What is the mathematical and quantum mechanical derivation of hybrid wavefunctions and bond angles for sp, sp², and sp³ hybridization?

### Quantum Mechanical Basis of Hybridization Hybridization involves the **Linear Combination of Atomic Orbitals (LCAO)** to construct new wavefunctions ($\psi$) that are orthonormal (normalized and mutually orthogonal). --- ### 1. $sp$ Hybridization Mixing one $s$ orbital ($\psi_s$) and one $p_x$ orbital ($\psi_{p_x}$) forms two equivalent hybrid wavefunctions, $\psi_1$ and $\psi_2$: $\psi_1 = a_1 \psi_s + b_1 \psi_{p_x}$ $\psi_2 = a_2 \psi_s + b_2 \psi_{p_x}$ #### Mathematical Derivation of Coefficients: 1. **s-character Distribution:** Since the $s$-orbital contribution (total electron density = 1) is distributed equally between 2 hybrid orbitals: $a_1^2 + a_2^2 = 1 \implies a_1^2 = a_2^2 = \frac{1}{2} \implies a_1 = a_2 = \frac{1}{\sqrt{2}}$ 2. **Normalization Condition ($\int \psi_1^2 \, d\tau = 1$):** $a_1^2 + b_1^2 = 1 \implies \frac{1}{2} + b_1^2 = 1 \implies b_1 = \frac{1}{\sqrt{2}}$ 3. **Orthogonality Condition ($\int \psi_1 \psi_2 \, d\tau = 0$):** $a_1 a_2 + b_1 b_2 = 0 \implies \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right) b_2 = 0 \implies b_2 = -\frac{1}{\sqrt{2}}$ #### Final Wavefunctions: $\psi_1 = \frac{1}{\sqrt{2}}\psi_s + \frac{1}{\sqrt{2}}\psi_{p_x}$ $\psi_2 = \frac{1}{\sqrt{2}}\psi_s - \frac{1}{\sqrt{2}}\psi_{p_x}$ #### Bond Angle ($\theta$): Using Coulson's Theorem ($\cos\theta = -\frac{1}{p}$, where $p$ is the $p$-character ratio): $\cos\theta = -\frac{1}{1} = -1 \implies \theta = 180^\circ \quad \text{(Linear Geometry)}$ --- ### 2. $sp^2$ Hybridization Mixing one $s$, $\psi_{p_x}$, and $\psi_{p_y}$ forms three equivalent orbitals ($\psi_1, \psi_2, \psi_3$). #### Mathematical Wavefunctions: 1. **Equal $s$-character contribution ($33.3\%$ each):** $a_1 = a_2 = a_3 = \frac{1}{\sqrt{3}}$ 2. Alignment along axes yields: $\psi_1 = \frac{1}{\sqrt{3}}\psi_s + \sqrt{\frac{2}{3}}\psi_{p_x}$ $\psi_2 = \frac{1}{\sqrt{3}}\psi_s - \frac{1}{\sqrt{6}}\psi_{p_x} + \frac{1}{\sqrt{2}}\psi_{p_y}$ $\psi_3 = \frac{1}{\sqrt{3}}\psi_s - \frac{1}{\sqrt{6}}\psi_{p_x} - \frac{1}{\sqrt{2}}\psi_{p_y}$ #### Bond Angle ($\theta$): $\cos\theta = -\frac{1}{p} = -\frac{1}{2} \implies \theta = \arccos\left(-\frac{1}{2}\right) = 120^\circ \quad \text{(Trigonal Planar)}$ --- ### 3. $sp^3$ Hybridization Mixing one $s$ and three $p$ orbitals ($\psi_{p_x}, \psi_{p_y}, \psi_{p_z}$) forms four equivalent orbitals ($\psi_1, \psi_2, \psi_3, \psi_4$). #### Mathematical Wavefunctions: 1. **Equal $s$-character contribution ($25\%$ each):** $a_1 = a_2 = a_3 = a_4 = \frac{1}{\sqrt{4}} = \frac{1}{2}$ 2. Constructing orthonormal combinations gives: $\psi_1 = \frac{1}{2}\left(\psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\right)$ $\psi_2 = \frac{1}{2}\left(\psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\right)$ $\psi_3 = \frac{1}{2}\left(\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\right)$ $\psi_4 = \frac{1}{2}\left(\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\right)$ #### Bond Angle ($\theta$): Using Coulson's Theorem with $p = 3$: $\cos\theta = -\frac{1}{3} \implies \theta = \arccos\left(-\frac{1}{3}\right) \approx 109.47^\circ \quad \text{(Tetrahedral)}$