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13.7 Electronic Spectra of Complexes: Term Symbols, Selection Rules, Orgel Diagrams, Charge Transfer

★★ 2021–2025: 10 parts, 88 marks (Civil Services 2 parts / 20 m · IFoS 2 / 13 m · Geo-Scientist 6 / 55 m). The questions ask why a complex has its colour and intensity: count the bands, apply the selection rules, spot charge transfer.

Syllabus lines. Civil Services and IFoS: “electronic spectra of transition metal complexes … term symbols, selection rules, charge transfer spectra”. Geo-Scientist: “electronic spectra of complexes, selection rules, colour of complexes, charge transfer”.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

13.7.1 Term symbols by L–S coupling ★★

(13.7.1)
  • Hund's rules for the ground term: (1) maximum S; (2) for that S, maximum L; (3) J = L − S if the shell is less than half full, L + S if more than half full. L = 0, 1, 2, 3, 4 → S, P, D, F, G.SN
  • Ground terms: d¹, d⁹ ²D · d², d⁸ ³F · d³, d⁷ ⁴F · d⁴, d⁶ ⁵D · d⁵ ⁶S. Hole rule: dn and d10−n have the same terms (the splitting is inverted). All free-ion terms of d²: ³F, ³P, ¹G, ¹D, ¹S.SN
Cr3+ d3+2↑+1↑0↑−1−2mlL = 2+1+0 = 3 → FS = 3/2 → 2S+1 = 4J = L − S = 3/24F3/2Mn3+ d4+2↑+1↑0↑−1↑−2mlL = 2+1+0−1 = 2 → DS = 2 → 2S+1 = 5J = L − S = 05D0
Fig. 13.7.1 Ground terms of Cr³⁺ and Mn³⁺ from the ml boxes.
Draw it in 60 s: five boxes labelled +2 … −2; fill electrons parallel from the left; add the ml values for L, count arrows for S, write 2ˢ+1LJ.

Solved PYQ Geo 2022 Q1(b) 5m

Ground terms of [Cr(NH₃)₆]³⁺ and [Mn(H₂O)₆]³⁺ by L–S coupling.

  1. Cr³⁺ is d³: ml = 2, 1, 0 → L = 3 (F), S = 3/2 → ⁴F3/2; in Oh it becomes ⁴A₂g (t₂g³).
  2. Mn³⁺ is d⁴ high spin: ml = 2, 1, 0, −1 → L = 2 (D), S = 2 → ⁵D₀; in Oh ⁵Eg (t₂g³eg¹), Jahn–Teller distorted.SN

13.7.2 How free-ion terms split in an octahedral field ★★

Free-ion termSPDFG
Oh termsA₁gT₁gEg + T₂gA₂g + T₁g + T₂gA₁g + Eg + T₁g + T₂g
  • Spin multiplicity does not change in the field. d¹ (²D → ²T₂g ground, ²Eg): one band; d⁹ is the hole mirror (²Eg ground). d⁸ (³F → ³A₂g ground) and d³ (⁴F → ⁴A₂g ground) give three spin-allowed bands with the P term; d² and d⁷ (high spin) have T₁g(F) ground.SN
  • In a tetrahedral field the order is reversed (no g subscripts): dn tetrahedral behaves like d10−n octahedral.SN

13.7.3 Selection rules and band intensity ★★★

(13.7.2)
  • Spin rule: ΔS = 0. Transitions that change multiplicity are very weak.SN
  • Laporte rule: in a centrosymmetric complex only g ↔ u is allowed; d → d is g → g, so all octahedral d–d bands are forbidden. They appear weakly through vibronic coupling (vibrations remove the centre of symmetry for an instant) and spin–orbit coupling.SN
  • Tetrahedral complexes have no centre of symmetry; d–p mixing lets d–d bands borrow intensity: ε about 100 times larger. Charge-transfer bands are allowed by both rules Geo 2022 Q4(b) 10m.SN
10−210−1100101102103104105ε / dm3 mol−1 cm−1spin- and Laporte-forbidden[Mn(H2O)6]2+, Fe3+ (high-spin d5)spin-allowed, Laporte-forbidden (Oₕ)[Co(H2O)6]2+ pink, [Ni(H2O)6]2+no centre of symmetry (tetrahedral)[CoCl4]2− blue, [NiBr4]2−charge transfer (fully allowed)MnO4−, CrO42−, Fe3+–Cl−/OH−
Fig. 13.7.2 Band intensity tells the kind of transition.
Draw it in 60 s: a log ε ruler 10−2 … 105; boxes: spin-forbidden (< 1), octahedral d–d (1–50), tetrahedral d–d (50–103), CT (103–105).

Solved PYQ Geo 2025 Q11(c) 10m

Octahedral Co²⁺ complexes are pale pink; tetrahedral Co²⁺ complexes are intensely blue. Why?

  1. [Co(H₂O)₆]²⁺: d⁷ high spin, ⁴T₁g(F) ground, centrosymmetric: spin-allowed but Laporte-forbidden, ε ≈ 5; it absorbs weakly near 515 nm (blue-green), so it looks pale pink.
  2. [CoCl₄]²⁻: no centre of symmetry, d–p mixing: ε ≈ 600, about 100 times stronger. Δt ≈ (4/9)Δo and Cl⁻ is a weak-field ligand, so the band moves to lower energy (≈ 690 nm, red); the transmitted colour is deep blue.SN
  • Colourless iron(II): [Fe(H₂O)₆]²⁺ (high-spin d⁶) has one spin-allowed band, ⁵T₂g → ⁵Eg, at ≈ 10 400 cm⁻¹ (near infrared) with ε ≈ 5: almost colourless (very pale green). [Fe(CN)₆]⁴⁻ (low-spin d⁶, ¹A₁g) has its spin-allowed bands at ≈ 31 000 and 37 000 cm⁻¹ (ultraviolet), because Δo is so large: almost colourless (pale yellow) IFoS 2025 Q7(c)(i) 5m.SN
  • High-spin d⁵ ([Mn(H₂O)₆]²⁺, Fe³⁺): ⁶A₁g ground and no other sextet term, so every d–d band is spin-forbidden (and Laporte-forbidden): ε ≈ 0.01–0.1, very pale pink/violet. [MnBr₄]²⁻ is tetrahedral, so its spin-forbidden bands are a little stronger (pale green) Geo 2022 Q4(b) 10m.SN
  • Bottle glass: Fe³⁺ (d⁵) bands are spin-forbidden, ε tiny. Through the wall (l ≈ 3 mm) A = εcl is too small to see; along the length (l ≈ 30 cm, 100 times more) A is 100 times larger and the green shows Geo 2024 Q7(c) 10m.SN

Solved PYQ Geo 2021 Q4(a) 15m

An “octahedral” Co(III) ammine–chlorido complex shows two bands with ε 60–80, one with ε = 2 and one with ε = 2 × 10⁴. Assign them.

  1. Co(III), low-spin d⁶, ¹A₁g ground. ε 60–80 (two bands): spin-allowed d–d, ¹A₁g → ¹T₁g and ¹A₁g → ¹T₂g; above the usual 1–10 because Cl lowers the symmetry (e.g. C4v in [Co(NH₃)₅Cl]²⁺, no centre of symmetry).
  2. ε = 2: spin-forbidden d–d, ¹A₁g → ³T₁g (singlet → triplet).
  3. ε = 2 × 10⁴: fully allowed LMCT, Cl⁻ (p) → Co(III), in the UV.SN

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