Crystal Field Theory (CFT) — Δ_oct, Δ_tet, splitting diagrams: Question bank

Before we start, one reminder of the vocabulary so nothing here is undefined:

== $Δ$ (splitting energy)== — the energy gap between the two orbital groups after ligands approach.
== $P$ (pairing energy)== — the energy cost of forcing two electrons into the same orbital.
== $e_{g} / t_{2 g}$ (octahedral) and $e / t_{2}$ == (tetrahedral) — the two groups the five $d$ orbitals split into.
CFSE — Crystal Field Stabilisation Energy, the net energy gained from splitting.
== $n$ — the number of unpaired electrons== a metal ion has after filling its split $d$ orbitals (count the singly-occupied orbitals).
== $μ$ (mu, magnetic moment)== — a number measuring how strongly the ion behaves like a tiny magnet in a magnetic field; more unpaired electrons → larger $μ$ .

Picture 1: where the two orbital groups come from

Before any question, look at the octahedron. The six ligands (magenta dots) sit exactly on the axes. Two $d$ orbitals aim at them, three aim between them — that single geometric fact is the seed of every answer below.

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

Picture 2: the splitting diagram and the –0.4 / +0.6 offsets

When people ask "why exactly $- 0.4Δ$ and $+ 0.6Δ$ ?", this is the answer. The dashed line is the barycentre — the average energy, which cannot move because splitting only redistributes energy, never creates or destroys it.

Set up a sign convention first: measure every orbital's energy relative to the barycentre, so up = positive, down = negative. The two $e_{g}$ orbitals go up by $+ y$ ; the three $t_{2 g}$ orbitals go down, i.e. by $- x$ (the minus sign is what makes "down" negative). The barycentre cannot move, so the total signed displacement of all five orbitals must be zero: $2 (+ y) + 3 (- x) = 0 \Rightarrow 2 y = 3 x .$ Combined with the fact that the gap between them is the whole $Δ$ , i.e. $x + y = Δ$ , this forces $y = \frac{3}{5} Δ = + 0.6Δ$ (the $e_{g}$ rise) and $x = \frac{2}{5} Δ$ , so the $t_{2 g}$ energy is $- x = - 0.4Δ$ — negative precisely because it dropped below the barycentre. That is why every CFSE sum below multiplies electron counts by $- 0.4Δ$ and $+ 0.6Δ$ .

Picture 3: why $Δ_{t e t} = \frac{4}{9} Δ_{oc t}$

The famous factor is not a rote number — it comes from geometry, and both factors below have a physical reason.

Factor (a) — fewer ligands: $\frac{4}{6}$ . The raw strength of the electrostatic field scales with how many point charges surround the metal. A tetrahedron has 4 ligands versus the octahedron's 6, so the whole field is weaker by $4/6$ .

Factor (b) — off-axis geometry: $\frac{2}{3}$ . In the octahedron a ligand sits on an axis, so an axis-pointing orbital feels the full head-on repulsion (call this "1 unit"). In the tetrahedron every ligand sits at a cube-corner, off every axis — the metal–ligand line makes an angle with the nearest orbital lobe rather than hitting it dead-on. Because the repulsion an orbital feels depends on how much of the ligand direction lines up with the orbital's axis, and a tetrahedral ligand's direction shares only two of its three coordinate components with any single orbital axis, each orbital captures on average $\frac{2}{3}$ of the head-on effect it would feel if the ligand were on-axis. No orbital ever gets the full punch — hence a further $\frac{2}{3}$ scaling.

Multiply the two independent weakenings: $\frac{4}{6} \times \frac{2}{3} = \frac{8}{18} = \frac{4}{9}$ . The diagram shows the same five orbitals, now with the order flipped and the gap visibly shrunk.

True or false — justify

True or false: In an octahedral field the

e_{g}

orbitals are lower in energy than the

t_{2 g}

orbitals.

False.

e_{g}

(

d_{x^{2} - y^{2}}, d_{z^{2}}

) point straight at the six axial ligands (Picture 1), feel more repulsion, and sit above

t_{2 g}

Δ_{oc t}

True or false: A tetrahedral complex can be low-spin if you use a strong-field ligand like CN

^{-}

False.

Δ_{t e t} = \frac{4}{9} Δ_{oc t}

(Picture 3) is so small it stays below

P

even for strong ligands, so tetrahedral complexes are treated as always high-spin at this level.

True or false: The tetrahedral orbital sets should be written

e_{g}

and

t_{2 g}

False. The subscript

g

(gerade) means a centre of inversion exists; a tetrahedron has none, so we write plain

e

and

t_{2}

True or false: CFT models the metal–ligand bond as a covalent, orbital-sharing interaction.

False. CFT treats it as purely electrostatic, with ligands as point negative charges — covalent character is what CFT ignores (and later Ligand Field Theory adds).

True or false: High-spin

d^{5}

octahedral has zero CFSE.

True. Using the offsets from Picture 2, config

t_{2 g}^{3} e_{g}^{2}

gives

3 (- 0.4Δ) + 2 (+ 0.6Δ) = - 1.2Δ + 1.2Δ = 0

, which is why such ions are pale and labile.

True or false: The colour you see for a complex is the colour it absorbs.

False. You see the complementary colour — a complex absorbing red light appears green.

True or false: For an octahedral

d^{3}

ion, whether it is high- or low-spin can differ depending on the ligand.

False. For octahedral

d^{3}

the filling is forced (

t_{2 g}^{3}

, all singly) regardless of

Δ

; the genuine high/low-spin choice only exists for octahedral

d^{4}

–

d^{7}

True or false: A larger

Δ_{oc t}

means the complex absorbs light of a longer wavelength.

False. Since

Δ_{oc t} = h c / λ

(with

h, c, λ

as defined above), a larger gap means higher-energy, shorter-wavelength absorption.

True or false: The barycentre rule says the total energy displacement of all five orbitals sums to zero.

True. Splitting only redistributes energy; the weighted average (the dashed line in Picture 2) stays put, so

3 (- x) + 2 (+ y) = 0

octahedrally.

Spot the error

Find the error: "

d_{x y}

points along the

x

and

y

axes, so in octahedral it is raised into

e_{g}

d_{x y}

points between the axes (into the quadrant), so it goes into the lowered

t_{2 g}

set, not

e_{g}

Find the error: "Because a tetrahedron has 4 ligands not 6,

Δ_{t e t}

is exactly

\frac{4}{6} Δ_{oc t}

The count

4/6

is only half the story; you must also multiply by

2/3

for the off-axis geometry (Picture 3), giving the true factor

\frac{4}{9}

Find the error: "For

[Fe(CN)_{6}]^{4 -}

, CN

^{-}

is weak-field so the ion is high-spin with 4 unpaired electrons."

^{-}

is a strong-field ligand, so

Δ_{oc t} > P

giving low-spin

t_{2 g}^{6} e_{g}^{0}

— zero unpaired electrons, diamagnetic.

Find the error: "CFSE for low-spin

d^{6}

- 2.4Δ

, so it is more stable than high-spin

d^{6}

with no strings attached."

You must also add the extra pairing cost

P

for the forced pairs; the

- 2.4Δ

is offset by pairing terms, though usually the net still favours low-spin for strong fields.

Find the error: "In tetrahedral geometry the

e

set is raised because it points at ligands."

Tetrahedrally,

e

(

d_{x^{2} - y^{2}}, d_{z^{2}}

) point away from the between-axis ligands and are lowered;

t_{2}

points closer and is raised (Picture 3).

Find the error: "Valence Bond Theory explains complex colour just as well as CFT."

VBT offers no mechanism for absorption at a specific

Δ

; CFT explains colour as a photon of energy

h c / λ = Δ

promoting a

t_{2 g}

electron to

e_{g}

(see the colour equation above) — VBT has no such gap.

Why questions

Why do the five

d

orbitals split into exactly two groups (not three or five) in an octahedron?

The octahedron's symmetry sorts them by geometry (Picture 1): three orbitals lie between axes (identical repulsion →

t_{2 g}

) and two lie along axes (identical repulsion →

e_{g}

Why is

Δ_{t e t}

always smaller than

Δ_{oc t}

Two compounding reasons (Picture 3) — only 4 ligands instead of 6 (factor

4/6

) and no

d

orbital pointing directly at a ligand (factor

2/3

), which multiply to

\frac{4}{9}

Why does the spin-state question only arise for

d^{4}

–

d^{7}

Only there does an electron face a genuine choice between climbing

Δ

or paying

P

; for

d^{1, 2, 3}

the lower set isn't full yet and

d^{8, 9, 10}

have no lower-set room to differ. The resulting unpaired count

n

feeds straight into

μ = n (n + 2)

Why does a bigger

Δ

tend to give a low-spin, diamagnetic-leaning complex?

A large gap makes climbing to

e_{g}

more expensive than pairing (

Δ > P

), so electrons pair in

t_{2 g}

first, minimising unpaired spins.

Why do we say Mn

^{2 +}

and Fe

^{3 +}

(

d^{5}

high-spin) are pale-coloured?

Every one of their five orbitals holds a single electron of the same spin, so any d–d jump would have to flip a spin — that violates the spin selection rule (defined above), making absorption very weak.

Why can CFT predict magnetism that VBT struggled with?

CFT ties the unpaired-electron count

n

directly to the

Δ

P

competition, letting one number decide high- vs low-spin and hence

μ = n (n + 2)

Edge cases

Edge case: What is the CFSE of a

d^{10}

octahedral ion?

Zero — config

t_{2 g}^{6} e_{g}^{4}

gives

6 (- 0.4Δ) + 4 (+ 0.6Δ) = - 2.4Δ + 2.4Δ = 0

(using the Picture 2 offsets); a full shell has no net stabilisation.

Edge case: Does a free (uncomplexed)

d^{5}

gas-phase ion have any splitting?

No. With no ligands the five

d

orbitals are degenerate; splitting requires the surrounding field of ligands.

Edge case: For octahedral

d^{3}

, does changing from a weak to a strong ligand change the number of unpaired electrons?

No. It stays

t_{2 g}^{3}

with 3 unpaired either way; the filling is forced, so only

Δ

's size (and thus colour) changes, not the spin count

n

Edge case: If

Δ_{oc t}

exactly equals

P

, is the complex high- or low-spin?

It sits at the borderline — both configurations have equal energy, so neither is strongly favoured; real ions rarely land precisely here and small perturbations tip the balance.

Edge case: Can a

d^{0}

ion (e.g. Ti

^{4 +}

, Sc

^{3 +}

) show a d–d colour?

No. With no

d

electrons there is nothing to promote across

Δ

, so d–d transitions are impossible and the ion is typically colourless.

Edge case: Is a high-spin

d^{6}

octahedral complex expected to distort as strongly as, say,

d^{9}

No. Its distortion comes from an unequal

t_{2 g}

occupation (

t_{2 g}^{4}

), but

t_{2 g}

orbitals point between ligands so the uneven repulsion is small — a weak Jahn–Teller effect. The classic strong distortion needs uneven

e_{g}

occupation, whose orbitals point straight at ligands (Picture 4).

Recall One-line survival summary

Octahedral: $e_{g}$ up, $t_{2 g}$ down, gap $= Δ_{oc t}$ ; spin choice only for $d^{4}$ – $d^{7}$ . Tetrahedral: order flips, no $g$ , $Δ_{t e t} = \frac{4}{9} Δ_{oc t}$ , always high-spin. Colour = complementary of absorbed d–d transition. See Stability of Complexes & CFSE.