13.2 Crystal Field Theory

13.2.1 The model and its assumptions ★★ 20 marks once

Crystal field theory (Bethe 1929, van Vleck) treats a complex as a metal ion sitting in the electric field of its ligands. It asks one question: which d orbitals feel the ligands' negative charge most? Those are raised in energy; the others are lowered. Everything else in this chapter (colour, magnetism, distortions, stability trends) follows from that split.

1. Ligands are point charges (anions) or point dipoles (neutral ligands, negative end towards the metal).SN
2. The metal–ligand bond is purely electrostatic. No orbital overlap, no covalency.SN
3. The five d orbitals keep their shapes. In the free ion they are degenerate; the field only shifts their energies.
4. Orbitals pointing at the ligands are repelled more than orbitals pointing between them. That lifts the degeneracy.SN
5. Barycentre rule: the weighted mean energy of the five orbitals stays at the spherical-field value. What one set gains, the other loses.SN
6. Electrons then fill the split orbitals by Aufbau and Hund's rule, with a contest between the splitting Δ and the pairing energy P (13.2.4).

Asked: Geo 2023 Q4(a) 20m assumptions plus splitting in octahedral, tetrahedral and square-planar fields. Write the six points above, then the three diagrams of Fig. 13.2.2.

13.2.2 Splitting in an octahedral field ★★★

• Put the six ligands on ±x, ±y, ±z. d_x²−y² and d_z² point straight at them: they rise and form e_g. d_xy, d_xz, d_yz point between them: they fall and form t_2g.SN
• Barycentre: 2(+0.6Δ_o) + 3(−0.4Δ_o) = 0. So t_2g at −0.4Δ_o, e_g at +0.6Δ_o, with Δ_o = 10 Dq.SN
• Labels: t = triply and e = doubly degenerate; g (gerade) because the octahedron has a centre of symmetry.

Measuring Δ_o

For a d¹ ion the only d–d band is t_2g¹ → e_g¹, so its energy is Δ_o. [Ti(H₂O)₆]³⁺ absorbs at 493 nm, i.e. 20 300 cm⁻¹:

It absorbs yellow-green light and transmits red and blue, hence the purple colour. Useful conversion: 1000 cm⁻¹ = 11.96 kJ mol⁻¹.SN

13.2.3 What decides the size of Δ ★★★

1. Charge on the metal. Higher charge → smaller ion, shorter M–L bonds → larger Δ. M³⁺ has roughly 1.5–2 times the Δ of M²⁺: aqua Co³⁺ 16 750 against Co²⁺ 8 400 cm⁻¹.SN
2. Row of the metal. Δ rises about 30–50 % from 3d to 4d and again to 5d: [Co(NH₃)₆]³⁺ 22 900 → [Rh(NH₃)₆]³⁺ 34 000 cm⁻¹. Bigger 4d/5d orbitals overlap better and pair more easily, so 4d and 5d complexes are almost always low spin.SN
3. The ligand, the spectrochemical series (same order for every metal):SN

   I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < py < NH₃ < en < bpy < phen < NO₂⁻ < PPh₃ < CN⁻ < CO

   Cr³⁺ shows it cleanly: Cl⁻ 13 700, H₂O 17 400, NH₃ 21 500, CN⁻ 26 600 cm⁻¹. Halides and O-donors are weak field; N-donors in the middle; C-donors (CN⁻, CO) strongest.

4. Geometry and number of ligands. Δ_t = 4/9 Δ_o for the same metal, ligand and distance (13.2.5).SN

Metal order for the same ligand: Mn²⁺ < Ni²⁺ < Co²⁺ < Fe²⁺ < V²⁺ < Fe³⁺ < Co³⁺ < Mo³⁺ < Rh³⁺ < Ru³⁺ < Pd⁴⁺ < Ir³⁺ < Pt⁴⁺.

Also asked: Geo 2022 Q3(b) 10m factors affecting Δ · Geo 2025 Q5(b) 10m order [CrCl₆]³⁻ < [Cr(dtp)₃] < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]³⁻ (S-donor dtp sits just above Cl⁻).

13.2.4 CFSE, pairing energy, high spin and low spin ★★★ most asked

Each t_2g electron is stabilised by 0.4Δ_o; each e_g electron destabilised by 0.6Δ_o. Every extra pair formed costs the pairing energy P (Coulomb repulsion in one orbital plus lost exchange energy).

• d¹–d³ and d⁸–d¹⁰ have only one way to fill. Only d⁴, d⁵, d⁶, d⁷ have a choice.SN
• Δ_o < P → high spin (weak field, electrons spread out). Δ_o > P → low spin (strong field, electrons pair in t_2g).SN
• Tetrahedral complexes are practically always high spin: Δ_t ≈ 4/9 Δ_o is almost never larger than P.SN

P in the low-spin column is counted relative to the high-spin arrangement.SN

Also asked: Geo 2021 Q3(a) 5m five Co complexes · Geo 2022 Q3(c) 10m and CSE 2023 Q8(b) 10m spin-only moments · Geo 2023 Q1(j) 5m V³⁺ (d², 2.83 BM) against V⁵⁺ (d⁰, diamagnetic).