Crystal Field Stabilization Energy (CFSE) — high-spin vs low-spin; spectrochemical series: Question bank

Every item leans on one of four skills: reading the barycentre, weighing $Δ_{o}$ against pairing energy $P$ , knowing which $d$ -counts even have a choice, and remembering that tetrahedral is a different geometry (Crystal Field Theory — Octahedral vs Tetrahedral Splitting).

True or false — justify

d^{3}

octahedral complex can be either high-spin or low-spin depending on the ligand.

False. The first three electrons each take a separate

t_{2 g}

orbital (Hund), so there is no fork;

d^{3}

is high-spin and low-spin at once — the configuration

t_{2 g}^{3}

is forced.

Low-spin complexes always have a larger (more negative) CFSE than high-spin ones for the same

d^{n}

The

Δ_{o}

term is indeed more negative for low-spin, but true stability must subtract the extra pairing cost $+ m P$ ; net stability only favours low-spin when

Δ_{o} > P

, so the raw CFSE is not the whole story.

A tetrahedral

d^{6}

complex is usually low-spin because

d^{6}

can be low-spin.

False. Tetrahedral

Δ_{t} \approx \frac{4}{9} Δ_{o}

is far too small to beat

P

, so tetrahedral complexes are essentially always high-spin — the "

d^{6}

can be low-spin" rule was an octahedral statement.

Splitting the

d

-orbitals raises the total electronic energy of the set.

False. Splitting only redistributes energy about a fixed barycentre; three orbitals fall by

0.4 Δ_{o}

and two rise by

0.6 Δ_{o}

, so an equally-filled set has the same average — no energy is created.

^{-}

produces a large

Δ_{o}

purely because it is a strong electrostatic point charge.

False. F

^{-}

carries the same charge yet sits far lower in the series. CN

^{-}

(and CO) rise high because of ==

π

-backbonding==: empty

π^{*}

orbitals accept metal

t_{2 g}

density, lowering

t_{2 g}

and enlarging

Δ_{o}

(see Ligand Field Theory & π-backbonding).

Going down a group (3d → 4d → 5d) tends to force low-spin behaviour.

True.

Δ_{o}

grows substantially down a group, so heavier congeners nearly always satisfy

Δ_{o} > P

and pair up.

A colourless (or diamagnetic)

d^{6}

complex must be high-spin.

False — it's the opposite for diamagnetism. Diamagnetic

d^{6}

means all six electrons are paired in

t_{2 g}^{6}

, which is the low-spin arrangement (see Magnetic Properties of Coordination Compounds (spin-only formula)).

The

\pm \frac{1}{2} Δ_{o}

split is a fine approximation of the octahedral energies.

False. There are 3 vs 2 orbitals, not 1 vs 1; barycentre balance (

3 x = 2 y

) forces

- 0.4 Δ_{o}

and

+ 0.6 Δ_{o}

, not a symmetric half-and-half.

Spot the error

"For high-spin

d^{6}

, CFSE

= - 0.4 Δ_{o}

and pairing costs nothing because no electrons are forced to pair."

The

- 0.4 Δ_{o}

is right, but

t_{2 g}^{4}

already has one forced pair — compared to the maximally-spread baseline you must add

+ 1 P

. Dropping

P

makes the high-spin/low-spin comparison inconsistent.

[FeF_{6}]^{4 -}

is low-spin because Fe

^{2 +}

d^{6}

and

d^{6}

likes to be low-spin."

^{-}

is a weak-field ligand (small

Δ_{o} < P

), so this complex is high-spin. The

d

-count permits low-spin; the ligand decides it, and F

^{-}

decides against.

"CFSE for octahedral

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

- 2.4 Δ_{o}

, so

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

is stabilized by exactly

2.4 Δ_{o}

The orbital term is

- 2.4 Δ_{o}

, but three extra electron pairs form (

+ 3 P

), so net CFSE is

- 2.4 Δ_{o} + 3 P

. Ignoring

P

overstates the stabilization.

"Tetrahedral splitting has

e

above

t_{2}

, just like octahedral

e_{g}

above

t_{2 g}

The tetrahedral order is inverted: the

e

set lies below

t_{2}

. No ligand points directly at any

d

-orbital in a tetrahedron, which also shrinks the gap to

Δ_{t} \approx \frac{4}{9} Δ_{o}

"Cr

^{3 +}

is inert because it has a small CFSE that keeps electrons loosely held."

Backwards.

d^{3}

^{3 +}

has a large CFSE (

- 1.2 Δ_{o}

) with a half-filled

t_{2 g}

; that stability, not weakness, is what makes it kinetically inert (Stability & Kinetic Inertness (Cr³⁺, Co³⁺)).

"Because splitting saves energy, every complex must be more stable than the free ion by its CFSE."

CFSE measures stabilization relative to the hypothetical spherical (barycentre) field, not to the bare gas-phase ion; it is a comparison within the crystal-field model, not a total bond energy.

Why questions

Why does the high-spin/low-spin choice only arise for octahedral

d^{4}

–

d^{7}

For

d^{1}

–

d^{3}

every electron fits into

t_{2 g}

singly with no upper-vs-pair decision; for

d^{8}

–

d^{10}

the

t_{2 g}

set is already full so extra electrons must enter

e_{g}

. Only

d^{4}

–

d^{7}

face the genuine "climb

e_{g}

or pair in

t_{2 g}

" fork.

Why do we compare

Δ_{o}

with

P

rather than with the total electron energy?

The fourth electron's only decision is a swap of one

e_{g}

occupancy (

+ Δ_{o}

) for one new pair (

+ P

); everything else is identical on both branches, so the tie-breaker is exactly the

Δ_{o}

P

contest.

Why does

π

-backbonding increase

Δ_{o}

rather than decrease it?

It pulls the

t_{2 g}

electrons into overlap with empty ligand

π^{*}

orbitals, lowering the

t_{2 g}

energy. Since

Δ_{o}

is the gap

e_{g} - t_{2 g}

, pushing

t_{2 g}

down widens the gap.

Why do

π

-donor ligands like halides sit at the weak-field end?

Their filled

π

orbitals push electron density into the metal

t_{2 g}

, raising

t_{2 g}

energy and thus shrinking

Δ_{o}

— the opposite of

π

-acceptors.

Why must the up-shift be

0.6 Δ_{o}

while the down-shift is only

0.4 Δ_{o}

Barycentre conservation demands total up = total down; with 2 raised orbitals balancing 3 lowered ones, each of the fewer

e_{g}

orbitals must move farther (

3 \times 0.4 = 2 \times 0.6

Why does a higher metal oxidation state raise

Δ_{o}

A more positive metal draws ligands closer and interacts more strongly with their charge/lone pairs, intensifying the repulsion felt by the

e_{g}

orbitals and thus enlarging the split.

Why can magnetism reveal the spin state without any energy calculation?

The number of unpaired electrons (via the spin-only moment) directly counts how the electrons distributed across

t_{2 g} / e_{g}

; diamagnetic means fully paired (low-spin), strongly paramagnetic means maximally spread (high-spin).

Edge cases

For octahedral

d^{5}

, how do high-spin and low-spin differ in unpaired electrons?

High-spin

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

has five unpaired electrons (a symmetric half-filled set, zero net CFSE before

P

); low-spin

t_{2 g}^{5}

has one unpaired and CFSE

- 2.0 Δ_{o}

. The magnetic moment cleanly distinguishes them.

What is special about high-spin

d^{5}

's CFSE?

It is zero (

- 0.4 \times 3 + 0.6 \times 2 = 0

): the equal, symmetric filling sits exactly at the barycentre. This weak stabilization is why high-spin Mn

^{2 +}

/Fe

^{3 +}

complexes are pale and labile.

Which octahedral

d

-configurations are Jahn–Teller active in the high-spin case, and why?

Uneven occupancy of the

e_{g}

set (e.g. high-spin

d^{4}

e_{g}^{1}

, and

d^{9}

e_{g}^{3}

) is unstable to distortion because the degenerate

e_{g}

orbitals are unequally filled — the geometry distorts to lower energy (see Jahn–Teller Distortion).

Does a

d^{0}

d^{10}

complex have any CFSE?

No — an empty set has nothing to stabilize, and a completely filled

d^{10}

set has equal occupancy across all five orbitals, sitting exactly at the barycentre. Both give CFSE

= 0

For

d^{7}

, why is low-spin only marginally favoured even under strong fields?

Low-spin

d^{7}

t_{2 g}^{6} e_{g}^{1}

— one electron is still forced into the high

e_{g}

orbital, so the extra stabilization over high-spin is modest and

d^{7}

low-spin (e.g. some Co

^{2 +}

) is comparatively rare and often Jahn–Teller distorted.

If two ligands give nearly equal

Δ_{o}

close to

P

, what physical behaviour can appear?

The complex sits at the spin-crossover boundary and can switch between high- and low-spin with temperature or pressure, since the

Δ_{o} \approx P

tie makes both configurations nearly degenerate.

Why is CO able to make even a neutral ligand give a bigger

Δ_{o}

than a charged F

^{-}

Because the splitting isn't purely electrostatic — CO's strong

π

-acceptance (empty

π^{*}

) lowers

t_{2 g}

so effectively that its field strength beats the simple point-charge expectation entirely.

Recall One-line takeaways

The forced-filling counts have no fork ::: $d^{1}$ – $d^{3}$ and $d^{8}$ – $d^{10}$ octahedral. The tie-breaker for spin state ::: compare $Δ_{o}$ against pairing energy $P$ . Tetrahedral spin state ::: essentially always high-spin ( $Δ_{t} \approx \frac{4}{9} Δ_{o} < P$ ). Zero-CFSE octahedral cases ::: $d^{0}$ , high-spin $d^{5}$ , $d^{10}$ .