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

We are going to answer one question: "When six ligands surround a metal ion, exactly how much energy does the ion save by rearranging its electrons — and when does it choose to pair them up?" That saved energy is the CFSE (Crystal Field Stabilization Energy).

Step 1 — Five identical "rooms" for electrons

WHAT: We draw the five $d$ -orbital clouds and note their shapes — three of them (called $d_{x y}, d_{y z}, d_{x z}$ ) point their lobes into the gaps between the coordinate axes; two of them ( $d_{z^{2}}$ and $d_{x^{2} - y^{2}}$ ) point their lobes straight along the axes.

WHY: The whole story hinges on direction. Some clouds aim where ligands will sit; others aim where they won't. Nothing splits until we know who points where.

PICTURE: Look at the two rows below — top row (blue) points along axes, bottom row (green) points between.

Recall Why exactly five?

Question: How many $d$ -orbitals are there and are they equal in a free ion? ::: Five, and in an isolated (free) ion they are all at the same energy — degenerate.

Step 2 — Six ligands arrive along the axes

WHAT: We place six negative point charges (that is Crystal Field Theory's simplification — treat each ligand as a small ball of negative charge) at the ends of the three axes.

WHY: Electrons repel electrons. So a ligand's negative charge pushes up the energy of any $d$ -orbital whose cloud reaches toward it. The closer the cloud points at a ligand, the harder the shove, the higher its energy climbs. This is the only physics we need.

PICTURE: The six orange ligands sit on the axes. The blue clouds ( $d_{z^{2}}, d_{x^{2} - y^{2}}$ ) crash head-on into them; the green clouds tuck safely between them.

Step 3 — The five rooms split into two floors

WHAT: Because the two blue orbitals point straight at ligands, they get shoved up. The three green orbitals point between ligands, so relative to the average they settle down. The five degenerate rooms become two groups.

WHY the labels? $e$ and $t$ are just symmetry names ( $e$ = a doubly-degenerate set of 2, $t$ = a triply-degenerate set of 3); the subscripts are group-theory bookkeeping. You only need: $e_{g}$ is the upstairs pair, $t_{2 g}$ is the downstairs trio, and $Δ_{o}$ is the height between floors.

PICTURE: The flat line of five (left) splits into an upstairs pair and a downstairs trio (right). The vertical arrow is $Δ_{o}$ .

For the full geometry of octahedral vs tetrahedral fields see Crystal Field Theory — Octahedral vs Tetrahedral Splitting.

Step 4 — Where exactly are the two floors? (the 0.4 / 0.6 rule)

Here is the beautiful part. We don't guess the floor heights — a conservation law fixes them.

Let each $t_{2 g}$ orbital drop by $x$ below the barycentre, and each $e_{g}$ orbital rise by $y$ above it.

Equation 1 — the two floors are $Δ_{o}$ apart: $\underbrace{y}_{\text{rise of each }e_g} + \underbrace{x}_{\text{drop of each }t_{2g}} = \underbrace{\Delta_o}_{\text{total gap}} \tag{1}$ (The gap from the top floor to the bottom floor is "how far up the $e_{g}$ went" plus "how far down the $t_{2 g}$ went".)

Equation 2 — the see-saw balances: $\underbrace{3x}_{\substack{3 \text{ orbitals} \\ \text{each drop } x}} \;=\; \underbrace{2y}_{\substack{2 \text{ orbitals} \\ \text{each rise } y}} \tag{2}$ (Total downward shift must equal total upward shift, or the average would move.)

WHAT we do now — solve the two equations. From (2), $y = \frac{3}{2} x$ . Put that into (1): $x + \frac{3}{2} x = Δ_{o} \Rightarrow \frac{5}{2} x = Δ_{o} \Rightarrow x = \frac{2}{5} Δ_{o} = 0.4 Δ_{o}, y = \frac{3}{5} Δ_{o} = 0.6 Δ_{o}$

WHY this matters: every CFSE number you will ever compute is just counting electrons on each floor and multiplying by these two heights.

PICTURE: the see-saw. Three green orbitals hang $0.4 Δ_{o}$ below the balance line; two blue orbitals sit $0.6 Δ_{o}$ above it; the dashed line is the barycentre.

Step 5 — Filling electrons: the fork in the road

Now drop electrons in, lowest floor first (nature is lazy — it minimises energy). For $d^{1}, d^{2}, d^{3}$ there is no drama: one electron per downstairs room, all unpaired (this is Hund's rule — spread out before pairing).

The fork appears at the 4th electron. Downstairs ( $t_{2 g}$ ) already has one electron in each of its three rooms. The 4th electron must choose:

WHY only $d^{4}$ – $d^{7}$ ? For $d^{1}$ – $d^{3}$ the downstairs isn't full yet, so climbing never tempts anyone. For $d^{8}$ – $d^{10}$ every arrangement forces the same pairs. Only in the middle band does the choice bite.

PICTURE: the same 4th electron, two destinies. Left: tall $Δ_{o}$ → it pairs (low-spin). Right: short $Δ_{o}$ → it climbs (high-spin).

Step 6 — Worked fork: $d^{6}$ (Fe $^{2 +}$ ), both roads costed

We count electrons on each floor and plug into the master formula from the parent: $CFSE = [- 0.4 n_{t_{2 g}} + 0.6 n_{e_{g}}] Δ_{o} + m P$ where $n_{t_{2 g}}, n_{e_{g}}$ are electron counts on each floor and $m$ = number of extra pairs the field forces, measured relative to the high-spin (maximally-spread) arrangement — that spread arrangement is our zero-of-pairing baseline.

High-spin $[FeF_{6}]^{4 -}$ : $t_{2 g}^{4} e_{g}^{2}$ (spread as much as possible → $m = 0$ ). $CFSE = (4 down - 0.4 \times 4 + 2 up 0.6 \times 2) Δ_{o} + 0 \cdot P = (- 1.6 + 1.2) Δ_{o} = - 0.4 Δ_{o}$

Low-spin $[Fe(CN)_{6}]^{4 -}$ : $t_{2 g}^{6} e_{g}^{0}$ (all crammed downstairs → $m = 2$ ). $CFSE = (6 down - 0.4 \times 6 + 0) Δ_{o} + 2 P = - 2.4 Δ_{o} + 2 P$

WHY the $+ 2 P$ : low-spin's $- 2.4 Δ_{o}$ looks far more stable, but it forced 2 extra pairs beyond high-spin. Low-spin truly wins only when $- 2.4 Δ_{o} + 2 P < - 0.4 Δ_{o} ⟺ Δ_{o} > P .$ That reproduces the Step-5 rule exactly. ✓

PICTURE: both $d^{6}$ diagrams side by side with their CFSE tallies underneath.

Which road nature took is measurable: a diamagnetic (no unpaired electrons) $d^{6}$ complex like $[Co(NH_{3})_{6}]^{3 +}$ must be low-spin — see Magnetic Properties of Coordination Compounds (spin-only formula).

Step 7 — Every $d^{n}$ case, including the tricky ones

Intuition The full classification (no gaps)

$d^{0}$ : floors empty → $CFSE = 0$ . No choice.
$d^{1}, d^{2}, d^{3}$ : electrons go one-per-room downstairs; high-spin = low-spin. $d^{3}$ (Cr $^{3 +}$ ) reaches $t_{2 g}^{3}$ with CFSE $= - 0.4 \times 3 Δ_{o} = - 1.2 Δ_{o}$ — a deep well that helps make Cr $^{3 +}$ so common and inert (see Stability & Kinetic Inertness (Cr³⁺, Co³⁺)).
$d^{4}$ – $d^{7}$ : the real fork — high-spin or low-spin depending on $Δ_{o}$ vs $P$ (Steps 5–6).
$d^{8}$ : $t_{2 g}^{6} e_{g}^{2}$ forced either way; no ambiguity. CFSE $= (- 0.4 \times 6 + 0.6 \times 2) Δ_{o} = - 1.2 Δ_{o}$ .
$d^{9}$ : $t_{2 g}^{6} e_{g}^{3}$ forced. CFSE $= (- 0.4 \times 6 + 0.6 \times 3) Δ_{o} = - 0.6 Δ_{o}$ . But the lone "hole" in the $e_{g}$ pair makes the two $e_{g}$ orbitals unequally occupied — an unstable, lopsided situation the ion relieves by stretching or squashing the octahedron. This is Jahn–Teller distortion (classic for Cu $^{2 +}$ , a $d^{9}$ ion); see Jahn–Teller Distortion.
$d^{10}$ : every room full → $CFSE = 0$ . No choice.

PICTURE: the $d^{3}$ forced case, the octahedral gap, the inverted tetrahedral gap, and the lopsided $d^{9}$ that triggers Jahn–Teller.

The one-picture summary

Everything above, compressed: five degenerate orbitals → the octahedral shove → the $0.4/0.6$ see-saw → fill electrons → the $Δ_{o}$ -vs- $P$ fork → high-spin or low-spin.

Recall Feynman: tell the whole walkthrough to a friend

Five kids sit on identical chairs — that's the five equal $d$ -orbitals. Six grumpy neighbours (ligands) march up along the six directions and shove. Two kids whose chairs face the neighbours get pushed onto tall stools (the $e_{g}$ pair); three kids sitting in the gaps sink into low comfy cushions (the $t_{2 g}$ trio). Because nobody was added or removed, the see-saw must balance: three kids drop by $0.4$ for every two who rise by $0.6$ . Now fill the room from the bottom. All goes smoothly until a kid must choose between climbing a lonely tall stool (cost $Δ_{o}$ ) or squeezing onto a cushion next to someone (cost $P$ , the annoyance of sharing — real electrons hate sharing a tiny room). Short stool → she climbs (high-spin, spread out); tall stool → she squeezes (low-spin, all paired). Bossy neighbours like CO build very tall stools, so everyone squeezes. Count who ended up on cushions ( $- 0.4$ each) versus stools ( $+ 0.6$ each), subtract the annoyance of any extra squeezes beyond the spread-out arrangement — and that final number is the CFSE. And if one upstairs stool ends up half-shared (like $d^{9}$ copper), the whole see-saw tilts sideways to get comfortable: that tilt is Jahn–Teller.

Flashcards

Why do

e_{g}

orbitals rise while

t_{2 g}

fall in an octahedral field?

e_{g}

(

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

) point straight at the axial ligands and are repelled up;

t_{2 g}

(

d_{x y}, d_{y z}, d_{x z}

) point between ligands and settle down.

Using the barycentre rule, derive the

t_{2 g}

and

e_{g}

energies.

Let each

t_{2 g}

drop

x

, each

e_{g}

rise

y

. Gap:

x + y = Δ_{o}

; balance:

3 x = 2 y

. Solve →

x = 0.4 Δ_{o}

y = 0.6 Δ_{o}

, so

E (t_{2 g}) = - 0.4 Δ_{o}

E (e_{g}) = + 0.6 Δ_{o}

When is a

d^{4}

–

d^{7}

octahedral complex low-spin?

When

Δ_{o} > P

(splitting exceeds pairing energy) — electrons pair downstairs rather than climb to

e_{g}

CFSE of high-spin vs low-spin

d^{6}

(relative to the high-spin pairing baseline)?

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

- 0.4 Δ_{o}

(

m = 0

). LS

t_{2 g}^{6}

- 2.4 Δ_{o} + 2 P

(

m = 2

). LS wins when

Δ_{o} > P

Which

d^{n}

case triggers Jahn–Teller distortion and why?

d^{9}

(e.g. Cu

^{2 +}

): the single hole in the

e_{g}

pair makes those two orbitals unequally occupied, so the octahedron stretches/squashes to lower energy.

Why are tetrahedral complexes almost always high-spin?

Splitting is inverted and small,

Δ_{t} \approx \frac{4}{9} Δ_{o}

, so

Δ_{t}

is essentially never bigger than

P

— pairing is never worth it.