Crystal Field Splitting Theory Explained | CFSE, t2g–eg, Octahedral, Tetrahedral, Square Planar

Crystal Field Splitting Theory (CFSE) — The Complete, Step-by-Step Student Guide

By Rizwan Ibn Ali Abdullah | Rizwan Chemistry Classes

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Introduction: Why Some Complexes Are Colorful, Others Magnetic — and Some Both

Transition-metal complexes surprise students: striking colours, unexpected magnetism, and different shapes for the “same” metal. The backbone theory that predicts these behaviours is the Crystal Field Theory (CFT) and its quantitative heart, the Crystal Field Splitting Energy (CFSE). In this long-form tutorial, we’ll go beyond definitions and show you exactly how to calculate CFSE for d¹ to d¹⁰ in octahedral, tetrahedral, and square planar fields — in both high-spin and low-spin cases where relevant.

Table of Contents

• Postulates of Crystal Field Theory
• Degeneracy, Barycenter & Why Splitting Happens
• Octahedral Field: t_2g/e_g, Δ_o, and CFSE
• Tetrahedral Field: t₂/e, Δ_t, and CFSE
• Square Planar Field: Ordering, Spin, and CFSE Approach
• Universal Algorithm to Calculate CFSE (Step-by-Step)
• CFSE Tables for d¹–d¹⁰ (High & Low Spin)
• Worked Examples & Mini Case Studies
• Colours, Magnetism & Stability — How CFSE Explains Them
• FAQ

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1) Postulates of Crystal Field Theory (CFT)

• Ligands are treated as point charges (or dipoles) that interact electrostatically with the metal ion.
• The five d orbitals of a free metal ion are degenerate (same energy). Ligand approach breaks this degeneracy.
• Orbitals pointing more directly toward ligands experience greater repulsion and become higher in energy.
• The net average (barycenter) energy of the five d orbitals remains the same as the free ion.
• Spin state (high vs low) depends on the competition between splitting (Δ) and pairing energy (P):
  • If Δ < P → high spin (maximize unpaired electrons)
  • If Δ > P → low spin (pair earlier in the lower set)

Key idea: CFT is an ionic model. It ignores covalency but still predicts geometry, colours, magnetic moments, and relative stabilities surprisingly well via CFSE.

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2) Degeneracy, the Barycenter, and Why Splitting Happens

In a spherical (isotropic) field the five d orbitals have the same energy. When specific numbers/positions of ligands approach, directional repulsions differ. The set of d orbitals splits into two (or more) levels, but the weighted average of energies equals the original average (the barycenter).

Barycenter rule: In octahedral fields, the upper set e_g lies at +3/5 Δ_o and the lower set t_2g lies at −2/5 Δ_o relative to the barycenter; net average = 0.

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3) Octahedral Field (O_h): t_2g/e_g, Δ_o, and CFSE

3.1 Orbital directions & splitting

Six ligands approach along the ±x, ±y, ±z axes. The orbitals pointing directly at axes (d_x²−y², d_z²) face maximum repulsion → higher energy set e_g at +3/5 Δ_o. Orbitals lying between axes (d_xy, d_yz, d_xz) feel less repulsion → lower set t_2g at −2/5 Δ_o.

3.2 CFSE formula (octahedral)

Let n_t2g = number of electrons in t_2g, and n_eg = number in e_g. Then

CFSE (octa) = [ (−0.4 × n_t2g) + ( +0.6 × n_eg ) ] Δ_o

When comparing high vs low spin, also account for pairing energy P for each additional electron pair formed.

Tip for students: Fill electrons following Hund’s rule within a set. Choose high vs low spin by comparing Δ_o with P for the ligand (use the spectrochemical series conceptually).

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4) Tetrahedral Field (T_d): t₂/e, Δ_t, and CFSE

4.1 Orbital directions & splitting

Four ligands approach between axes. Now the three (d_xy, d_yz, d_xz) are higher (call this set t₂) and the two (d_x²−y², d_z²) are lower (set e).

4.2 Relative magnitude

Δ_t ≈ (4/9) Δ_o (for the same metal/ligand set).

Since Δ_t is smaller, tetrahedral complexes are almost always high spin.

4.3 CFSE formula (tetrahedral)

Let n_e = electrons in e (lower), and n_t2 = electrons in t₂ (higher). Then

CFSE (tetra) = [ (−0.6 × n_e) + ( +0.4 × n_t2 ) ] Δ_t

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5) Square Planar Field (D_4h): Ordering, Spin, and CFSE Approach

5.1 How square planar arises

Start from octahedral and remove the two axial ligands (strong tetragonal distortion). This pushes d_x²−y² very high in energy, while d_z² drops. Typical qualitative order:

d_x²−y² (highest) » d_xy > d_z² > d_xz ≈ d_yz (lowest)

5.2 Spin tendencies

Square planar fields (especially for d⁸ like Ni(II), Pd(II), Pt(II)) are usually low spin because the gap to d_x²−y² is very large, keeping it empty.

5.3 CFSE in square planar

Precise CFSE uses tetragonal parameters (D_q, D_s, D_t) beyond the basic course. For problem-solving, you can:

• Fill electrons bottom→up using the order above.
• Keep d_x²−y² empty unless forced (rare).
• If your question provides level spacings, multiply occupancies by their offsets from the barycenter (exactly like octa/tetra).

Exam-safe rule of thumb: For d⁸ (e.g., [Ni(CN)₄]²⁻), square planar is very stable and diamagnetic: d_xz, d_yz (4e⁻), d_z² (2e⁻), d_xy (2e⁻), d_x²−y² (0).

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6) A Universal Step-by-Step Algorithm to Calculate CFSE

1. Find d-count: Determine the oxidation state → electron configuration → number of d electrons (d¹…d¹⁰).
2. Pick geometry: Octahedral, tetrahedral, or square planar (usually given or infer from ligand/metal trends).
3. Estimate field strength: Use spectrochemical intuition (CN⁻, CO, NO₂⁻ strong; H₂O, F⁻ weak). Decide high vs low spin by comparing Δ with pairing energy P.
4. Distribute electrons:
   • Octa: fill t_2g (3 orbitals), then e_g (2 orbitals).
   • Tetra: fill e (2 orbitals), then t₂ (3 orbitals).
   • Square planar: fill from lowest upward; keep d_x²−y² empty if possible.
5. Compute CFSE: Multiply occupancies by their energy factors:
   • Octa: each t_2g e⁻ = −0.4Δ_o; each e_g e⁻ = +0.6Δ_o.
   • Tetra: each e e⁻ = −0.6Δ_t; each t₂ e⁻ = +0.4Δ_t.
   • Square planar: use given level separations (or do qualitative analysis if not provided).
6. Add pairing penalties when comparing alternative spin states: each extra pair formed costs P.
7. Report magnetism: Count unpaired electrons n → spin-only μ ≈ √(n(n+2)) BM.

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7) CFSE Tables for d¹–d¹⁰ (High & Low Spin)

7.1 Octahedral CFSE (include pairing notes)

d-count	High-spin occupancy (t2g, eg)	CFSE (Δo)	Unpaired e⁻	Low-spin occupancy (t2g, eg)	CFSE (Δo)	Unpaired e⁻
d¹	t2g1 eg0	−0.4Δo	1	same	same	1
d²	t2g2 eg0	−0.8Δo	2	same	same	2
d³	t2g3 eg0	−1.2Δo	3	same	same	3
d⁴	t2g3 eg1	−0.6Δo	4	t2g4 eg0	−1.6Δo (+1 pairing P)	2
d⁵	t2g3 eg2	0	5	t2g5 eg0	−2.0Δo (+2P)	1
d⁶	t2g4 eg2	−0.4Δo	4	t2g6 eg0	−2.4Δo (+2P)	0
d⁷	t2g5 eg2	−0.8Δo	3	t2g6 eg1	−1.8Δo (+3P)	1
d⁸	t2g6 eg2	−1.2Δo	2	t2g6 eg2	−1.2Δo (+3P vs +4P depends on distribution)	2 or 0*
d⁹	t2g6 eg3	−0.6Δo	1	same	same	1
d¹⁰	t2g6 eg4	0	0	same	same	0

*Low-spin d⁸ in octahedra is uncommon; many d⁸ prefer square planar (strong field) or high-spin octahedral depending on metal/ligand.

How those CFSE numbers were obtained (example derivation)

For octa d⁶ high-spin: t_2g⁴ e_g² → (4 × −0.4 + 2 × +0.6)Δ_o = (−1.6 + 1.2)Δ_o = −0.4Δ_o.

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7.2 Tetrahedral CFSE (usually high spin)

d-count	Occupancy (e lower, t2 higher)	CFSE (Δt)	Unpaired e⁻
d¹	e¹ t20	−0.6Δt	1
d²	e² t20	−1.2Δt	2
d³	e² t21	−0.8Δt	3
d⁴	e² t22	−0.4Δt	4
d⁵	e² t23	0	5
d⁶	e³ t23	−0.6Δt	4
d⁷	e⁴ t23	−1.2Δt	3
d⁸	e⁴ t24	−0.8Δt	2
d⁹	e⁴ t25	−0.4Δt	1
d¹⁰	e⁴ t26	0	0

Remember: Δ_t ≈ (4/9)Δ_o for the same metal/ligand set.

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8) Worked Examples & Mini Case Studies

Example 1 — Octahedral, High Spin: [Fe(H₂O)₆]²⁺

• Fe²⁺ → d⁶.
• H₂O = weak field → high spin likely.
• Occupancy: t_2g⁴ e_g².
• CFSE = (4×−0.4 + 2×+0.6)Δ_o = −0.4Δ_o.
• Unpaired e⁻ = 4 → μ ≈ √(4×6) = √24 ≈ 4.90 BM.

Example 2 — Octahedral, Low Spin: [Fe(CN)₆]⁴⁻

• Fe²⁺ → d⁶; CN⁻ strong field → low spin.
• Occupancy: t_2g⁶ e_g⁰.
• CFSE = 6×(−0.4)Δ_o = −2.4Δ_o.
• Pairing cost: +2P (because 2 extra pairs vs high-spin).
• Unpaired e⁻ = 0 → diamagnetic.

Example 3 — Octahedral d⁴ (Compare Spin States)

• d⁴ high spin: t_2g³ e_g¹ → CFSE = −0.6Δ_o; pairs = 0; n = 4.
• d⁴ low spin: t_2g⁴ e_g⁰ → CFSE = −1.6Δ_o; +P; n = 2.
• Decision: Low spin wins if (extra stabilization) 1.0Δ_o > P.

Example 4 — Tetrahedral: [CoCl₄]²⁻

• Co²⁺ → d⁷; Cl⁻ weak; tetrahedral Δ small → high spin.
• Occupancy e (lower): 4 e⁻; t₂ (higher): 3 e⁻.
• CFSE = (4×−0.6 + 3×+0.4)Δ_t = (−2.4 + 1.2)Δ_t = −1.2Δ_t.
• n = 3 → μ ≈ √(3×5) = √15 ≈ 3.87 BM.

Example 5 — Square Planar d⁸: [Ni(CN)₄]²⁻

• Ni²⁺ → d⁸; CN⁻ strong; square planar favoured.
• Fill (lowest→highest): d_xz, d_yz (4e⁻ total), d_z² (2e⁻), d_xy (2e⁻), d_x²−y² (0).
• All paired → diamagnetic (n = 0).
• CFSE is large and stabilizing; exact value requires given level spacings for square planar (exam questions sometimes supply them).

Example 6 — Octahedral d⁵ Crossover

• d⁵ high spin: t_2g³ e_g² → CFSE = 0; n = 5.
• d⁵ low spin: t_2g⁵ e_g⁰ → CFSE = −2.0Δ_o; +2P; n = 1.
• Decision: Low spin wins if 2.0Δ_o > 2P → Δ_o > P.

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9) Applications: Colours, Magnetism, Stability

Colours

Light promotes t_2g → e_g (or e → t₂) transitions. The absorbed wavelength corresponds to Δ; the complementary colour is observed.

Magnetism

Unpaired electrons (n) determine magnetic moment (spin-only): μ ≈ √(n(n+2)) BM. Low spin reduces n; high spin increases n.

Stability

Greater (more negative) CFSE → greater thermodynamic stabilization. Low-spin t_2g⁶ (d⁶) is notably stable in octahedra.

Jahn–Teller Notes

d⁹ (octa) and high-spin d⁴ often distort to lower symmetry to remove degeneracy — useful clue in spectroscopy and EPR.

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FAQ — Quick Concept Checks

Q1. What is the “average energy when ligands appear”?

The barycenter: after splitting, the weighted average energy of all five d orbitals equals the unsplit average. That’s why e_g is +3/5Δ_o and t_2g is −2/5Δ_o.

Q2. Why is Δ_t smaller than Δ_o?

Because tetrahedral ligands approach between axes (less direct repulsion). Empirically Δ_t ≈ (4/9)Δ_o.

Q3. When do we add pairing energy (P)?

When comparing different spin states. Each additional paired electron (relative to the alternative) adds +P to the total energy.

Q4. Can tetrahedral complexes be low spin?

Very rarely; Δ_t is usually too small to overcome P, so they’re almost always high spin.

Q5. How do I handle square planar CFSE in exams?

Use the standard ordering (d_xz/d_yz < d_z² < d_xy ≪ d_x²−y²). Fill and discuss spin; compute CFSE only if the problem provides level separations.

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Takeaway: Decide geometry and spin first, then fill orbitals and multiply by the correct Δ-factors. Report CFSE, pairing differences, and unpaired electrons. That’s the entire game.

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