Huckel Molecular Orbital Theory - Quantum Mechanics for Chemistry

Conjugated systems¶

Molecules with extensive $\pi$ bonding systems, such as benzene, are not described very well by valence bond theory because the $\pi$ electrons are delocalized over the whole molecule. The $\sigma$ and $\pi$ bonds are illustrated below for ethylene ( $C_2H_4$ , with $sp^2$ carbons):

Sigma and pi bonding in ethylene — Fig. 1 The $\sigma$ framework and the delocalized $\pi$ system of ethylene ( $C_2H_4$ ).

We have chosen the $z$ -axis along the internuclear axis. Because both $\sigma$ and $\pi$ bonding occur between the two carbon atoms, we say that this is a double bond. The hybrid orbitals here also explain the geometry. For triple bonds, one $\sigma$ and two $\pi$ bonds are formed.

Huckel MO theory¶

Huckel molecular orbital theory assumes that the $\pi$ electrons, which are responsible for the special properties of conjugated and aromatic hydrocarbons, do not interact with one another, and that the total wavefunction is just a product of the one-electron molecular orbitals. The $\pi$ molecular orbital of the two carbons in $C_2H_4$ can be written approximately as:

{\psi = c_1\phi_1 + c_2\phi_2}

(1)

where $\phi_1$ and $\phi_2$ are the $2p_y$ atomic orbitals for carbons 1 and 2, respectively. Using the variational principle gives the following secular determinant:

{\begin{vmatrix} H_{11} - ES_{11} & H_{12} - ES_{12}\\ H_{21} - ES_{21} & H_{22} - ES_{22}\\ \end{vmatrix} = 0 \textnormal{ with } H_{ij} = \int\phi_i^*H\phi_jd\tau \textnormal{ and } S_{ij} = \int\phi_i^*\phi_jd\tau}

(2)

With these rules, the secular determinant for ethylene becomes:

{\begin{vmatrix}\alpha - E & \beta\\ \beta & \alpha - E\\ \end{vmatrix} = 0}

(3)

In Huckel theory, the Coulomb integral $\alpha$ and the resonance integral $\beta$ are regarded as empirical parameters. They can be obtained, for example, from experimental data. Thus, in Huckel theory it is not necessary to specify the Hamiltonian operator. Expansion of the determinant leads to a quadratic equation for $E$ , with solutions $E = \alpha \pm \beta$ . In general, it can be shown that $\beta < 0$ , which implies that the lowest orbital energy is $E_1 = \alpha + \beta$ . There are two $\pi$ electrons, and therefore the total energy is $E_{tot} = 2E_1 = 2\alpha + 2\beta$ . Do not confuse $\alpha$ and $\beta$ here with electron spin.

Solving for the Huckel MOs¶

The wavefunctions (the coefficients $c_1$ and $c_2$ ) can be obtained by substituting the two values of $E$ into the original linear equations:

{c_1(\alpha - E) + c_2\beta = 0}

(4)

{c_1\beta + c_2(\alpha - E) = 0}

(5)

For the lowest energy orbital ( $E_1 = \alpha + \beta$ ), we get (including normalization):

{\psi_1 = \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) \hspace{0.5cm}(\textnormal{i.e. }c_1 = c_2 = \frac{1}{\sqrt{2}})}

(6)

and for the highest energy orbital ( $E_2 = \alpha - \beta$ ) (including normalization):

{\psi_2 = \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) \hspace{0.5cm}(\textnormal{i.e. }c_1 = \frac{1}{\sqrt{2}}, c_2 = -\frac{1}{\sqrt{2}})}

(7)

These orbitals resemble the $H_2^+$ LCAO MOs discussed previously. This also gives us an estimate for one of the excited states, where one electron is promoted from the bonding to the antibonding orbital. The excitation energy is found to be $2|\beta|$ , which allows, for example, the estimation of $\beta$ from UV/VIS absorption spectroscopy.

HOMO orbital	The highest occupied molecular orbital.
LUMO orbital	The lowest unoccupied molecular orbital.

Example: 1,3-butadiene¶

Let us calculate the $\pi$ electronic energy for 1,3-butadiene ( $CH_2=CHCH=CH_2$ ) using Huckel theory. First we write the secular determinant using the rules given earlier. To do this, it is convenient to number the carbon atoms in the molecule:

\overset{1}{\textnormal{CH}_2} = \overset{2}{\textnormal{CH}} - \overset{3}{\textnormal{CH}} = \overset{4}{\textnormal{CH}_2}

(8)

In this case there are two scenarios to consider:

A localized solution where the $\pi$ electrons are shared either between atoms 1 and 2 or between atoms 3 and 4. This implies that the $\beta$ parameter should not be written between nuclei 2 and 3.
A delocalized solution where the $\pi$ electrons are delocalized over all four carbons. This implies that the $\beta$ parameter should be written between nuclei 2 and 3.

Here it turns out that scenario 2 gives a lower energy solution, and we study that in more detail. In general, however, both cases should be considered. The energy difference between scenarios 1 and 2 is called the resonance stabilization energy. The secular determinant is:

{\begin{matrix} 1\\2\\3\\4\\ \end{matrix}\begin{vmatrix} \alpha - E & \beta & 0 & 0\\ \beta & \alpha - E & \beta & 0\\ 0 & \beta & \alpha - E & \beta\\ 0 & 0 & \beta & \alpha - E\\ \end{vmatrix} = 0}

(9)

To simplify the notation, we divide each row by $\beta$ and denote $x = (\alpha - E) / \beta$ :

{\begin{vmatrix} x & 1 & 0 & 0\\ 1 & x & 1 & 0\\ 0 & 1 & x & 1\\ 0 & 0 & 1 & x\\ \end{vmatrix} = 0}

(10)

Expansion of this determinant gives $x^4 - 3x^2 + 1 = 0$ . There are four solutions, $x = \pm 0.618$ and $x = \pm 1.618$ . Thus there are four possible orbital energy levels:

{E_1 = \alpha + 1.618\beta \hspace{0.5cm}\textnormal{(lowest energy)}} \\ {E_2 = \alpha + 0.618\beta} \\ {E_3 = \alpha - 0.618\beta} \\ {E_4 = \alpha - 1.618\beta \hspace{0.5cm}\textnormal{(highest energy)}}

(11)

Huckel molecular orbitals of butadiene — Fig. 2 The four Huckel $\pi$ molecular orbitals of 1,3-butadiene, ordered by energy.

There are four $\pi$ electrons, which occupy the two lowest energy orbitals. This gives the total $\pi$ electronic energy of the molecule:

{E_\pi = 2(\alpha + 1.618\beta) + 2(\alpha + 0.618\beta) = 4\alpha + 4.472\beta}

(12)

and the lowest excitation energy is $1.236|\beta|$ .

The four Huckel MO wavefunctions are (calculations not shown):

{\psi_1 = 0.372\phi_1 + 0.602\phi_2 + 0.602\phi_3 + 0.372\phi_4}

(13)

{\psi_2 = 0.602\phi_1 + 0.372\phi_2 - 0.372\phi_3 - 0.602\phi_4}

(14)

{\psi_3 = 0.602\phi_1 - 0.372\phi_2 - 0.372\phi_3 + 0.602\phi_4}

(15)

{\psi_4 = 0.372\phi_1 - 0.602\phi_2 + 0.602\phi_3 - 0.372\phi_4}

(16)

Example: the benzene molecule¶

Let us now apply the Huckel method to the benzene molecule. The secular determinant for benzene is (electrons delocalized):

{ \begin{vmatrix} \alpha - E & \beta & 0 & 0 & 0 & \beta\\ \beta & \alpha - E & \beta & 0 & 0 & 0\\ 0 & \beta & \alpha - E & \beta & 0 & 0\\ 0 & 0 & \beta & \alpha - E & \beta & 0\\ 0 & 0 & 0 & \beta & \alpha - E & \beta\\ \beta & 0 & 0 & 0 & \beta & \alpha - E\\ \end{vmatrix} = 0 }

(17)

The solutions are (showing where the six $\pi$ electrons should be placed):

{E_1 = \alpha + 2\beta~~\textnormal{(lowest energy)}} \\ {E_2 = E_3 = \alpha + \beta} \\ {E_4 = E_5 = \alpha - \beta} \\ {E_6 = \alpha - 2\beta~~\textnormal{(highest energy)}}

(18)

Huckel molecular orbitals of benzene — Fig. 3 The six Huckel $\pi$ molecular orbitals of benzene. The doubly degenerate pairs reflect the high symmetry of the ring.

The six $\pi$ electrons fill the lowest three levels ( $E_1$ and the degenerate pair $E_2 = E_3$ ), giving a total $\pi$ energy of $6\alpha + 8\beta$ . Compared with three isolated ethylene double bonds ( $6\alpha + 6\beta$ ), benzene is stabilized by an extra $2\beta$ , which is the origin of its aromatic stability.