§ C-2 Orbital Hybridization & 3D Molecular Geometry (VSEPR) (Grade 9–11 · Chemistry)
Linear sp (180°), trigonal planar sp² (120°), tetrahedral sp³ (109.5°) & lone pairs. Valence electron pairs around a central atom arrange themselves in 3D space so as to minimize mutual Coulomb repulsion.
Theory Summary & Scientific Mechanism
The Valence Shell Electron Pair Repulsion (VSEPR) model (Gillespie–Nyholm) predicts the 3D geometry of covalent molecules AXₙEₘ. Electron domains in the valence shell of central atom A, both bonding pairs (X) and lone pairs (E), orient themselves to minimize Pauli and Coulomb repulsion by maximizing their angular separation.
The steric number S = n + m dictates the electron-domain geometry: S = 2 linear (180^°, sp hybridization, e.g. CO₂), S = 3 trigonal planar (120^°, sp², BF₃), and S = 4 tetrahedral (109.5^°, sp³, CH₄). Because a lone pair E is localized near a single nucleus rather than stretched between two nuclei, it occupies a wider angular domain: repulsion follows E–E > E–X > X–X. Consequently, the bond angle contracts to 107.3^° in NH₃ (AX₃E₁) and 104.5^° in H₂O (AX₂E₂).
Molecular polarity is governed by the vector sum of individual bond dipoles μₙet = ∑ᵢ=1^n μᵢ. In symmetric geometries without lone pairs (AX₂, AX₃, AX₄), identical bond dipoles cancel out (|μₙet| = 0, nonpolar molecule), whereas lone pairs break inversion symmetry and yield a strong permanent dipole moment (such as μ(H₂O) = 1.85 D).
Everyday Analogy
Tie 2, 3, or 4 inflated balloons together at their knots: they naturally push each other apart into a line (180^°), a flat triangle (120^°), or a 3D tetrahedron (109.5^°)! A lone electron pair E acts like an extra-bulky balloon that squeezes the neighboring bonds even closer.
Interactive Question & Derivation
Why are the ∠H–O–H bond angle in water H₂O (104.5^°) and the ∠H–N–H angle in ammonia NH₃ (107.3^°) smaller than the ideal tetrahedral angle of methane CH₄ (109.5^°)?
- A: Because non-bonding lone pairs (E) are held by only one nucleus, occupy a wider angular domain near the central atom, and repel bonding pairs more strongly (LP–LP > LP–BP > BP–BP)
- B: Because Hydrogen atoms carry positive charges and attract one another
- C: Because NH₃ and H₂O have no orbital hybridization
Study Hint: In symmetric methane CH₄ (n=4, m=0), all four domains are identical (109.5^°). When lone pairs E replace bonding pairs (as in NH₃ and H₂O), their wider electron clouds repel bonding pairs more strongly, compressing the bond angle.
Proof & Explanation: In VSEPR theory, CH₄, NH₃, and H₂O all share steric number SN = 4 (sp³ hybridization). However, a lone pair (E) sits closer to the central nucleus than a shared bonding pair (X), compressing the bond angles progressively from 109.5^° → 107.3^° → 104.5^°.
Key Equations & 3D Simulation Parameters
cos θₜetr = -(1)/(3) ⇒ θ₀ = 109.47^°, μₙet = ∑ᵢ=1^N μᵢ
- Molecule type (1=BeCl₂, 2=BF₃, 3=CH₄, 4=NH₃, 5=H₂O): 1 … 5 (default: 3)
- Internuclear bond length d(A–X): 1.1 … 2 Å (default: 1.5 Å)
- Electronegativity difference Δχ: 0.4 … 1.8 (default: 1)
- Thermal vibrational amplitude: 0 … 1 × (default: 0.3 ×)