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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper II

Q8(e). Charge-transfer spectra of donor–acceptor complexes

  • 10 marks
  • Answered
Question

Explain charge-transfer spectra by taking a suitable donor-acceptor type complex.

Answer in brief

A CT spectrum is the absorption of light that moves an electron from a donor to an acceptor in their complex; its energy follows hν_(CT) ≈ I_D - E_A - C. Benzene–I₂ (brown, new UV band) and TCNE–arene complexes (coloured) are typical; lower donor ionisation energy gives longer CT wavelength.

Key points

  • ground state Ψ_N = a ψ(D,A) + b ψ(D⁺A⁻) with a ≫ b (mostly no-bond); excited state Ψ_E is mostly the ion pair D⁺A⁻.
  • The CT transition Ψ_N→Ψ_E transfers an electron from the HOMO of D to the LUMO of A. Its energy is hν_(CT) ≈ I_D - E_A - C (I_D ionisation energy of D, E_A electron affinity of A, C Coulomb attraction in D⁺A⁻).
  • Iodine is violet in CCl₄ but brown in benzene. Benzene (π donor) and I₂ (σ* acceptor) form a 1: 1 complex with a strong new band in the near UV (near 290–300 nm), absent in both components.
  • With tetracyanoethylene (strong π acceptor) and methylbenzenes the CT band moves into the visible and the solutions are coloured; with more methyl groups (lower I_D) the band shifts to longer wavelength: a straight line of…

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. When an electron donor D (low ionisation energy, high-lying HOMO) meets an electron acceptor A (high electron affinity, low-lying LUMO), a weak charge-transfer (CT) complex D·A forms. It shows a new, broad, intense absorption band that neither D nor A has alone: the CT band, in which light moves an electron from D to A.

  1. Mulliken theory: ground state Ψ_N = a ψ(D,A) + b ψ(D⁺A⁻) with a ≫ b (mostly no-bond); excited state Ψ_E is mostly the ion pair D⁺A⁻.
  2. The CT transition Ψ_N→Ψ_E transfers an electron from the HOMO of D to the LUMO of A. Its energy is hν_(CT) ≈ I_D - E_A - C (I_D ionisation energy of D, E_A electron affinity of A, C Coulomb attraction in D⁺A⁻).

Figure: The CT transition (red) needs less energy than either partner's own transition (dashed).

Example: benzene (or hexamethylbenzene) with iodine

  1. Iodine is violet in CCl₄ but brown in benzene. Benzene (π donor) and I₂ (σ* acceptor) form a 1: 1 complex with a strong new band in the near UV (near 290–300 nm), absent in both components.
  2. With tetracyanoethylene (strong π acceptor) and methylbenzenes the CT band moves into the visible and the solutions are coloured; with more methyl groups (lower I_D) the band shifts to longer wavelength: a straight line of hν_(CT) against I_D is the proof of charge transfer.
  3. CT bands are broad (the complex has many geometries) and intense (ε of the order of 10³–10⁴), because the electron moves a long distance (large transition dipole).

Conclusion. A CT spectrum is the absorption of light that moves an electron from a donor to an acceptor in their complex; its energy follows hν_(CT) ≈ I_D - E_A - C. Benzene–I₂ (brown, new UV band) and TCNE–arene complexes (coloured) are typical; lower donor ionisation energy gives longer CT wavelength.

Application. CT complexes explain the colours of picrates used to identify arenes, starch–iodine, organic conductors (TTF–TCNQ) and many inorganic colours (MnO₄⁻, ligand-to-metal CT); CT bands are used to measure ionisation energies of donors.

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Model answer to Q8(e), 2018 Paper II: solution page 22
Solution page 22 of 24