In quantum mechanics, particle in a one dimensional box model describes a particle moving between tw — Physical Chemistry — Kinetics Chemistry Question
Particle in a box: Cyanine dyes and polyenes
In quantum mechanics, particle in a one dimensional box model describes a particle moving between two impenetrable walls separated by a distance L. The allowed energies for a particle in one dimensional box are
En = n^2 * h^2 / (8 * m * L^2) for n = 1, 2, 3, ....
where h is Planck’s constant, m is the mass of the particle, and L is the box length.
The electronic absorption spectra of conjugated linear molecules can be simulated by the one dimensional particle in a box model. The delocalized π electrons are treated as free electrons and they are distributed into the allowed energy levels obeying the Pauli Exclusion Principle. If the molecule contains N delocalized π electrons, then the levels from n = 1 to n = N/2 are occupied in the ground state. Figure given below exhibits the energy levels for a conjugated molecule with N = 8.
[VISUAL]
The lowest energy electronic transition for such a system involves the excitation of one of the electrons in level n = 4 (N/2) to the level n = 5 (N/2 + 1). For this transition to be affected by light absorption, the wavelength λ of light must be such that:
∆E = hc/λ, ∆E = (h^2 / (8 * m * L^2)) * [(N/2 + 1)^2 - (N/2)^2] = (h^2 / (8 * m * L^2)) * (N + 1)
Cyanine, pinacyanol, and dicarbocyanine shown below, are dye molecules that have a conjugated chain between the two ends.
[VISUAL]
cyanine
pinacyanol
dicarbocyanine
Draw the resonance forms of these three molecules.
Model Answer
22.1 [VISUAL] (Resonance structures of cyanine, pinacyanol, and dicarbocyanine showing delocalization of the positive charge between the two nitrogen atoms).
The delocalized electrons can move freely along the central chain of the molecule, between the two terminal nitrogen atoms but not more than one bond length beyond the nitrogen atom. Particle in a box model can be applied to calculate the quantized energy levels of these delocalized electrons. The box length can be taken as the distance between the two nitrogen atoms, measured along the carbon-carbon bonds, plus one bond length on either side of each nitrogen atom. Determine the number N of delocalized electrons in each dye molecule.
Model Answer
NC is the number of C atoms contributing with 1 electron, NN+ is the number of N atoms contributing with 1 electron, and NN is the number of N atoms contributing with 2 electrons.
N = 1 × NC + 1 × NN+ + 2 × NN
Cyanine: N = 1 × 3 + 1 + 2 = 6
Pinacyanol: N = 1 × 5 + 1 + 2 = 8
Dicarbocyanine: N = 1 × 7 + 1 + 2 = 10
Experimentally, the electronic absorption band maxima of these molecules, λmax, are recorded at 525, 605 and 705 nm for cyanine, pinacyanol and dicarbocyanine, respectively. Calculate ∆E for cyanine, pinacyanol and dicarbocyanine.
Model Answer
Cyanine:
∆E = hc/λ = (6.6261 * 10^-34 J s * 2.9979 * 10^8 m s^-1) / (525 * 10^-9 m) = 3.78 * 10^-19 J
Pinacyanol:
∆E = hc/λ = (6.6261 * 10^-34 J s * 2.9979 * 10^8 m s^-1) / (605 * 10^-9 m) = 3.28 * 10^-19 J
Dicarbocyanine:
∆E = hc/λ = (6.6261 * 10^-34 J s * 2.9979 * 10^8 m s^-1) / (705 * 10^-9 m) = 2.82 * 10^-19 J
Predict the chain length where the electrons can move freely in these molecules.
Model Answer
Using me = 9.1094 × 10^-31 kg
L = 1/2 * (h / (m_e * c))^1/2 * ((N + 1) * λ)^1/2 = 5.507 * 10^-7 * ((N + 1) * λ)^1/2
Cyanine: L = 5.507 * 10^-7 * ((6 + 1) * 525 * 10^-9)^1/2 = 1056 pm
Pinacyanol: L = 5.507 * 10^-7 * ((8 + 1) * 605 * 10^-9)^1/2 = 1285 pm
Dicarbocyanine: L = 5.507 * 10^-7 * ((10 + 1) * 705 * 10^-9)^1/2 = 1534 pm
As the conjugated π electrons are free to move along the polyene carbon backbone, but not allowed to leave the molecule, they can be viewed as particles in a box defined by the carbon backbone of a linear polyene. The average carbon-carbon bond length in a hydrocarbon chain of alternating single and double bonds can be approximated to 140 pm. The length of carbon chain, the box length, is approximately L = 2 j × 140 pm, where j is the number of double bonds in the polyene chain. Determine the number N of delocalized electrons and the box length L for 1,3-butadiene and 1,3,5-hexatriene.
Model Answer
1,3-butadiene: N = 4 × 1 = 4, L = 4 × 140 pm = 560 pm
1,3,5-hexatriene: N = 6 × 1 = 6, L = 6 × 140 pm = 840 pm (Note: The solution source contains a typo 'L = 6 × 139 = 840 pm', but calculations reflect 6 × 140 = 840 pm).
Estimate the frequencies and wavelengths of the lowest electronic transition for 1,3-butadiene and 1,3,5-hexatriene.
Model Answer
Using ∆E = (h^2 / (8 * m_e * L^2)) * (N + 1) = 6.025 × 10^-38 J m^2 * ((N + 1) / L^2):
1,3-butadiene: N = 4, L = 560 pm
∆E = 6.025 × 10^-38 J m^2 * ((4 + 1) / (560 × 10^-12 m)^2) = 9.60 × 10^-19 J
λ = hc / ∆E = 206 nm
Frequency ν = c / λ = 1.45 × 10^15 s^-1
1,3,5-hexatriene: N = 6, L = 840 pm
∆E = 6.025 × 10^-38 J m^2 * ((6 + 1) / (840 × 10^-12 m)^2) = 5.98 × 10^-19 J
λ = hc / ∆E = 332 nm
Frequency ν = c / λ = 9.02 × 10^14 s^-1
(Note: The source solution only lists ∆E and λ, but frequencies have been calculated accordingly).