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Theme 3 — Postulates and Applications

Hückel model of the allyl radical

Keywords: Hückel model · quantum chemistry · diagonalisation · molecular orbitals · delocalisation energy · electron populations

Exercise 1 : Hückel model of the allyl radical

The allyl radical C3H5\mathrm{C_3H_5} has three carbon atoms arranged in a chain and numbered 11, 22, 33, each carrying a pzp_z orbital and one π\pi electron. In the Hückel model, a π\pi electron is described in the space spanned by three orthonormal states ∣1⟩,∣2⟩,∣3⟩\ket1,\ket2,\ket3 localised on the three carbon atoms, with the Hamiltonian

H^=(α−J0−Jα−J0−Jα),\begin{aligned} \hat H=\begin{pmatrix}\alpha&-J&0\\-J&\alpha&-J\\0&-J&\alpha\end{pmatrix}, \end{aligned}

where α\alpha is the site energy and J>0J>0 the coupling between neighbouring carbon atoms. Carbon atoms 11 and 33 are not neighbours. Assume that each orbital can accommodate at most two electrons with opposite spins, and that the total energy in the independent-electron model is the sum of the orbital energies.

Question 1
Determine the eigenenergies εk\varepsilon_k and the normalised eigenvectors of H^\hat H.

Hint
Write H^=α1−JM\hat H=\alpha\mathbf{1}-JM and find the eigenvalues of the matrix MM by solving its characteristic equation.

Solution
We write H^=α1−JM\hat H=\alpha\mathbf{1}-JM with

M=(010101010).\begin{aligned} M=\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}. \end{aligned}

The eigenvectors of H^\hat H are those of MM, and if M∣v⟩=λ∣v⟩M\ket v=\lambda\ket v, then H^∣v⟩=(α−Jλ)∣v⟩\hat H\ket v=(\alpha-J\lambda)\ket v. The characteristic polynomial of MM is

det⁡(M−λ1)=−λ(λ2−1)−1⋅(−λ)=−λ3+2λ=−λ(λ2−2),\det(M-\lambda\mathbf{1})=-\lambda(\lambda^2-1)-1\cdot(-\lambda)=-\lambda^3+2\lambda=-\lambda(\lambda^2-2),

whose roots are λ=2\lambda=\sqrt2, 00, −2-\sqrt2. For each one, we solve Mv=λvM\mathbf v=\lambda\mathbf v, that is, v2=λv1v_2=\lambda v_1, v1+v3=λv2v_1+v_3=\lambda v_2, v2=λv3v_2=\lambda v_3. For λ=±2\lambda=\pm\sqrt2, we obtain v1=v3v_1=v_3 and v2=±2v1v_2=\pm\sqrt2v_1; for λ=0\lambda=0, v2=0v_2=0 and v3=−v1v_3=-v_1. After normalisation:

orbitalcoefficients on (1,2,3)energyχ112(1, 2, 1)ε1=α−2Jχ212(1, 0, −1)ε2=αχ312(1, −2, 1)ε3=α+2J\begin{aligned} \begin{array}{c|c|c} \text{orbital}&\text{coefficients on }(1,2,3)&\text{energy}\\ \hline \chi_1&\frac12(1,\ \sqrt2,\ 1)&\varepsilon_1=\alpha-\sqrt2J\\[2pt] \chi_2&\frac1{\sqrt2}(1,\ 0,\ -1)&\varepsilon_2=\alpha\\[2pt] \chi_3&\frac12(1,\ -\sqrt2,\ 1)&\varepsilon_3=\alpha+\sqrt2J \end{array} \end{aligned}

The orbital χ1\chi_1 has coefficients that are all of the same sign: it is bonding. The orbital χ3\chi_3 changes sign between each pair of neighbours: it is antibonding. The orbital χ2\chi_2, with energy α\alpha, has a node on the central carbon atom: it is said to be non-bonding. The sum of the energies is 3α3\alpha, equal to the trace of H^\hat H.

Question 2
Write the ground-state configuration of the three π\pi electrons in the allyl radical, and its energy EπE_\pi. Compare it with the energy of a system consisting of an isolated double bond, described by two carbon atoms coupled by −J-J, and an uncoupled electron of energy α\alpha. Hence deduce the delocalisation energy.

Solution
The orbitals are filled in order of increasing energy: two electrons in χ1\chi_1 and one electron in χ2\chi_2. The energy is

Eπ=2(α−2J)+α=3α−22J.E_\pi=2(\alpha-\sqrt2J)+\alpha=3\alpha-2\sqrt2J .

For the localised reference system, a pair of coupled carbon atoms is described by the matrix (α−J−Jα)\begin{pmatrix}\alpha&-J\\-J&\alpha\end{pmatrix}, with eigenvalues α∓J\alpha\mp J. Its two electrons occupy the level α−J\alpha-J, and the isolated electron has energy α\alpha: the total energy is 2(α−J)+α=3α−2J2(\alpha-J)+\alpha=3\alpha-2J. The delocalisation energy is the difference

Eπ−(3α−2J)=−2(2−1)J≃−0,83 J.E_\pi-(3\alpha-2J)=-2(\sqrt2-1)J\simeq-0{,}83\,J .

The delocalisation of the electrons over the three carbon atoms lowers the energy, which explains the particular stability of the allyl radical.

Question 3
Calculate the mean π\pi-electron population of each carbon atom, defined by Nj=∑knk∣⟨j|χk⟩∣2N_j=\sum_kn_k|\braket{j}{\chi_k}|^2, where nkn_k is the number of electrons in the orbital χk\chi_k. Answer the same question for the allyl cation C3H5+\mathrm{C_3H_5^+}, which has two π\pi electrons. Comment on the result.

Solution
For the radical, n1=2n_1=2, n2=1n_2=1, n3=0n_3=0. The squared moduli of the coefficients are (14,12,14)(\frac14,\frac12,\frac14) for χ1\chi_1 and (12,0,12)(\frac12,0,\frac12) for χ2\chi_2. Thus

N1=2⋅14+1⋅12=1,N2=2⋅12+0=1,N3=1.N_1=2\cdot\frac14+1\cdot\frac12=1,\qquad N_2=2\cdot\frac12+0=1,\qquad N_3=1 .

On average, each carbon atom carries one π\pi electron. For the cation, only χ1\chi_1 is occupied, by two electrons:

N1=12,N2=1,N3=12.N_1=\frac12,\qquad N_2=1,\qquad N_3=\frac12 .

Carbon atoms 11 and 33 have each lost half an electron: the cation's positive charge is distributed equally over the two terminal carbon atoms, rather than over the central carbon atom. This result is well known to chemists, who represent it by two limiting resonance structures. The model reproduces it simply: the removed electron occupied the non-bonding orbital χ2\chi_2, which has a node on the central carbon atom.

Question 4
Calculate the π\pi bond order between carbon atoms 11 and 22, defined by p12=∑knk⟨1|χk⟩⟨χk|2⟩p_{12}=\sum_kn_k\braket{1}{\chi_k}\braket{\chi_k}{2}, for the allyl radical. Compare it with that of ethylene, consisting of two coupled carbon atoms with two electrons in the bonding orbital. Interpret the result.

Solution
Only χ1\chi_1 contributes, since the coefficient of χ2\chi_2 on carbon atom 22 vanishes:

p12=2⋅12⋅22+1⋅12⋅0=22≃0,71.p_{12}=2\cdot\frac12\cdot\frac{\sqrt2}{2}+1\cdot\frac{1}{\sqrt2}\cdot0=\frac{\sqrt2}{2}\simeq0{,}71 .

By symmetry, p23=p12p_{23}=p_{12}. For ethylene, the bonding orbital is (∣1⟩+∣2⟩)/2(\ket1+\ket2)/\sqrt2, and p12=2⋅12⋅12=1p_{12}=2\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}=1, which corresponds to a complete double bond. In allyl, the π\pi electrons are distributed over the two bonds, each of which has a character intermediate between a single and a double bond: the two carbon—carbon bonds are equivalent, as confirmed experimentally by their equal lengths.