low.

Step 3 — Fill. All 6 pile into the three $t_{2 g}$ orbitals (3 pairs): $t_{2 g}^{6} e_{g}^{0}$ . Every electron is paired → $n = 0$ → diamagnetic. Why this step? When $Δ > P$ , filling the low set fully (even at the cost of pairing) minimises total energy — this is the defining move of the low-spin branch.

Step 4 — CFSE. Free-ion baseline for $d^{6}$ is 1 pair; low-spin has 3 pairs → 2 extra pairs → add $2 P$ . $CFSE = 6 (- 0.4Δ) + 0 + 2 P = - 2.4 Δ_{oc t} + 2 P (kJ mol^{- 1}) .$ Why this step? Every pair created beyond the free-ion baseline of 1 costs one $P$ ; here 2 new pairs, so $+ 2 P$ — this bookkeeping is what separates low-spin CFSE from high-spin.

Verify: $μ = 0 (0 + 2) = 0$ BM — diamagnetic, matching $[Fe(CN)_{6}]^{4 -}$ . Compare Ex. 2(b): same ion, magnetism flipped by ligand. ✓

Example 4 — Cell D: $d^{8}$ , forced filling

Step 1 — $d^{n}$ . Ni is $[Ar] 3 d^{8} 4 s^{2}$ ; Ni $^{2 +}$ → $3 d^{8}$ . Why this step? Everything downstream is counted from $n$ ; establish it before filling.

Step 2 — Fill. Lower $t_{2 g}$ (3 orbitals) fills completely = 6 electrons (3 pairs). Remaining 2 go one each into the two $e_{g}$ orbitals: $t_{2 g}^{6} e_{g}^{2}$ . Why this step? With 8 electrons, $t_{2 g}$ is full no matter what, and the last two must occupy $e_{g}$ singly (Hund). Weak or strong field gives the same picture → cell D has no choice.

Step 3 — Unpaired. The six $t_{2 g}$ electrons are all paired; only the two $e_{g}$ singles remain unpaired → $n = 2$ . Why this step? Only singly-occupied orbitals contribute to magnetism, so we isolate them before computing $μ$ .

Step 4 — CFSE. Free-ion baseline for $d^{8}$ is $max (0, 8 - 5) = 3$ pairs; complex also has 3 → no extra $P$ . $CFSE = 6 (- 0.4Δ) + 2 (+ 0.6Δ) = - 2.4Δ + 1.2Δ = - 1.2 Δ_{oc t} (kJ mol^{- 1}) .$ Why this step? Occupancy × energy, summed — and since no pairs were created beyond the free-ion baseline, no $P$ term appears.

Verify: $μ = 2 (2 + 2) = 8 \approx 2.83$ BM — standard Ni $^{2 +}$ value. ✓

Example 5 — Cell E: tetrahedral, order flipped, always high-spin

Figure 1 (below) shows the two energy-level diagrams side by side so you can literally watch the order flip. On the left (octahedral) the blue $t_{2 g}$ line sits below the barycentre at $- 0.4Δ$ and the pink $e_{g}$ line above at $+ 0.6Δ$ . On the right (tetrahedral) the roles swap: the pink $e$ line drops to $- 0.6 Δ_{t e t}$ (lower) and the blue $t_{2}$ line rises to $+ 0.4 Δ_{t e t}$ (upper). The yellow double-arrows mark the gaps; the tetrahedral gap is drawn narrower to remind us $Δ_{t e t} = \frac{4}{9} Δ_{oc t}$ (derived in the box just after the figure). Read the figure as: "in a tetrahedron the axis-pointing orbitals become the comfortable low set."

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

Step 1 — $d^{n}$ . Co is $[Ar] 3 d^{7} 4 s^{2}$ ; Co $^{2 +}$ → $3 d^{7}$ . Why this step? Seven electrons is what forces the interesting $e^{4} t_{2}^{3}$ split; fix $n$ first.

Step 2 — Geometry. Tetrahedral → lower set is $e$ (2 orbitals), upper is $t_{2}$ (3 orbitals) — the flip drawn on the right of Figure 1. No $g$ subscript (no centre of inversion). Why this step? In tetrahedral geometry the axis-pointing $e$ orbitals now avoid the between-axis ligands, so they drop; the $t_{2}$ orbitals point closer to ligands, so they rise — the labels and order invert versus octahedral.

Step 3 — Fill (always high-spin). $Δ_{t e t} \approx \frac{4}{9} Δ_{oc t}$ is tiny → $Δ_{t e t} < P$ → spread out. Filling singly in energy order: $e^{2}$ (2 singles) then $t_{2}^{3}$ (3 singles) = 5 electrons all unpaired; the remaining 2 must pair, filling the lower $e$ set first → $e^{4} t_{2}^{3}$ . In $e^{4}$ both orbitals are paired (0 unpaired); in $t_{2}^{3}$ each orbital has one electron (3 unpaired). So $n = 3$ . Why this step? Because $Δ_{t e t} < P$ always, there is no low-spin option — we fill singly as far as possible, exactly as cell E demands, then pair into the lowest orbitals.

Step 4 — CFSE. $e$ at $- 0.6 Δ_{t e t}$ , $t_{2}$ at $+ 0.4 Δ_{t e t}$ : $CFSE = 4 (- 0.6 Δ_{t e t}) + 3 (+ 0.4 Δ_{t e t}) = - 2.4 Δ_{t e t} + 1.2 Δ_{t e t} = - 1.2 Δ_{t e t} (kJ mol^{- 1}) .$ Why this step? Same occupancy × energy rule, but with the tetrahedral energies and the flipped set sizes — a good habit-check that you used $- 0.6/ + 0.4$ , not the octahedral $- 0.4/ + 0.6$ .

Verify: $μ = 3 (3 + 2) = 15 \approx 3.87$ BM — Co $^{2 +}$ tetrahedral is indeed ~3.9–4.8 BM (spin-only floor 3.87). ✓ There is no low-spin alternative — cell E confirmed.

Example 6 — Cell F: $d^{5}$ high-spin, the zero-CFSE case

Step 1 — $d^{n}$ . Mn is $[Ar] 3 d^{5} 4 s^{2}$ ; Mn $^{2 +}$ → $3 d^{5}$ . Why this step? Five electrons is exactly the "half-filled" case that produces the zero-CFSE surprise; confirm it first.

Step 2 — Fill. Weak field, high-spin: one electron in each of the five orbitals: $t_{2 g}^{3} e_{g}^{2}$ . All five unpaired → $n = 5$ . Why this step? Half-filling every orbital singly maximises unpaired spins (Hund) and is the arrangement whose energies happen to cancel.

Step 3 — CFSE. Three electrons in $t_{2 g}$ at $- 0.4Δ$ each, two in $e_{g}$ at $+ 0.6Δ$ each; free-ion baseline for $d^{5}$ is 0 pairs and the complex has 0 pairs, so no $P$ : $CFSE = 3 (- 0.4Δ) + 2 (+ 0.6Δ) = - 1.2Δ + 1.2Δ = 0 (kJ mol^{- 1}) .$ The $- 1.2Δ$ stabilisation from the three lower electrons is exactly cancelled by the $+ 1.2Δ$ destabilisation from the two upper electrons. Why this step? This term-by-term cancellation is the whole point of cell F — the balanced half-filled shell sits right on the barycentre, so its net crystal-field stabilisation is genuinely zero.

Verify: $μ = 5 (5 + 2) = 35 \approx 5.92$ BM — matches Mn $^{2 +}$ . And CFSE $= 0$ explains why such ions are pale and kinetically labile (see Stability of Complexes & CFSE). ✓

Example 7 — Cell G: $d^{10}$ , the full-shell limiting case

Step 1 — $d^{n}$ . Zn is $[Ar] 3 d^{10} 4 s^{2}$ ; Zn $^{2 +}$ → $3 d^{10}$ . Why this step? Ten electrons means a completely full $d$ shell — the limiting case we want to test.

Step 2 — Fill. All five orbitals doubly occupied: $t_{2 g}^{6} e_{g}^{4}$ . Every electron paired → $n = 0$ . Why this step? A full shell has no vacancies and no unpaired spins — this is what makes it both diamagnetic and (as we'll see) colourless.

Step 3 — CFSE. Free-ion baseline for $d^{10}$ is $max (0, 10 - 5) = 5$ pairs; the complex also has 5, so no extra $P$ : $CFSE = 6 (- 0.4Δ) + 4 (+ 0.6Δ) = - 2.4Δ + 2.4Δ = 0 (kJ mol^{- 1}) .$ Why this step? A full shell is symmetric — lower and upper contributions cancel exactly, just like $d^{5}$ (but here everything is paired).

Step 4 — Colour. For a d–d transition an electron must jump $t_{2 g} \to e_{g}$ , but $e_{g}$ is full — no empty slot to receive it. Why this step? No vacancy → no d–d transition → no visible absorption → colourless. This is the limiting case of Colour & d–d Transitions.

Verify: $μ = 0 (0 + 2) = 0$ BM, diamagnetic, colourless — exactly Zn $^{2 +}$ complexes. ✓

Example 8 — Cell H: colour from $Δ$

Step 1 — Per-molecule energy. Divide by Avogadro's number $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ : $E = \frac{250 000 J mol ^{- 1}}{6.022 \times 1 0 ^{23}} = 4.152 \times 1 0^{- 19} J .$ Why this step? $Δ_{oc t} = h ν$ is a per-photon relation, but $Δ$ was quoted per mole ( $kJ mol^{- 1} \to J mol^{- 1}$ ); we must convert to energy per photon before using $h ν$ , keeping units consistent.

Step 2 — Wavelength. From $E = \frac{h c}{λ}$ , solve $λ = \frac{h c}{E}$ with $h = 6.626 \times 1 0^{- 34}$ J·s, $c = 3.0 \times 1 0^{8}$ m/s: $λ = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{4.152 \times 1 0 ^{- 19}} = 4.79 \times 1 0^{- 7} m = 479 nm .$ Why this step? Wavelength (not energy) is what tells us which colour band is absorbed, so we convert $E$ to $λ$ .

Step 3 — Perceived colour. $479$ nm is blue. We see the complementary colour → orange. Why this step? We observe what is not absorbed — the complement of the absorbed band — never the absorbed colour itself (a classic trap).

Verify: $λ \approx 479$ nm sits in the blue band (450–495 nm); complement of blue is orange. Units: $\frac{J\cdots \cdot m/s}{J} = m$ . ✓

Example 9 — Cell I: the exam twist (same ion, two ligands)

Step 1 — $d^{n}$ . Co is $[Ar] 3 d^{7} 4 s^{2}$ ; Co $^{3 +}$ removes $4 s^{2}$ + one $3 d$ → $3 d^{6}$ . Why this step? Both complexes share the same electron count, so any difference in magnetism must come from the ligand, not from $n$ — worth pinning down first.

Step 2 — Rank ligands. NH $_{3}$ is strong field (right of H $_{2}$ O); F $^{-}$ is weak. Why this step? Field strength sets whether $Δ_{oc t}$ beats $P$ , which decides high- vs low-spin — the crux of the twist.

Step 3 — Assign spin.

$[Co(NH_{3})_{6}]^{3 +}$ : strong → low-spin → $t_{2 g}^{6} e_{g}^{0}$ → $n = 0$ → diamagnetic.
$[CoF_{6}]^{3 -}$ : weak → high-spin → $t_{2 g}^{4} e_{g}^{2}$ → $n = 4$ → paramagnetic. Why this step? Once the field is ranked, the electron arrangement follows mechanically, and the unpaired count gives the magnetic verdict.

Step 4 — CFSE. The free-ion baseline for $d^{6}$ is 1 pair (from the box above: $max (0, 6 - 5) = 1$ ).

Low-spin ( $t_{2 g}^{6}$ ) has 3 pairs → $3 - 1 = 2$ extra pairs → $+ 2 P$ : $CFSE = 6 (- 0.4Δ) + 2 P = - 2.4 Δ_{oc t} + 2 P$ .
High-spin ( $t_{2 g}^{4} e_{g}^{2}$ ) has just 1 pair → $1 - 1 = 0$ extra pairs → no $P$ : $CFSE = 4 (- 0.4Δ) + 2 (+ 0.6Δ) = - 0.4 Δ_{oc t}$ . The strong-field complex has by far the larger (more negative) stabilisation. Why this step? We measure extra pairs against the free-ion baseline of 1 — that consistent reference is what makes the $+ 2 P$ term correct and avoids the double-counting trap.

Verify: $μ_{NH_{3}} = 0 \cdot 2 = 0$ BM (diamagnetic); $μ_{F} = 4 \cdot 6 = 24 \approx 4.90$ BM. Matches: $[Co(NH_{3})_{6}]^{3 +}$ diamagnetic, $[CoF_{6}]^{3 -}$ paramagnetic. ✓

Recall check

Recall Which cells give CFSE exactly zero, and why?

$d^{5}$ high-spin (cell F) and $d^{10}$ (cell G) — the lower-set stabilisation and upper-set destabilisation cancel in each symmetric fill. Cells with a genuine spin choice ::: only octahedral $d^{4}$ – $d^{7}$ (cells B and C) Tetrahedral spin state ::: always high-spin, because $Δ_{t e t} = \frac{4}{9} Δ_{oc t} < P$ Where does the 4/9 come from? ::: two-thirds fewer effective pushers (4 vs 6 ligands) times two-thirds for no direct hits — $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$ Free-ion pairs for $d^{n}$ ::: $max (0, n - 5)$ Colour of a $d^{10}$ complex ::: colourless — $e_{g}$ is full, so no d–d transition We see which colour, relative to what is absorbed? ::: the complementary colour of the light absorbed, never the absorbed colour itself What does BM stand for? ::: Bohr magneton, the unit of magnetic moment

Related depth: Jahn–Teller Distortion (why some of these configs distort), Valence Bond Theory of Complexes (the rival model CFT replaced), Coordination Compounds — Nomenclature (naming these ions).