Chemical Bonding & Molecular Structure

Chemical bonding explains how atoms combine to form stable molecules. Ionic bonding involves electron transfer from metals to non-metals, forming a crystal lattice stabilized by electrostatic attraction. Lattice enthalpy quantifies this stability and can be calculated using the Born-Haber cycle: Δ$H_{f}$ = Δ$H_{sub}$ + ½Δ$H_{diss}$ + IE + EA + U (lattice energy). Fajan's rules predict covalent character in ionic bonds: covalent character increases with small cation size, large anion size, and high cation charge (greater polarization).

Covalent bonding involves electron sharing. Dipole moment μ = q × d (units: Debye; 1 D = 3.336 × 10⁻³⁰ C·m). Symmetric molecules have μ = 0 (CO₂ linear, BF₃ trigonal planar, CCl₄ tetrahedral) because individual bond dipoles cancel by vector addition. Dimensional analysis: μ = q × d → C × m = C·m ✓

VSEPR theory predicts molecular geometry from electron pair repulsion around the central atom. Lone pair repulsion order: lp-lp > lp-bp > bp-bp, which compresses bond angles. Key shapes: linear (AB₂, 180°), trigonal planar (AB₃, 120°), tetrahedral (AB₄, 109.5°), trigonal bipyramidal (AB₅), octahedral (AB₆, 90°). With lone pairs: AB₃E = trigonal pyramidal, AB₂E₂ = bent, AB₄E₂ = square planar.

Hybridization is determined by the steric number ($\sigma$ bonds + lone pairs): sp (2, linear), sp² (3, trigonal planar), sp³ (4, tetrahedral), sp³d (5, trigonal bipyramidal), sp³d² (6, octahedral). VBT describes $\sigma$ bonds (head-on overlap) and $\pi$ bonds (lateral overlap): single = 1σ, double = 1σ + 1π, triple = 1σ + 2π.

Solved Example 1: Shape and hybridization of XeF₄. Xe valence electrons = 8; bonds to 4 F atoms; lone pairs = (8 − 4)/2 = 2 (considering expanded octet: 4 bond pairs + 2 lone pairs = 6 pairs). Steric number = 6 → hybridization = sp³d². Shape = square planar (lone pairs occupy axial positions to minimize lp-lp repulsion).

Solved Example 2: Bond order of O₂⁻ (superoxide). Electronic configuration: σ1s² σ1s² σ2s² σ2s² σ2p² π2p⁴ π2p³ BO = (Nb − Na)/2 = (10 − 7)/2 = 1.5. O₂⁻ is paramagnetic (1 unpaired electron in π2p).

Solved Example 3: Lattice energy of NaCl using Born-Haber cycle. Given: Δ$H_{f}$ = −411, Δ$H_{sub}$(Na) = 108, IE(Na) = 496, ½Δ$H_{diss}$(Cl₂) = 121, EA(Cl) = −349 kJ/mol. U = Δ$H_{f}$ − Δ$H_{sub}$ − IE − ½Δ$H_{diss}$ − EA = −411 − 108 − 496 − 121 − (−349) = −787 kJ/mol

MOT uses LCAO to form bonding (σ, π) and antibonding (σ*, π*) MOs. Bond order = (Nb − Na)/2. For Z ≤ 7 (B₂, C₂, N₂): σ2p lies ABOVE π2p (mixing). For Z > 7 (O₂, F₂): σ2p lies BELOW π2p (normal order). O₂ is paramagnetic (2 unpaired e⁻ in π*2p) — a key MOT success that VBT fails to predict.

Resonance involves delocalization of electrons (e.g., benzene, SMILES: c1ccccc1; ozone, SMILES: [O-][O+]=O). Hydrogen bonding (H bonded to F, O, N) affects boiling points; metallic bonding (electron sea model) explains conductivity.

The key testable concept is VSEPR shape prediction using steric number and MOT bond order calculations for diatomic species.