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

WHAT is happening (the model)

In an octahedral field (6 ligands along $\pm x, \pm y, \pm z$ ):

Set	Orbitals	Point...	Energy
$e_g$	$d_{z^2},\, d_{x^2-y^2}$	at ligands	raised
$t_{2g}$	$d_{xy},\, d_{yz},\, d_{xz}$	between ligands	lowered

The energy gap between them is $==\Delta_o==$ (octahedral splitting parameter, "10 Dq").

HOW to derive the orbital energies (first principles)

Let the down-shift of each $t_{2g}$ orbital be $x$ (below barycentre) and the up-shift of each $e_g$ orbital be $y$ (above barycentre). By definition of the gap:

$y - (-x) = \Delta_o \quad\Rightarrow\quad x + y = \Delta_o \tag{1}$

Barycentre conservation: total down = total up.

$3x = 2y \tag{2}$

Why this step? 3 orbitals drop by $x$ each, 2 orbitals rise by $y$ each; for the centre of gravity to be unchanged the two must balance.

Solve (1) and (2): from (2), $y = \tfrac{3}{2}x$ . Substitute into (1):

$x + \tfrac{3}{2}x = \Delta_o \Rightarrow \tfrac{5}{2}x = \Delta_o \Rightarrow x = \tfrac{2}{5}\Delta_o,\quad y = \tfrac{3}{5}\Delta_o$

High-spin vs Low-spin: the competition

This choice only matters for $d^4$ – $d^7$ (octahedral). For $d^1$ – $d^3$ and $d^8$ – $d^{10}$ the filling is forced — no ambiguity.

Worked example

$d^6$ — Fe $^{2+}$ , the classic split decision High-spin (weak field, e.g. $[\text{FeF}_6]^{4-}$ ): $t_{2g}^4 e_g^2$ $\text{CFSE} = (-0.4\times4 + 0.6\times2)\Delta_o = (-1.6+1.2)\Delta_o = -0.4\,\Delta_o$ Extra pairs vs free ion = 1, so add $+1P$ → CFSE $= -0.4\Delta_o + P$ .

Low-spin (strong field, e.g. $[\text{Fe(CN)}_6]^{4-}$ ): $t_{2g}^6 e_g^0$ $\text{CFSE} = (-0.4\times6 + 0)\Delta_o = -2.4\,\Delta_o$ Extra pairs vs free ion = 3, so add $+3P$ → CFSE $= -2.4\Delta_o + 3P$ .

Why this step? The free $d^6$ ion already has 1 forced pair; we only count additional pairs created by the field, since $P$ for the unavoidable pair is the same on both sides. Low-spin is favoured when $-2.4\Delta_o+3P < -0.4\Delta_o + P$ , i.e. when $\Delta_o > P$ . ✓ consistent.

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

The Spectrochemical Series (what sets $\Delta_o$ ?)

$\Delta_o$ also grows with higher oxidation state and going down a group (3d < 4d < 5d), so 4d/5d complexes are almost always low-spin.

Common mistake Steel-manning the common errors

(1) "Low-spin always has bigger CFSE so it's always more stable." Why it feels right: low-spin packs electrons into the deep $t_{2g}$ , giving a big negative $\Delta_o$ term. The fix: you must subtract the pairing-energy cost $+mP$ . Net stability only wins if $\Delta_o > P$ . Forget $P$ and you'll "predict" low-spin for every fluoride complex — wrong.

(2) Forgetting the barycentre and using $\pm\frac{1}{2}\Delta_o$ . Why it feels right: two sets, so split symmetrically by half each. The fix: there are 3 vs 2 orbitals, not 1 vs 1. Balance forces $-0.4$ and $+0.6$ , not $-0.5/+0.5$ .

(3) Counting pairing energy for electrons that were already paired in the free ion. The fix: only count extra pairs the ligand field forces. For $d^6$ HS, $m=1$ (not 0), because forming $t_{2g}^4$ requires one pair beyond the 0 pairs of a hypothetical fully-spread arrangement — but compare consistently to the free-ion baseline.

(4) Using octahedral numbers for tetrahedral. The fix: tetrahedral splitting is inverted ( $e$ below $t_2$ ) and $\Delta_t \approx \tfrac{4}{9}\Delta_o$ , so tetrahedral complexes are essentially always high-spin.

Recall Feynman: explain it to a 12-year-old

Imagine five kids on identical chairs (the $d$ -orbitals). Now six grumpy neighbours (ligands) come and shove. Two kids whose chairs face the neighbours get pushed up onto tall stools (uncomfortable, $e_g$ ); three kids who sit between the neighbours sink into comfy low cushions ( $t_{2g}$ ). Kids prefer comfy cushions, so the group is happier (more stable) than before — that "happiness saved" is the CFSE. Now: a sixth kid arrives. Does she squeeze onto a comfy cushion next to someone (annoying = pairing energy $P$ ) or climb a lonely tall stool (height cost = $\Delta_o$ )? If the stool is too high she squeezes (low-spin); if the stool is low she climbs (high-spin). Bossy neighbours like CO make very tall stools → everyone squeezes.

Flashcards

What does CFSE stand for and what does it measure?

Crystal Field Stabilization Energy; the net energy a complex is stabilized by because

d

-electrons preferentially occupy the lower-energy split orbitals (relative to the unsplit barycentre).

In an octahedral field, which orbitals form

t_{2g}

and which form

e_g

t_{2g} = d_{xy},d_{yz},d_{xz}

(point between ligands, lower);

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

(point at ligands, higher).

Derive the energies of

t_{2g}

and

e_g

relative to barycentre.

Let down-shift

x

, up-shift

y

. Gap:

x+y=\Delta_o

; barycentre:

3x=2y

. Solve →

x=0.4\Delta_o

y=0.6\Delta_o

. So

t_{2g}=-0.4\Delta_o

e_g=+0.6\Delta_o

What determines high-spin vs low-spin?

Compare

\Delta_o

with pairing energy

P

\Delta_o<P

→ high-spin (weak field);

\Delta_o>P

→ low-spin (strong field).

For which

d^n

counts does the HS/LS distinction exist (octahedral)?

d^4

through

d^7

. (

d^1

–

d^3

d^8

–

d^{10}

are forced.)

CFSE of high-spin

d^6

(ignoring

P

t_{2g}^4e_g^2

(-0.4\times4+0.6\times2)\Delta_o=-0.4\Delta_o

CFSE of low-spin

d^6

(ignoring

P

t_{2g}^6

-0.4\times6\,\Delta_o=-2.4\Delta_o

Why does

[\text{Co(NH}_3)_6]^{3+}

being diamagnetic prove low-spin?

Diamagnetic = no unpaired electrons =

t_{2g}^6e_g^0

, which requires

\Delta_o>P

(low-spin).

Order three of these by field strength: CN⁻, H₂O, I⁻.

I⁻ < H₂O < CN⁻ (weak → strong).

Why is CO a stronger-field ligand than F⁻ despite being neutral?

CO has empty π* orbitals that accept metal

t_{2g}

density (π-backbonding), lowering

t_{2g}

and increasing

\Delta_o

; F⁻ is a π-donor that does the opposite.

How does

\Delta_t

compare to

\Delta_o

, and why are tetrahedral complexes high-spin?

\Delta_t\approx\tfrac{4}{9}\Delta_o

; it's so small that

\Delta_t<P

almost always, so electrons spread out (high-spin).

How does oxidation state and metal row affect

\Delta_o

\Delta_o

increases with higher oxidation state and down a group (3d<4d<5d), so 4d/5d complexes are usually low-spin.

Connections

Crystal Field Theory — Octahedral vs Tetrahedral Splitting
Magnetic Properties of Coordination Compounds (spin-only formula)
Colour of Transition-Metal Complexes (d-d transitions, $\Delta_o$ and λ)
Jahn–Teller Distortion
Pairing Energy and Hund's Rule
Ligand Field Theory & π-backbonding
Stability & Kinetic Inertness (Cr³⁺, Co³⁺)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab ek metal ion ke paas 6 ligands aate hain (octahedral), to woh negative point charges ki tarah metal ke $d$ -electrons ko repel karte hain. Jo orbitals seedhe ligand ki taraf point karte hain ( $d_{z^2}, d_{x^2-y^2}$ = $e_g$ ) unki energy badh jaati hai, aur jo beech mein hote hain ( $d_{xy}, d_{yz}, d_{xz}$ = $t_{2g}$ ) unki energy ghat jaati hai. Is gap ko hum $\Delta_o$ bolte hain. Ab kyunki electrons neeche wale comfy $t_{2g}$ orbitals ko prefer karte hain, complex thoda zyada stable ho jaata hai — yahi extra stability CFSE hai. Numbers yaad rakho: $t_{2g}$ neeche $-0.4\Delta_o$ , $e_g$ upar $+0.6\Delta_o$ (barycentre balance se aate hain, kyunki 3 orbital neeche, 2 upar).

High-spin vs low-spin ka pura khel ek choice hai. Jab 4th electron daalna ho, to do raaste: ya to upar wale $e_g$ mein jao (cost $=\Delta_o$ ), ya neeche pair bana lo (cost $=P$ , pairing energy). Agar $\Delta_o$ chhota hai (weak ligand jaise F⁻, Cl⁻, H₂O) to electron upar spread ho jaata hai → high-spin. Agar $\Delta_o$ bada hai (strong ligand jaise CN⁻, CO, NH₃) to pair ban jaata hai → low-spin. Yeh distinction sirf $d^4$ se $d^7$ tak matter karta hai.

Spectrochemical series basically ligands ko unke field strength ke order mein lagaata hai: weak (I⁻, Br⁻, Cl⁻, F⁻, H₂O) se strong (NH₃, en, NO₂⁻, CN⁻, CO) tak. CO aur CN⁻ itne strong isliye hain kyunki unme empty π* orbitals hote hain jo metal se electron density wapas le lete hain (π-backbonding), jisse $t_{2g}$ aur neeche chala jaata hai aur $\Delta_o$ bada ho jaata hai. Exam tip: agar complex diamagnetic hai (saare electron paired) to samajh jao low-spin,

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

WHAT is happening (the model)

HOW to derive the orbital energies (first principles)

High-spin vs Low-spin: the competition

The Spectrochemical Series (what sets $\Delta_o$ ?)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Coordination Chemistry

Connections