Q4(e). Why d–d transitions are forbidden, weak, and yet observed
Why are d-d electronic transition forbidden? Why are they weakly absorbing and why do they occur at all?
Answer in brief
d–d transitions break the Laporte (parity) rule (d → d is g → g) and some also the spin rule; so they are weak, ε≈20–100 dm³ mol⁻¹ cm⁻¹ in octahedral complexes. They are seen because the rules are relaxed by vibronic coupling, by d–p mixing in complexes without a centre of symmetry, and (for the spin rule) by spin–orbit coupling.
Key points
- in a centrosymmetric molecule the only allowed transitions are those with a change of parity, g ↔ u. The transition dipole ∫ψ_f μ ψᵢ dτ vanishes unless the product changes sign on inversion; μ is u, so ψᵢ and ψ_f must differ.…
- light does not act on spin, so Δ S=0. A d–d transition that changes multiplicity is spin forbidden as well (all transitions of high-spin d⁵, Mn²⁺).
- because they are forbidden, the transition dipole is close to zero: spin-allowed but Laporte-forbidden bands have εₘₐₓ about 20–100 dm³ mol⁻¹ cm⁻¹, compared with 1000–50 000 for allowed charge-transfer bands. Hence the pale…
- an asymmetric vibration momentarily removes the centre of symmetry, so the d orbitals mix slightly with p (u) character and the transition borrows intensity. The bands grow stronger on heating.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Why forbidden
- Laporte (parity) rule: in a centrosymmetric molecule the only allowed transitions are those with a change of parity, g ↔ u. The transition dipole ∫ψ_f μ ψᵢ dτ vanishes unless the product changes sign on inversion; μ is u, so ψᵢ and ψ_f must differ. d orbitals are all g, so d → d in an octahedral complex is Laporte forbidden (for an isolated atom, Δ l=0 instead of ±1).
- Spin rule: light does not act on spin, so Δ S=0. A d–d transition that changes multiplicity is spin forbidden as well (all transitions of high-spin d⁵, Mn²⁺).
Why weak, and why they occur at all
- Weak: because they are forbidden, the transition dipole is close to zero: spin-allowed but Laporte-forbidden bands have εₘₐₓ about 20–100 dm³ mol⁻¹ cm⁻¹, compared with 1000–50 000 for allowed charge-transfer bands. Hence the pale colours of octahedral complexes such as [Ti(H₂O)₆]³⁺ and [Co(H₂O)₆]²⁺.
- Vibronic coupling: an asymmetric vibration momentarily removes the centre of symmetry, so the d orbitals mix slightly with p (u) character and the transition borrows intensity. The bands grow stronger on heating.
- No centre of symmetry: in tetrahedral complexes metal d and p orbitals (both t₂) mix; the Laporte rule no longer applies strictly and the bands are stronger (about 250): [CoCl₄]²⁻ is intensely blue, [Co(H₂O)₆]²⁺ pale pink.
- Spin–orbit coupling mixes states of different multiplicity, so spin-forbidden bands appear very weakly (ε1), more so for heavy metals: [Mn(H₂O)₆]²⁺ is very pale pink.
Application. the intensity ladder in the figure is used to assign spectra: a band of ε≈10⁴ is charge transfer, of ≈10² is a spin-allowed d–d band, and of 1 is spin forbidden.
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All parts of Question 4
- 4(a) LCAO; bonding, non-bonding and antibonding orbitals of H₂10 marks
- 4(b) Capillary action, contact angle and γ=12hρ gr10 marks
- 4(c) Does a KCl salt bridge eliminate the liquid junction?10 marks
- 4(d) Diamagnetic [Ni(CN)₄]²⁻ and paramagnetic [NiCl₄]²⁻10 marks
- 4(e) Why d–d transitions are forbidden, weak, and yet observed10 marks