Coordination Compounds · Lesson 8 of 12
Crystal Field Splitting and Spin
“Build orbital-energy diagrams and predict occupancy from geometry, splitting and pairing energy.”
• Explain octahedral and tetrahedral splitting from donor directions. • Fill split d levels using the exclusion principle and Hund’s rule. • Compare Δ and pairing energy for d⁴–d⁷ octahedral cases. • Calculate the orbital-energy contribution to crystal field stabilisation.
Why five equal d energies become unequal
An isolated gaseous metal ion has five d orbitals of equal energy in the absence of external fields. A perfectly spherical ligand field would shift their average energy without differentiating them. Actual donors occupy particular directions, so the repulsion experienced by d-electron density depends on orbital orientation. Crystal field theory models anionic ligands as point charges and neutral ligands as oriented dipoles. It is deliberately an electrostatic approximation, useful despite its neglect of covalent orbital overlap.
In an octahedron, six donors approach along ±x, ±y and ±z. The d(x²−y²) and d(z²) orbitals direct substantial density along these axes and experience greater repulsion. They form the higher-energy e_g set. The d(xy), d(xz) and d(yz) orbitals lie between axes and form the lower-energy t₂g set. Lower means relative to the average energy of the spherical reference field, not necessarily below the energy of the isolated ion.
The splitting Δo changes with metal identity, oxidation state, ligand identity and metal–ligand distance. A ligand’s charge alone does not rank its splitting effect. The experimentally determined spectrochemical series orders ligands by their typical ability to produce splitting. For a comparable metal environment, the following sequence is a useful qualitative guide; its donor labels and chemical context matter.
| Increasing field strength in a commonly used series |
|---|
| I⁻ < Br⁻ < SCN⁻ (S-bound) < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ (N-bound) < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO |
The fourth electron introduces a choice
Fill the lowest available levels while respecting the exclusion principle: at most two opposite-spin electrons per orbital. Within an equal-energy set, electrons occupy separate orbitals with parallel spins before pairing. Octahedral d¹, d² and d³ therefore put one, two and three unpaired electrons into t₂g. For d⁴, the fourth electron can pair in t₂g or enter the higher e_g level. Pairing has an energy cost P, while promotion costs Δo. The lower total-energy choice determines high or low spin.
| d count | High-spin occupancy; n | Low-spin occupancy; n |
|---|---|---|
| d⁴ | t₂g³e_g¹; 4 | t₂g⁴e_g⁰; 2 |
| d⁵ | t₂g³e_g²; 5 | t₂g⁵e_g⁰; 1 |
| d⁶ | t₂g⁴e_g²; 4 | t₂g⁶e_g⁰; 0 |
| d⁷ | t₂g⁵e_g²; 3 | t₂g⁶e_g¹; 1 |
| d⁸ | t₂g⁶e_g²; 2 | No separate ordinary octahedral high/low-spin alternatives |
Problem
Compare [Mn(H₂O)₆]²⁺ and [Mn(CN)₆]⁴⁻.
- 1.Both contain Mn(II), d⁵: in the cyanido ion x − 6 = −4 gives +2.
- 2.Water yields the usual high-spin t₂g³e_g² arrangement: one electron in each d orbital, n = 5 and μso ≈ 5.92 BM.
- 3.Cyanido produces a larger splitting; low-spin t₂g⁵e_g⁰ has two paired t₂g orbitals and one singly occupied one, n = 1 and μso ≈ 1.73 BM.
- 4.The different charge of the whole entity does not cause the spin difference directly; the local ligand field and pairing energetics do.
Stabilisation is an orbital-energy sum
Problem
Find the orbital-energy contributions for high-spin and low-spin octahedral d⁶ when Δo = 20 000 cm⁻¹.
- 1.High spin has t₂g⁴e_g². The coefficient is 4(−0.4)+2(0.6)=−0.4; the contribution is −8 000 cm⁻¹.
- 2.Low spin has t₂g⁶. Its coefficient is 6(−0.4)=−2.4; the contribution is −48 000 cm⁻¹.
- 3.Low spin gains 2Δo in orbital energy relative to high spin, but has two additional electron pairs. Add 2P to that comparison. The low-spin form is favoured in this simplified model when 2Δo exceeds 2P.
- 4.A larger negative orbital sum alone is not a complete stability argument across different compounds; ligand bonding, pairing, solvation and other energy terms also matter.
Tetrahedral fields reverse the ordering
Four tetrahedral donors approach between the Cartesian axes. The d(xy), d(xz) and d(yz) set now points more toward donor directions and forms the higher t₂ level. The d(z²) and d(x²−y²) set forms the lower e level. Their energies relative to the barycentre are −0.6Δt and +0.4Δt. The labels omit g because tetrahedral symmetry lacks inversion; this is not an absence of all symmetry.
The tetrahedral gap is usually small compared with pairing energy, so ordinary first-row tetrahedral complexes are predominantly high spin. For tetrahedral Ni(II), d⁸ fills e⁴t₂⁴, leaving two unpaired electrons. For tetrahedral Mn(II), d⁵ fills e²t₂³, leaving five. Square-planar splitting is different again and can produce paired d⁸ species such as [Ni(CN)₄]²⁻; do not apply an octahedral t₂g/e_g diagram to a square plane.
What the electrostatic model misses
The observed spectrochemical series includes neutral CO among strong-field ligands and anionic halides among weaker ones. Charge alone cannot explain it. Covalent σ and π interactions require ligand-field or molecular-orbital treatment. CFT explains major patterns of spin and colour, but it does not prove that real metal–ligand bonds are purely ionic or make all total-stability comparisons automatic.
Quick Quiz
Quiz
Which orbitals form the lower octahedral set?
These d orbitals lie between donor axes and experience less repulsion than e_g.
For octahedral d⁴ with Δo > P, what is the occupancy?
Pairing in the lower set costs less than occupying the upper set.
How many unpaired electrons occur in high-spin octahedral d⁶?
t₂g⁴ has two unpaired electrons, and e_g² adds two more.
Which set is lower in a tetrahedral field?
The e set lies at −0.6Δt; tetrahedral labels omit g.
Why must pairing energy be included in a high/low-spin comparison?
The gain from occupying lower orbitals competes with the cost of additional pairing.
Practice Problems
- Draw orbital-box occupations for octahedral d⁴ and d⁵ in strong and weak fields.
- Calculate the crystal field contribution for octahedral d³, high-spin d⁵ and low-spin d⁵.
- For Δo = 18 000 cm⁻¹ and P = 22 000 cm⁻¹, predict octahedral d⁶ spin and μso.
- An ideal matched octahedral field has Δo = 18 000 cm⁻¹. Estimate Δt and explain tetrahedral spin.
- Explain the spectrochemical distinction between SCN⁻ and NCS⁻.
- Why does d⁸ have two unpaired electrons in an ordinary octahedron but can be paired in a square plane?
1. For weak-field d⁴, t₂g has ↑,↑,↑ and e_g has ↑,empty: four unpaired. Strong-field d⁴ has t₂g ↑↓,↑,↑ and empty e_g: two unpaired. 2. Weak-field d⁵ adds one ↑ to each e_g orbital, giving five unpaired. Strong-field d⁵ has t₂g ↑↓,↑↓,↑ and empty e_g, giving one unpaired. 3. The change follows Δo versus P.
Key Takeaways
• Octahedral donors split d orbitals into lower t₂g and higher e_g. • The barycentre is an average-energy reference. • Δo competes with pairing energy for octahedral d⁴–d⁷. • High-spin and low-spin configurations have different unpaired counts. • Crystal field stabilisation is an orbital-energy contribution, not total stability. • Tetrahedral e lies below t₂ and Δt is usually smaller. • Spectrochemical ordering is experimental and donor-dependent. • CFT neglects covalent bonding and has defined limits.