The energy levels of π-electrons in molecules with conjugated bonds can be calculated with varying d — Physical Chemistry — Kinetics Chemistry Question
Particles in a Box Problem and Conjugated Polyenes
The energy levels of π-electrons in molecules with conjugated bonds can be calculated with varying degrees of accuracy, depending on the complexity of the model. The most sophisticated and accurate approaches involve complex theoretical methods for solving the multi-particle Schrödinger equation. A simplified yet still powerful approach is to treat the π-electrons as independent “particles in a box.” This model is useful for determining the energies of π-electrons and the electronic spectra of such molecules as ethylene or molecules with conjugated double bonds. In this problem, use the “particle in a box” model to describe the π-electron states of ethylene as well as linear and cyclic conjugated molecules.
The particle in a box model yields the energy levels for π-electrons by treating them as moving freely along the length of the conjugated π-bonds. An example of a hydrocarbon with a non-branched chain of conjugated π-bonds is trans-1,3,5-hexatriene shown below.
[VISUAL]
The allowed quantum states occur for electronic wavefunctions that have wavelengths of λ = n L /2, where n is an integer starting with n = 1 and L is the length of the molecule. The effective molecule lengths are L = 289 pm for ethylene and L = 867 pm for trans-1,3,5-hexatriene. The allowed energy states for the π-electrons are given by Eq. 1.
En = (n^2 * h^2) / (8 * me * L^2) (Eq. 1)
In Eq. 1, n is the quantum number for the energy state and is an integer between 1 and ∞, h is the Planck’s constant in J⋅s, me is the mass of the electron in kilograms and L is the length of the box in meters. Use two significant figures for your calculations.
Some molecules have cyclic conjugated π-systems. Benzene and coronene are examples of such molecules.
For molecules with “circular” π-electron distributions, the quantized energy levels are given by Eq. 2.
En = (n^2 * h^2) / (8 * me * R^2 * π^2) (Eq. 2)
In this case, the quantum number n has integer values between 0 and ±∞ and R is the radius of the ring in meters. Unlike the linear particle in a box problem, the circular problem allows for both positive and negative integer values for n for clockwise and counterclockwise motion. Also, for the circular problem, n = 0 is an eligible quantum state.
For this problem, assume that the ring radii are 139 pm for benzene and 368 pm for coronene.
Use the particle in a box model to determine the following:
i. the first two energy levels for the π-electrons in ethylene;
ii. the first four energy levels for the π-electrons in 1,3,5-hexadiene.
Model Answer
i. Ethylene states are En = n2 × 7.2 · 10–19 J
E1 = 7.2 · 10–19 J
E2 = 4 × 7.2 · 10–19 J = 2.9 · 10–18 J
ii. 1,3,5-hexatriene states are En = n2 × 8.0 · 10–20 J
E1 = 8.0 · 10–20 J
E2 = 4 × 8.0 · 10–20 J = 3.2 · 10–19 J
E3 = 9 × 8.0 · 10–20 J = 7.2 · 10–19 J
E4 = 16 × 8.0 · 10–20 J = 1.3 · 10–18 J
For each species, fill the energy levels with the π-electrons, keeping in mind the Pauli principle for electron pairing. Identify the quantum number n of the highest occupied energy level of each species.
Model Answer
Ethylene has 2 π-electrons, so the n = 1 state is the highest occupied level.
Hexatriene has 6 π-electrons and the n = 3 state is the highest occupied level.
Use the highest occupied and lowest unoccupied energy levels to predict the wavelength of light that can be used to excite a π-electron from the highest energy occupied state to the lowest energy unoccupied state for each species.
Model Answer
The energy gap between the highest occupied state and the lowest unoccupied state corresponds to the energy of the maximum wavelength of the light that can excite the π-electrons. For ethylene, ∆E = 2.2 ⋅ 10–18 J, which corresponds to λ = hc / ∆E = 92 nm. For hexatriene, ∆E = 5.6 ⋅ 10–19 J and λ = 350 nm.
The molecule in carrots that makes them appear orange is β-carotene. Use the particle in a box model to predict the energy gap between the highest occupied state and the lowest unoccupied state. Use this energy to determine the maximum wavelength for absorption of light by β-carotene. Use a length for β-carotene of L = 1850 pm.
trans-β-carotene
[VISUAL]
Model Answer
The π-electron energy levels of β-carotene have values of En = n2×1.8 ⋅ 10–20 J. β-carotene has 22 p-electrons so the energy gap for absorption of light is ∆E = E12 – E11. The wavelength for this absorption is λ = 490 nm, which is in the blue region of the visible spectrum. Absorption of blue light gives carrots their orange appearance.
Describe the benzene’s π-electron system using the particle-in-the-ring equation for energy levels. Draw a diagram depicting all occupied energy levels as well as the lowest-unoccupied energy level. When building the energy levels, keep in mind the Pauli principle for electron pairing and that there may be several states with the same energy referred to as degenerate states. Make sure that you use the right number of π electrons. Use two significant figures in your answers.
Model Answer
The quantum number n could be zero, then ±1, ±2, etc. Benzene has 6 π electrons, so the first three energy levels are occupied.
En = n2 ħ2 / 2 me R 2 = n2 × 3.2 ⋅ 10–19 J.
The energy levels will be
E0 = n2 ħ2 / 2me R2 = 0 (single),
E1 = ħ2 / 2 me R2 = 3.2 ⋅ 10–19 J (double-degenerate), the highest occupied energy level,
E2 = 4 ħ2 / 2me R2 = 1.3 ⋅ 10–18 J (double-degenerate), the lowest unoccupied energy level.
Now, draw a similar energy level diagram for coronene and calculate the quantized energy values for the occupied energy levels and the lowest unoccupied energy level. Use two significant figures in your answers.
Model Answer
En = n2 ħ2 / 2 me R 2 = n2 × 4.5 ⋅ 10–20 J.
Coronene has 24 π electrons, so the first 12 orbitals are occupied. The energy levels are
E0 = ħ2 / 2meR 2 = 0 (single),
E1 = ħ2 / 8meR 2 = 4.5 ⋅ 10–20 J (double-degenerate),
E2 = 4 ħ2 / 8meR 2= 1.8 ⋅ 10–19 J (double-degenerate),
E3 = 9 ħ2 / 8meR 2= 4.1 ⋅ 10–19 J (double-degenerate),
E4 = 16 ħ2 / 8meR 2= 7.2 ⋅ 10–19 J (double-degenerate),
E5 = 25 ħ2 / 8meR 2= 1.1 ⋅ 10–18J (double-degenerate),
E6 = 36 ħ2 / 8meR 2= 1.6 ⋅ 10–18J (double-degenerate), the highest occupied energy level,
E7 = 49 ħ2 / 8meR 2= 2.2 ⋅ 10–18J (double-degenerate), the lowest unoccupied energy level.
Calculate the energy gaps between the highest occupied and lowest unoccupied energy levels for benzene and coronene.
Model Answer
Benzene: ∆E = 9.5 ⋅ 10–19 J; coronene: ∆E = 5.9 ⋅ 10–19 J.
Predict whether benzene or coronene is colored. The recommended way is to determine the longest wavelength of light absorption in nanometers (with two significant figures) for each molecule assuming that the electronic transition responsible for it is one between highest occupied and lowest unoccupied energy levels of each particular molecule.
Model Answer
λ = c/ν = hc / ∆E for each species, yielding λ = 210 nm for benzene and λ = 340 nm for coronene. Benzene is colorless because the absorption is in the UV range (λ < 300 -350 nm). Coronene absorbs near the visible range, so it is colored.