Chemistry · Ch 5 — Coordination Chemistry
Spectrochemical Series and Distribution of d Electrons
Spectrochemical Series and Distribution of d Electrons
The size of the crystal field splitting energy depends not only on whether the field is octahedral or tetrahedral, but also on the identity of the ligand, the identity of the central metal, and its charge. This is demonstrated by comparing octahedral titanium(III) complexes of three different ligands using their absorption spectral data: [TiBr₆]³⁻, [TiF₆]³⁻ and [Ti(H₂O)₆]³⁺ absorb at 12500, 19000 and 20000 cm⁻¹ respectively. Using Δ = hcν̄ (h = Planck's constant, c = speed of light, ν̄ = absorption wavenumber) and converting to a per-mole basis with Avogadro's number gives Δ values of 149.4, 227.7 and 239.7 kJ/mol respectively -- so for Ti³⁺, the crystal field splitting strength runs Br⁻ < F⁻ < H₂O. Extending this kind of spectral comparison across many metal-ligand combinations gives the general spectrochemical series (in increasing splitting strength): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ ≈ urea < ox²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO. Ligands toward the right (like CO) are strong-field ligands causing a large splitting, while ligands toward the left (like I⁻) are weak-field ligands causing a small splitting. This series governs how d electrons fill the split orbitals in octahedral complexes with d⁴ through d⁷ configurations (which have a genuine choice of filling pattern, per Hund's rule for d1-d3 and d8-d10): if the octahedral splitting energy Δ₀ exceeds the electron-pairing energy P, an electron is forced to pair up in a lower t2g orbital rather than occupy a higher eg orbital, giving a low-spin configuration; if Δ₀ is smaller than P, the electron instead occupies the higher eg orbital unpaired, giving a high-spin configuration. Comparing [Fe(H₂O)₆]³⁺ (weak field, Δ₀ = 14000 cm⁻¹, less than the Fe³⁺ pairing energy of 30000 cm⁻¹) against [Fe(CN)₆]³⁻ (strong field, Δ₀ = 35000 cm⁻¹, greater than the pairing energy) shows exactly this contrast: the aqua complex is high spin with configuration t2g³eg², while the cyanido complex is low spin with configuration t2g⁵eg⁰, even though both start from the same Fe³⁺, d⁵ ion. The actual electron distribution is quantified by the crystal field stabilisation energy, CFSE = ΔEo = {ELF} - {Eiso} = {[nt2g(-0.4) + neg(0.6)]Δ₀ + npP} - {n'pP}, where nt2g and neg are the electron counts in each set, np is the number of electron pairs in the ligand field and n'p is the number of pairs in the isotropic (baryc …
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ ≈ urea < ox²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO. Ligands on the right (carbonyl, cyanide) are strong-field ligands producing a large Δ; ligands on the left (halides, thiocyanate) are weak-field ligands producing a small Δ. This ordering is empirical, built from absorption-spectral data across many metal-ligand combinations, and it is what deter …
Complex | Absorption wavenumber (cm⁻¹) | Δ (kJ/mol)
[TiBr₆]³⁻ | 12500 | 149.4
[TiF₆]³⁻ | 19000 | 227.7
[Ti(H₂O)₆]³⁺ | 20000 | 239.7
Each Δ is calculated as Δ = hcν̄, then multiplied by Avogadro's number to express it on a per-mole basis. The increasing Δ from bromide to fluoride to water for the same Ti³⁺ ion directly demonstrates t …
Complex | Δ₀ (cm⁻¹) | Pairing energy P (cm⁻¹) | Low-spin CFSE | Actual ground state
[Fe(H₂O)₆]³⁺ | 14000 | 30000 | -2Δ₀+2P = +32000 cm⁻¹ (unfavourable) | High spin, t2g³eg², paramagnetic, μs = 5.916 BM
[Fe(CN)₆]³⁻ | 35000 | 30000 | -2Δ₀+2P = -10000 cm⁻¹ (favourable) | Low spin, t2g⁵eg⁰, paramagnetic, μs = 1.732 BM …
Worked out. Three self-check problems applying the Δ₀-versus-P comparison. Q11: [Mn(CN)₆]³⁻ has mean pairing energy 28800 cm⁻¹ and Δ₀ = 38500 cm⁻¹ -- since Δ₀ > P, the complex is stable in the low-spin state. Q12: asks the student to draw the energy-level (t2g/eg) diagram for [Cu(H₂O)₆]²⁺ (Cu²⁺, d⁹, only one possible filling regardless of field strength: t2g⁶eg³, one unpaired electron) and state that it is paramagnetic. Q13: for [CoF₆]³⁻, mean pairing energy 21000 cm⁻¹ and Δ₀ = 13000 cm⁻¹ (so Δ₀ < P, high spin favoured), the student must calculate CFSE for both the low-spin and high-spin states using the CFSE formula, the same calculation worked through …