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Absorption spectra of conjugated dyes: a particle in a box

Measure the absorption maxima of nine cyanine dyes in three families, calculate them with a one-dimensional particle in a box, then fit the box length by least squares — and see how well the simplest quantum model describes nine measured peaks.

Difficulty ●●○ ≈ 30 min

Aim Determine the absorption maxima of cyanine dyes, test the free-electron (particle-in-a-box) model, and fit the box length per bond ℓ and the end correction γ.
Test the free-electron model against nine measured absorption maxima

The electron box inside a dye

SSNNLUMO (n = 5)HOMO (n = 4)
Thiacarbocyanine (3-carbon bridge): 5 chain carbons, 8 π electrons; box 11.1 Å long
model400500600700800Wavelength / nm00.20.40.6Absorbance
Model
453 nm
Measured
559 nm
Error
−106 nm

Root-mean-square error over this family’s three dyes: 82.5 nm

Thiacarbocyanine (3-carbon bridge): 559 nmThiadicarbocyanine (5-carbon bridge): 655.8 nmThiatricarbocyanine (7-carbon bridge): 763.2 nmOxacarbocyanine (3-carbon bridge): 485 nmOxadicarbocyanine (5-carbon bridge): 582 nmOxatricarbocyanine (7-carbon bridge): 687.5 nmIndocarbocyanine (3-carbon bridge): 544.2 nmIndodicarbocyanine (5-carbon bridge): 646 nmIndotricarbocyanine (7-carbon bridge): 742.2 nm68101214π electrons N3004005006007008009001000Absorption maximum λmax / nm
Absorption maxima and molar absorptivities: the PhotochemCAD database (Du et al. 1998; Dixon et al. 2005, compiled at omlc.org), from Sims et al. (1974), Tredwell & Keary (1979) and Waggoner et al. (1989). Solution colours are computed from the absorption spectrum through CIE 1931 colour matching; the band shape (main band plus vibronic shoulder) is a teaching fit, while peak positions and heights are measured. Model: λ = 8mₑcL²/(h(N+1)), L = (p + 3 + γ)ℓ.

Procedure

  1. Make each dye about 5 µmol/L in alcohol, zero the instrument on the same solvent, and scan the absorption spectrum from 400 to 850 nm.
  2. Read each band’s wavelength of maximum absorption, λmax (the main peak, not the shoulder on its short-wavelength side), and note the colour.
  3. For each dye count the chain carbons p between the nitrogens and the π electrons N = p + 3; calculate λ with the textbook box (ℓ = 1.39 Å, γ = 0) and compare with the measurement.
  4. Plot √(λ(N + 1)/K) against p + 3 (K = 8mc/h = 32.97 nm Å⁻²) and find the slope ℓ and intercept ℓγ by least squares; fit each family separately.
  5. Use the fitted parameters to predict the absorption maximum of each family’s next member (a nine-carbon bridge), and discuss what ℓ and γ mean physically.

What to notice

  • Each extra –CH=CH– unit moves the peak about 100 nm to longer wavelength: the solutions go from magenta to blue to a pale green as the band moves into the near infrared.
  • The textbook box predicts exactly the same wavelengths for all three families (453, 579 and 706 nm) — the model knows nothing about the end groups. It falls up to 19 % short for the thiacyanines, and is about 3 % too long for the longest oxacyanine.
  • After linearising, the points fall on straight lines; all three families give a slope ℓ of about 1.24 Å, close to the projection of a zigzag bond on the chain axis, 1.39 × cos 30° = 1.20 Å.
  • The end correction γ depends on the end group: about 1.3 for the oxacyanines, 1.9 for the thia- and indocyanines.

In a real lab

  • Cyanine dyes stain intensely and are light-sensitive: wear gloves, make solutions fresh, and keep them in the dark.
  • Keep the concentration low and use alcohol: in water or at high concentration the dye molecules aggregate (H-aggregates), shifting the band to shorter wavelength and broadening it.
  • Maxima shift by a few nanometres between solvents; when comparing model and data, measure every dye in the same solvent.
  • The shoulder on the short-wavelength side of the main band is vibrational structure (the 0–1 transition), not a second electronic transition.

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