CSIR-NET Inorganic — Crystal Field Theory and Spectra Questions
Coordination chemistry questions in CSIR-NET rarely ask you to state crystal field theory. They hand you a colour, a molar absorptivity, a magnetic moment or a band position and ask you to work backwards to the electronic structure. That is a different skill from reciting the splitting diagram, and it is trainable. This article covers the results you need to have at fingertip level, and the reasoning chains that connect them.
Start from the free-ion ground term
Every electronic spectrum question begins with the ground term of the free dn ion, obtained from Hund's rules. Learn this row; it is short and it is used constantly.
| dn | d¹ | d² | d³ | d⁴ | d⁵ | d⁶ | d⁷ | d⁸ | d⁹ |
|---|---|---|---|---|---|---|---|---|---|
| Ground term | ²D | ³F | ⁴F | ⁵D | ⁶S | ⁵D | ⁴F | ³F | ²D |
Notice the symmetry: d¹ and d⁹ share ²D, d⁴ and d⁶ share ⁵D, d² and d⁷ share ³F, d³ and d⁸ share ⁴F and ³F respectively. This hole formalism is the reason Orgel diagrams pair these configurations, and it is worth quoting explicitly in a descriptive answer.
In an octahedral field a D term splits into T2g + Eg, giving one spin-allowed d–d band. An F term splits into A2g + T1g + T2g and, with the P term of the same multiplicity lying nearby, gives three spin-allowed bands. An S term does not split at all, so a high-spin d⁵ ion has no spin-allowed d–d transition — which is exactly why aqueous Mn²⁺ is almost colourless.
Reading Δo straight off a spectrum
For a d¹ octahedral ion the single band is ²T2g → ²Eg and its energy is Δo directly. For d³ and d⁸ octahedral ions the lowest-energy band (ν1) also equals Δo. Beyond that you need Tanabe–Sugano diagrams, because the higher bands depend on the Racah parameter B as well as on Δo.
Worked example 1 — from wavelength to Δo in kJ/mol. [Ti(H₂O)₆]³⁺ (d¹) absorbs with λmax ≈ 500 nm.
ν̄ = 10⁷ ÷ 500 = 20 000 cm⁻¹, so Δo ≈ 20 000 cm⁻¹.
In kJ/mol: 20 000 × 0.011963 = 239 kJ/mol.
Absorbing green light at 500 nm leaves red and violet to be transmitted, which is why the ion is violet. A frequent follow-up asks why the band has a shoulder: the excited ²Eg state is orbitally degenerate and therefore Jahn–Teller distorted, splitting the upper level and giving an asymmetric band.
Worked example 2 — Δo for a d³ ion. [Cr(H₂O)₆]³⁺ shows its lowest-energy band at 17 400 cm⁻¹. For a d³ octahedral complex the ⁴A2g → ⁴T2g transition energy is Δo.
Δo = 17 400 cm⁻¹ = 17 400 × 0.011963 = 208 kJ/mol.
The two higher bands are ⁴A2g → ⁴T1g(F) and ⁴A2g → ⁴T1g(P). If a question gives you all three, the third one is the route to the Racah parameter B in the complex, and comparing that with the free-ion B gives the nephelauxetic ratio β = Bcomplex ÷ Bfree ion. Values of β below 1 measure how much the d electrons have expanded onto the ligands, i.e. how covalent the bonding is.
Selection rules explain intensity — and intensity is the clue
Two rules govern whether an electronic transition is allowed:
- Spin rule: ΔS = 0. A transition that changes the number of unpaired electrons is spin-forbidden and very weak.
- Laporte rule: in a centrosymmetric molecule, g ↔ g and u ↔ u are forbidden; only g ↔ u is allowed. All d orbitals are g, so every d–d transition in an octahedral complex is Laporte-forbidden.
Both rules are relaxed in practice. Vibronic coupling — an odd vibration that momentarily destroys the centre of symmetry — gives octahedral d–d bands a small but real intensity. A tetrahedral complex has no centre of symmetry at all, so the Laporte rule does not apply and d–p mixing makes its bands roughly one to two orders of magnitude stronger.
| Transition type | Typical ε (dm³ mol⁻¹ cm⁻¹) | Example |
|---|---|---|
| Spin-forbidden and Laporte-forbidden | ~0.01–1 | [Mn(H₂O)₆]²⁺, very pale pink |
| Spin-allowed, Laporte-forbidden (octahedral d–d) | ~1–100 | [Co(H₂O)₆]²⁺, pale pink |
| Spin-allowed, no centre of symmetry (tetrahedral d–d) | ~100–1000 | [CoCl₄]²⁻, intense blue |
| Charge transfer (fully allowed) | ~1000–50 000 | MnO₄⁻, deep purple |
This table answers a whole family of questions in one line. Why is [CoCl₄]²⁻ far more intensely coloured than [Co(H₂O)₆]²⁺, even though its Δ is smaller? Because the tetrahedral ion is not centrosymmetric. Why is permanganate so intensely purple when Mn(VII) is d⁰ and can have no d–d transition at all? Because the colour is ligand-to-metal charge transfer, an allowed transition.
The two series you must be able to write out
Spectrochemical series (increasing Δ): I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < CH₃CN < py < NH₃ < en < bipy < phen < NO₂⁻ < CN⁻ < CO. The trend is not electrostatic: the strongest field ligands are π acceptors (CN⁻, CO, phen) and the weakest are π donors (halides). This is a molecular orbital result that crystal field theory cannot explain on its own, and saying so earns marks.
Nephelauxetic series (increasing reduction of B, i.e. increasing covalency): F⁻ < H₂O < NH₃ < en < ox²⁻ < NCS⁻ < Cl⁻ < CN⁻ < Br⁻ < I⁻. Note that it is not the same order as the spectrochemical series — a classic trap.
Δ also rises with metal oxidation state (roughly 50% higher for M³⁺ than M²⁺) and rises sharply down a group, by about 30–50% from 3d to 4d and again from 4d to 5d. That is why second- and third-row complexes are almost invariably low spin.
Magnetic moments — where the spin-only formula fails
n = 1 → 1.73 · n = 2 → 2.83 · n = 3 → 3.87 · n = 4 → 4.90 · n = 5 → 5.92 BM
Worked example 3 — predicting spin state from the ligand. Compare [CoF₆]³⁻ and [Co(NH₃)₆]³⁺, both Co(III), d⁶.
F⁻ sits low in the spectrochemical series, so Δo < P and the ion is high spin: t2g⁴eg², n = 4, μ = √(4 × 6) = 4.90 BM, paramagnetic.
NH₃ is a strong-field ligand, so Δo > P and the ion is low spin: t2g⁶eg⁰, n = 0, μ = 0 BM, diamagnetic.
The same reasoning explains their colours: the high-spin fluoride absorbs at low energy and is blue, the low-spin ammine absorbs at higher energy and is yellow-orange.
The spin-only formula works well when the ground term in the ligand field is A or E, because the orbital angular momentum is quenched. It fails when the ground term is T (T1g or T2g), where an unquenched orbital contribution raises the observed moment. Octahedral high-spin Co²⁺ has a ⁴T1g ground term and typically shows 4.7–5.2 BM against a spin-only prediction of 3.87 BM — a favourite Part C discrepancy to explain.
Traps that catch good candidates
- Assuming Δo is the first band for every ion. True for d¹, d³ and d⁸ octahedral; not true for d² or d⁷, where B enters and you need a Tanabe–Sugano analysis.
- Confusing the colour absorbed with the colour seen. The observed colour is the complement of the absorbed band.
- Calling a tetrahedral complex low spin. Δt ≈ (4/9)Δo for the same metal and ligands, so Δt is essentially never larger than the pairing energy — tetrahedral complexes of the first row are high spin.
- Attributing MnO₄⁻ or CrO₄²⁻ colour to d–d transitions. Both metals are d⁰; the colour is charge transfer.
- Using the spin-only formula for a T ground term without commenting on the orbital contribution.
- Reversing the nephelauxetic and spectrochemical orders. Iodide is a weak field ligand but strongly nephelauxetic.
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