Huckel Molecular Orbital (HMO) Calculations for Ethylene

Huckel Calculations for Ethylene ($C_2H_4$)

Ethylene is considered a simple conjugated system containing two interacting $p$-orbitals perpendicular to the molecular plane. The interaction of these atomic orbitals forms a $\pi$-electron system. The determinantal form of the secular equation for this system is given by:

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

The expansion of this secular determinant yields a direct quadratic algebraic equation:

$$x^2 - 1 = 0 \quad \therefore \quad x = \pm 1 \tag{1}$$

In Huckel theory, these roots correspond to the energy levels of the molecular systems. Solving for energy ($E$), we find two distinct states:

$$\begin{aligned} E_1 &= \alpha + \beta \quad \text{(Bonding Molecular Orbital)} \\ E_2 &= \alpha - \beta \quad \text{(Antibonding Molecular Orbital)} \end{aligned} \tag{2}$$

The two available $\pi$-electrons in ethylene occupy the lower energy molecular orbital ($E_1$) in the ground state. The total $\pi$-electron energy ($E_\pi$) is computed as follows:

$$E_\pi = 2(\alpha + \beta) = 2\alpha + 2\beta$$

Since the baseline energy of two isolated electrons residing in unhybridized carbon $2p$ orbitals is $2\alpha$, the net $\pi$-bond stabilization energy (delocalization/resonance energy) in ethylene is exactly $2\beta$.

Derivation of Coefficients and Wave Functions

The coefficients of the atomic orbitals in each molecular orbital are obtained by substituting the values of $x$ back into the secular equations:

$$\begin{aligned} C_1 x + C_2 &= 0 \\ C_1 + C_2 x &= 0 \end{aligned} \tag{3}$$

For the ground state where $x = 1$, substituting this value into equation (3) gives:

$$C_1 + C_2 = 0 \quad \text{or,} \quad C_1 = -C_2 \tag{4}$$

Since any valid molecular orbital wave function must be normalized, the normalisation requirement dictates that:

$$\psi = C_1 \phi_1 + C_2 \phi_2$$

Integrating the square of the wave function over all space gives:

$$C_1^2 \int \phi_1^2 d\tau + C_2^2 \int \phi_2^2 d\tau + 2C_1C_2 \int \phi_1 \phi_2 d\tau = 1$$

When orbital overlap ($S_{12} = \int \phi_1 \phi_2 d\tau$) is neglected ($S_{12} = 0$) and individual atomic orbitals are normalized ($\int \phi_i^2 d\tau = 1$), this reduces seamlessly to:

$$C_1^2 + C_2^2 = 1 \tag{5}$$

Combining equations (4) and (5), we can solve for the unique normalization coefficients:

$$C_1 = \frac{1}{\sqrt{2}}, \quad C_2 = -\frac{1}{\sqrt{2}}$$

Hence, the molecular orbital wave function corresponding to the higher antibonding energy level $E_2 = \alpha - \beta$ is given by:

$$\psi_2 = \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) \tag{6}$$

Similarly, evaluating the system for the bonding condition ($x = -1$) yields matching coefficients, meaning the molecular orbital whose energy is $E_1 = \alpha + \beta$ is given by:

$$\psi_1 = \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) \tag{7}$$

Electron Densities, Bond Orders, and Free Valence

Since both $\pi$-electrons occupy the ground-state molecular orbital $\psi_1$, the individual $\pi$-electron densities ($q_1, q_2$) and the resulting net $\pi$-bond order ($p_{12}$) evaluate uniformly to unity:

$$q_1 = q_2 = 1; \quad p_{12} = 1 \tag{8}$$

In an ethylene molecule, each carbon atom is structurally bound to two hydrogen atoms via $\sigma$-bonds. If we assign a local bond order value of unity to each distinct $C-H$ covalent linkage, the total bond order value ($N$) associated with each carbon atom is four ($2 \times \sigma_{\text{C-H}} + 1 \times \sigma_{\text{C-C}} + 1 \times \pi_{\text{C-C}} = 4$). Following the standard mathematical definition of free valence index ($F$), we find:

$$F_1 = F_2 = 0.732 \tag{9}$$

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