Coordination Chemistry — CFSE & Crystal Field Theory

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Crystal field theory answers three exam questions with one picture: why a complex has the colour it has, how many unpaired electrons it carries, and how much extra stability its d electrons gain. Learn the splitting diagram once and the rest is counting.

What you will be able to do

• Draw the octahedral, tetrahedral and square planar splitting diagrams and label the orbital sets
• Use the spectrochemical series and Δ versus P to decide high spin or low spin
• Calculate CFSE for any dⁿ configuration in octahedral and tetrahedral fields
• Predict the number of unpaired electrons and the spin-only magnetic moment of a complex
• Explain colour from d–d transitions and calculate Δ from the absorbed wavelength

The big picture

One splitting diagram, three answers.

In one line: Ligands push the five d orbitals apart in energy; how far apart (Δ) compared with the pairing energy (P) decides where the electrons go, and that decides colour, magnetism and CFSE.

• Crystal field theory (CFT) treats ligands as point negative charges (or dipoles) and the metal–ligand bond as purely electrostatic.
• In a free ion the five d orbitals have the same energy (degenerate). Ligands approaching along particular directions repel some d orbitals more than others, and the set splits.
• The size of the split, Δ, depends on the ligand, the metal and its oxidation state, and the geometry.

Think of it like this: Picture five friends sitting in a room. Six people walk in and stand along the x, y and z axes. The two friends sitting on those axes (dz² and dx²−y²) are crowded and uncomfortable — higher energy. The three sitting between the axes (dxy, dyz, dxz) are left in peace — lower energy.

Octahedral splitting

Six ligands along ±x, ±y, ±z.

Set	Orbitals	Points	Energy relative to the average
eg	dz², dx²−y²	Directly at the ligands	+0.6 Δo (raised)
t₂g	dxy, dyz, dxz	Between the ligands	−0.4 Δo (lowered)

Key idea: The barycentre rule: the average energy is unchanged, so 2 × (+0.6) + 3 × (−0.4) = 0. Each electron in t₂g stabilises by 0.4 Δo; each in eg destabilises by 0.6 Δo.

Spectrochemical series

Ligands ranked by how hard they split the d orbitals.

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO

Weak field (small Δ) on the left → strong field (large Δ) on the right

• Halides are weak field; CN⁻ and CO are the strongest (they also accept electron density back from the metal).
• Oxidation state: a higher charge on the metal pulls ligands closer → larger Δ. Δo for M³⁺ is roughly 1.5–2 times that for M²⁺.
• Down a group: Δ increases about 30–50% from 3d to 4d to 5d. Complexes of 4d and 5d metals are almost always low spin.
• SCN⁻ binding through S is weak field; NCS⁻ binding through N is stronger.

Memory hook: “I Bring Some Cold Sandwiches For Our Christmas Holiday, Nobody Eats Nuts Except Cranky Cooks” — I⁻, Br⁻, SCN⁻, Cl⁻, S²⁻, F⁻, OH⁻, C₂O₄²⁻, H₂O, NCS⁻, edta, NH₃, en, CN⁻, CO.

High spin and low spin

Pair up, or move upstairs?

For d¹, d², d³ the electrons simply go into t₂g. For d⁴ to d⁷ there is a choice for the next electron: pay the pairing energy P and stay in t₂g, or pay Δo and move up to eg.

Condition	Ligands	Result	Name
Δo < P	Weak field	Electrons spread out into eg before pairing	High spin (maximum unpaired)
Δo > P	Strong field	Electrons pair in t₂g before entering eg	Low spin (minimum unpaired)

dⁿ	High spin (weak field)	Unpaired	Low spin (strong field)	Unpaired
d⁴	t₂g³ eg¹	4	t₂g⁴	2
d⁵	t₂g³ eg²	5	t₂g⁵	1
d⁶	t₂g⁴ eg²	4	t₂g⁶	0
d⁷	t₂g⁵ eg²	3	t₂g⁶ eg¹	1

Key idea: Only d⁴–d⁷ octahedral complexes have a high-spin/low-spin choice. d¹–d³ and d⁸–d¹⁰ have one arrangement whatever the ligand.

Solved example 01 — Fe²⁺ with two ligands

Compare [Fe(H₂O)₆]²⁺ and [Fe(CN)₆]⁴⁻: configuration and unpaired electrons.

1. Fe²⁺ is d⁶ in both.
2. H₂O is weak field → high spin: t₂g⁴ eg², four unpaired.
3. CN⁻ is strong field → low spin: t₂g⁶, all paired.

Answer: [Fe(H₂O)₆]²⁺: 4 unpaired, paramagnetic. [Fe(CN)₆]⁴⁻: 0 unpaired, diamagnetic.

Calculating CFSE

Count the electrons in each set, multiply, add.

CFSE (octahedral) = (−0.4 x + 0.6 y) Δo + (extra pairs) P

x = electrons in t₂g; y = electrons in eg

Key idea: Convention on P: here P is added only for each pair the strong field forces beyond the high-spin arrangement. Some books add P for every pair in the complex. Both give the same Δo part — check which convention your exam uses before writing the P term.

dⁿ	High spin CFSE	Low spin CFSE
d¹	−0.4 Δo	—
d²	−0.8 Δo	—
d³	−1.2 Δo	—
d⁴	−0.6 Δo	−1.6 Δo + P
d⁵	0	−2.0 Δo + 2P
d⁶	−0.4 Δo	−2.4 Δo + 2P
d⁷	−0.8 Δo	−1.8 Δo + P
d⁸	−1.2 Δo	—
d⁹	−0.6 Δo	—
d¹⁰	0	—

Solved example 02 — CFSE of a d⁶ low-spin complex

Find the CFSE of [Co(NH₃)₆]³⁺.

1. Co³⁺ is d⁶; NH₃ with a 3+ ion is strong enough → low spin: t₂g⁶ eg⁰.
2. Δo part: 6 × (−0.4) = −2.4 Δo.
3. High-spin d⁶ has 1 pair; low-spin has 3 → 2 extra pairs.

Answer: CFSE = −2.4 Δo + 2P; diamagnetic.

Solved example 03 — CFSE of a d⁵ high-spin complex

Find the CFSE of [Mn(H₂O)₆]²⁺.

1. Mn²⁺ is d⁵; H₂O is weak field → high spin: t₂g³ eg².
2. 3 × (−0.4) + 2 × (+0.6) = −1.2 + 1.2.

Answer: CFSE = 0 — five unpaired electrons, no field stabilisation.

Check yourself

• C1. How many unpaired electrons are in [CoF₆]³⁻?
• C2. Write the CFSE of a d⁸ octahedral complex.

Answers at the end.

Tetrahedral and square planar

Different shapes, different splitting.

	Octahedral	Tetrahedral	Square planar
Lower set	t₂g (3 orbitals)	e (dz², dx²−y²)	dxz, dyz lowest; then dz², dxy
Upper set	eg (2 orbitals)	t₂ (dxy, dyz, dxz)	dx²−y² far above all others
Splitting	Δo	Δt = (4/9) Δo	Largest overall
Spin	Depends on ligand	Almost always high spin	Usually low spin (d⁸)

CFSE (tetrahedral) = (−0.6 x + 0.4 y) Δt

x = electrons in e (lower); y = electrons in t₂ (upper)

• The tetrahedral pattern is inverted compared with octahedral, and there is no g label (no centre of symmetry).
• Δt is small (four ligands, none pointing straight at an orbital), so it is almost always smaller than P → high spin.
• Square planar is typical of d⁸ ions with strong-field ligands — Ni²⁺, Pd²⁺, Pt²⁺: [Ni(CN)₄]²⁻, [PtCl₄]²⁻, cisplatin. The eight electrons fill the four lower orbitals and leave dx²−y² empty → diamagnetic.

Solved example 04 — Two Ni²⁺ complexes

[NiCl₄]²⁻ is paramagnetic but [Ni(CN)₄]²⁻ is diamagnetic. Explain.

1. Ni²⁺ is d⁸.
2. Cl⁻ is weak field → tetrahedral: e⁴ t₂⁴ → two unpaired electrons.
3. CN⁻ is strong field → square planar: all eight paired below the empty dx²−y².

Answer: Tetrahedral [NiCl₄]²⁻ has 2 unpaired; square planar [Ni(CN)₄]²⁻ has none.

Magnetism

Unpaired electrons, counted from the splitting diagram.

μ (spin-only) = √[n(n + 2)] BM

n = unpaired electrons: 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92 BM

Complex	Metal ion	Field	Arrangement	Unpaired	μ / BM
[Fe(H₂O)₆]³⁺	Fe³⁺ d⁵	Weak	t₂g³ eg²	5	5.92
[Fe(CN)₆]³⁻	Fe³⁺ d⁵	Strong	t₂g⁵	1	1.73
[CoF₆]³⁻	Co³⁺ d⁶	Weak	t₂g⁴ eg²	4	4.90
[Co(NH₃)₆]³⁺	Co³⁺ d⁶	Strong	t₂g⁶	0	0
[Cr(H₂O)₆]³⁺	Cr³⁺ d³	Either	t₂g³	3	3.87
[Cu(NH₃)₄]²⁺	Cu²⁺ d⁹	—	one unpaired	1	1.73

Key idea: Working back from μ: if a question gives μ, find n first (4.90 BM → 4 unpaired), then decide whether the complex must be high or low spin. This is the fastest way to identify the field strength in assertion–reason questions.

Colour

What we see is what is not absorbed.

An electron absorbs a photon of energy Δo and jumps from t₂g to eg — a d–d transition. The complex shows the complementary colour of the light absorbed.

Δo = hc / λ (per ion); multiply by Nₐ for per mole

A larger Δo absorbs shorter wavelengths

• No d–d transition, no colour: d⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Zn²⁺, Cu⁺) complexes are colourless.
• Changing the ligand changes Δo and so the colour: [Ni(H₂O)₆]²⁺ is green; [Ni(en)₃]²⁺ is violet (stronger field, larger Δo, absorbs shorter wavelength).
• Anhydrous CuSO₄ is white; with water ligands it becomes blue — no ligand field, no splitting, no colour.

Solved example 05 — Δo from absorption

[Ti(H₂O)₆]³⁺ absorbs most strongly at 498 nm. Find Δo in kJ mol⁻¹. (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, Nₐ = 6.022 × 10²³)

1. Per ion: E = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (498 × 10⁻⁹) = 3.99 × 10⁻¹⁹ J.
2. Per mole: 3.99 × 10⁻¹⁹ × 6.022 × 10²³ = 2.40 × 10⁵ J.

Answer: Δo ≈ 240 kJ mol⁻¹. It absorbs blue-green light and looks purple.

Check yourself

• C3. Why is [Sc(H₂O)₆]³⁺ colourless?
• C4. Which absorbs at a shorter wavelength: [Co(NH₃)₆]³⁺ or [CoF₆]³⁻?

Answers at the end.

Limits of crystal field theory

Know what the model cannot do.

• It treats ligands as point charges, so it cannot explain why neutral CO is a stronger ligand than charged F⁻ or OH⁻.
• It ignores covalent character in metal–ligand bonds; ligand field theory and MO theory correct this.
• Jahn–Teller distortion: octahedral complexes with unequal filling of eg — high-spin d⁴ and d⁹ (Cu²⁺) — distort, usually by lengthening the two axial bonds. This is why Cu²⁺ complexes are rarely perfect octahedra.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	In an octahedral field, dxy, dyz, dxz are raised.	They form the lower t₂g set; dz² and dx²−y² (eg) are raised.	eg orbitals point straight at the ligands.
2	Tetrahedral complexes are low spin with strong-field ligands.	Tetrahedral complexes are almost always high spin.	Δt = 4/9 Δo is usually smaller than P.
3	d³ complexes can be high or low spin.	Only d⁴–d⁷ octahedral complexes have the choice.	Three electrons fit singly in t₂g either way.
4	CFSE of high-spin d⁵ is −2.0 Δo.	High-spin d⁵ is 0; low-spin d⁵ is −2.0 Δo + 2P.	3(−0.4) + 2(+0.6) = 0.
5	Tetrahedral CFSE uses −0.4 and +0.6.	Tetrahedral uses −0.6 (e) and +0.4 (t₂) with Δt.	The pattern is inverted.
6	F⁻ is a strong-field ligand because F is very electronegative.	F⁻ is weak field; CN⁻ and CO are strong.	Use the spectrochemical series, not electronegativity.
7	[Ni(CN)₄]²⁻ is tetrahedral and paramagnetic.	It is square planar and diamagnetic.	Strong-field d⁸ → square planar.
8	A complex absorbing red light looks red.	It looks the complementary colour (green-blue).	We see the light that is not absorbed.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. In an octahedral field the eg set consists of: (Easy)

• (a) dxy, dyz
• (b) dz², dx²−y²
• (c) dxy, dxz, dyz
• (d) dxz, dz²

Q02. The strongest-field ligand among these is: (Easy)

• (a) Cl⁻
• (b) H₂O
• (c) NH₃
• (d) CN⁻

Q03. Δt in terms of Δo is: (Easy)

• (a) (4/9) Δo
• (b) (9/4) Δo
• (c) (1/2) Δo
• (d) 2 Δo

Q04. Which complex is colourless? (Easy)

• (a) [Ti(H₂O)₆]³⁺
• (b) [Cu(H₂O)₆]²⁺
• (c) [Zn(H₂O)₆]²⁺
• (d) [Ni(H₂O)₆]²⁺

Q05. Each electron in t₂g of an octahedral complex contributes: (Easy)

• (a) +0.6 Δo
• (b) −0.4 Δo
• (c) −0.6 Δo
• (d) +0.4 Δo

Q06. The number of unpaired electrons in [Fe(CN)₆]³⁻ is: (Moderate)

• (a) 5
• (b) 3
• (c) 1
• (d) 0

Q07. CFSE of high-spin d⁴ octahedral is: (Moderate)

• (a) −1.6 Δo
• (b) −0.6 Δo
• (c) −0.4 Δo
• (d) 0

Q08. [CoF₆]³⁻ is: (Moderate)

• (a) diamagnetic, low spin
• (b) paramagnetic, high spin
• (c) diamagnetic, high spin
• (d) paramagnetic, low spin

Q09. Which has the largest Δo? (Moderate)

• (a) [Co(H₂O)₆]²⁺
• (b) [Co(H₂O)₆]³⁺
• (c) [Rh(H₂O)₆]³⁺
• (d) [Co(NH₃)₆]²⁺

Q10. The geometry of [Ni(CN)₄]²⁻ is: (Moderate)

• (a) tetrahedral
• (b) square planar
• (c) octahedral
• (d) trigonal bipyramidal

Q11. Spin-only μ of [Fe(H₂O)₆]²⁺ is about: (Moderate)

• (a) 0
• (b) 2.83 BM
• (c) 4.90 BM
• (d) 5.92 BM

Q12. CFSE of tetrahedral d² (in Δt) is: (Moderate)

• (a) −0.8 Δt
• (b) −1.2 Δt
• (c) −0.6 Δt
• (d) −0.4 Δt

Q13. Which shows Jahn–Teller distortion? (Tough)

• (a) [Cr(H₂O)₆]³⁺
• (b) [Cu(H₂O)₆]²⁺
• (c) [Fe(CN)₆]⁴⁻
• (d) [Zn(NH₃)₆]²⁺

Q14. [Ni(H₂O)₆]²⁺ is green; [Ni(en)₃]²⁺ is violet. This means en: (Tough)

• (a) is a weaker field ligand
• (b) gives a larger Δo
• (c) gives no d–d transition
• (d) makes the complex tetrahedral

Q15. CFSE of low-spin d⁶ octahedral (counting extra pairs only) is: (Tough)

• (a) −2.4 Δo + 2P
• (b) −0.4 Δo
• (c) −2.4 Δo + 3P
• (d) −1.8 Δo + P

Answer key with explanations

1. (b) dz², dx²−y² — They point at the ligands.
2. (d) CN⁻ — Near the end of the series.
3. (a) (4/9) Δo — Fewer ligands, none pointing at orbitals.
4. (c) [Zn(H₂O)₆]²⁺ — Zn²⁺ is d¹⁰.
5. (b) −0.4 Δo — Lower set.
6. (c) 1 — Low-spin d⁵: t₂g⁵.
7. (b) −0.6 Δo — t₂g³ eg¹: −1.2 + 0.6.
8. (b) paramagnetic, high spin — F⁻ weak field; d⁶ high spin, 4 unpaired.
9. (c) [Rh(H₂O)₆]³⁺ — Higher charge and 4d metal.
10. (b) square planar — Strong-field d⁸.
11. (c) 4.90 BM — High-spin d⁶: 4 unpaired.
12. (b) −1.2 Δt — e²: 2 × (−0.6).
13. (b) [Cu(H₂O)₆]²⁺ — d⁹: unequal eg filling.
14. (b) gives a larger Δo — Absorbs shorter wavelength.
15. (a) −2.4 Δo + 2P — t₂g⁶; two more pairs than high spin.

Check-yourself answers

• C1. Four (Co³⁺ d⁶, F⁻ weak field → t₂g⁴ eg²).
• C2. −1.2 Δo (t₂g⁶ eg²: −2.4 + 1.2).
• C3. Sc³⁺ is d⁰ — no d electron, so no d–d transition.
• C4. [Co(NH₃)₆]³⁺ — NH₃ is stronger field, so Δo is larger and the absorbed light has a shorter wavelength.

One-page revision

Read this page the night before the exam.

Octahedral

• t₂g (dxy, dyz, dxz): −0.4 Δo
• eg (dz², dx²−y²): +0.6 Δo
• Choice only for d⁴–d⁷
• Δo > P → low spin

Series

• I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O
• < NH₃ < en < CN⁻ < CO
• Higher charge → larger Δ
• 4d, 5d → almost always low spin

CFSE

• (−0.4x + 0.6y) Δo + extra pairs × P
• HS d⁵ = 0; LS d⁵ = −2.0 Δo + 2P
• LS d⁶ = −2.4 Δo + 2P
• d⁸ = −1.2 Δo; d¹⁰ = 0

Tetrahedral and sq. planar

• e (lower), t₂ (upper); −0.6, +0.4
• Δt = 4/9 Δo → high spin
• Sq. planar: d⁸ + strong field
• [Ni(CN)₄]²⁻ diamagnetic; [NiCl₄]²⁻ 2 unpaired

Magnetism

• μ = √[n(n+2)] BM
• [Fe(CN)₆]⁴⁻ 0; [Fe(H₂O)₆]²⁺ 4
• [CoF₆]³⁻ 4; [Co(NH₃)₆]³⁺ 0
• [Fe(CN)₆]³⁻ 1; [Fe(H₂O)₆]³⁺ 5

Colour and limits

• Δo = hc/λ; seen = complementary
• d⁰, d¹⁰ colourless
• [Ti(H₂O)₆]³⁺: 498 nm, 240 kJ mol⁻¹
• Jahn–Teller: HS d⁴, d⁹ (Cu²⁺)