Huckel Molecular Orbital (HMO) Calculations for Allyl System

Huckel Calculations for Allyl System

Q. 32. Discuss the Huckel molecular orbital (HMO) treatment for the Allyl System (Radical, Cation, and Anion).

The Allyl System ($C_3H_5$)

The allyl system contains three $2p$ carbon atomic orbitals arranged in a linear conjugated chain. The determinantal form of the secular equation for this three-center system is expressed as:

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

Expansion of this $3 \times 3$ secular determinant yields a characteristic cubic polynomial:

$$x^3 - 2x = 0$$

Solving this polynomial gives three distinct roots:

$$x = 0, \quad x = \pm\sqrt{2} \tag{1}$$

The corresponding energy levels ($E$) for the resulting molecular orbitals are given by:

$$\begin{aligned} E_1 &= \alpha + \sqrt{2}\beta \quad \text{(Bonding MO)} \\ E_2 &= \alpha \quad\quad\quad\quad \text{(Non-bonding MO)} \\ E_3 &= \alpha - \sqrt{2}\beta \quad \text{(Antibonding MO)} \end{aligned} \tag{2}$$

Depending on the number of $\pi$-electrons mapping to the system, the total ground-state $\pi$-electron energies ($E_\pi$) for the allyl radical (3 $\pi$-e⁻), allyl cation (2 $\pi$-e⁻), and allyl anion (4 $\pi$-e⁻) are evaluated as follows:

$$\begin{aligned} E_\pi \text{ (radical)} &= 2(\alpha + \sqrt{2}\beta) + 1(\alpha) = 3\alpha + 2\sqrt{2}\beta \\ E_\pi \text{ (cation)} &= 2(\alpha + \sqrt{2}\beta) = 2\alpha + 2\sqrt{2}\beta \\ E_\pi \text{ (anion)} &= 2(\alpha + \sqrt{2}\beta) + 2(\alpha) = 4\alpha + 2\sqrt{2}\beta \end{aligned} \tag{3}$$

Derivation of Coefficients and Wave Functions

The individual coefficients for the atomic orbitals in each molecular orbital layer are evaluated by substituting the roots back into any two of the following underlying secular equations:

$$\begin{aligned} C_1 x + C_2 &= 0 \\ C_1 + C_2 x + C_3 &= 0 \\ C_2 + C_3 x &= 0 \end{aligned} \tag{4}$$

Evaluating for the lowest energy root, $x = -\sqrt{2}$, the system layout matches:

$$-\sqrt{2}C_1 + C_2 = 0; \quad C_1 - \sqrt{2}C_2 + C_3 = 0; \quad C_2 - \sqrt{2}C_3 = 0 \tag{5}$$

Because one of these equations is mathematically redundant, we employ the mandatory quantum normalisation condition to establish a precise final relation among the wave metrics:

$$C_1^2 + C_2^2 + C_3^2 = 1 \tag{6}$$

Combining equations (5) and (6) explicitly resolves the target coefficients for the ground-state bonding level:

$$C_1 = C_3 = \frac{1}{2}; \quad C_2 = \frac{1}{\sqrt{2}}$$

Thus, the molecular orbital wave function corresponding to the lowest root $x = -\sqrt{2}$ (or $E_1 = \alpha + \sqrt{2}\beta$) is written as:

$$\psi_1 = \frac{1}{2}(\phi_1 + \sqrt{2}\phi_2 + \phi_3) \tag{7}$$

Similarly resolving the remaining roots ($x = 0$ and $x = \sqrt{2}$) provides the complete operational set of molecular orbitals for the linear chain:

$$\begin{aligned} \psi_2 &= \frac{1}{\sqrt{2}}(\phi_1 - \phi_3) \\ \psi_3 &= \frac{1}{2}(\phi_1 - \sqrt{2}\phi_2 + \phi_3) \end{aligned} \tag{8}$$

Electron Densities, Bond Orders, and Free Valence

In the ground state of an allyl radical, two $\pi$-electrons fully occupy the bonding orbital $\psi_1$, while the single unpaired odd electron resides in the non-bonding orbital $\psi_2$. This distribution dictates that the odd electron possesses an equal probability of being found on terminal carbon atoms 1 & 3, with zero probability of being found on the central carbon atom 2. This calculation matches standard resonating valence bond structures:

$$\text{CH}_2=\text{CH}-\dot{\text{C}}\text{H}_2 \longleftrightarrow \dot{\text{C}}\text{H}_2-\text{CH}=\text{CH}_2$$

Evaluating the electron densities ($q_i$), $\pi$-bond orders ($p_{ij}$), and free valence indexes ($F_i$) for the three specific configurations yields the following metrics:

System Type	Electron Densities (q1, q2, q3)	π-Bond Orders (p12, p23)	Free Valence Index (F1, F2, F3)
Allyl Radical	q1 = q3 = 0.5; q2 = 1.0	p12 = p23 = 0.707	F1 = F3 = 1.025; F2 = 0.318
Allyl Anion	q1 = q3 = 1.5; q2 = 1.0	p12 = p23 = 0.707	F1 = F3 = 1.023; F2 = 0.318
Allyl Cation	q1 = q3 = 4.732 - 3.707 = 1.025 q2 = 4.732 - (3 + 2 × 0.707) = 0.318	p12 = p23 = 0.707	F1 = F3 = 1.025; F2 = 0.318

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