§ C-25 Electronegativity, Bond Polarity & Molecular Dipole Moment Vector μ (Grade 8–10 · Molecular Polarity)
Partial charges δ⁺/δ⁻, vector addition of bond dipoles, and molecular torque in an electric field E. The dipole moment of a molecule is its 3D geometry multiplied by the asymmetry of its electron cloud.
Theory Summary & Scientific Mechanism
When two bonded atoms differ in Pauling electronegativity (Δχ = |χ_A - χ_B| > 0), shared electron density shifts toward the more electronegative atom, generating partial charges ±δ q and a bond dipole vector μ_bond = δ q d (1 D = 3.336× 10⁻³⁰ C·m).
The net polarity of a polyatomic molecule is the vector sum of its individual bond dipoles: μₙet = ∑ᵢ μᵢ. For a triatomic AB₂ molecule with bond angle θ, μₙet = 2μ_bondcos (θ/2).
Consequently, bent H₂O (θ = 104.5^°) has a large permanent dipole (1.85 D) and experiences an aligning torque τ = μ×E in an external electric field, whereas linear CO₂ (θ = 180^°) has μₙet = 0 because its two polar C=O vectors cancel.
Everyday Analogy
Think of a tug-of-war! In linear CO₂ (180^°), two oxygen atoms pull electrons equally in opposite directions so the vectors cancel (μ = 0). In bent water (104.5^°), both pulls angle toward oxygen, creating a strong net dipole!
Interactive Question & Derivation
Why does the bent water molecule H₂O (θ = 104.5^°) possess a large permanent dipole moment (μ = 1.85 D), whereas linear carbon dioxide CO₂ (θ = 180^°) is nonpolar (μₙet = 0)?
- A: Molecular dipole moment is the vector sum μₙet = μ₁ + μ₂, yielding μₙet = 2μ_bondcos (θ/2): at θ = 180^° (cos 90^° = 0) the two polar bond vectors cancel exactly
- B: Carbon and oxygen have identical electronegativities (Δχ = 0), so individual C=O bonds are completely nonpolar
- C: Dipole moment is a scalar sum of bond lengths and never depends on the bond angle θ
Study Hint: Molecular dipole moment is the vector sum μₙet = μ₁ + μ₂; at θ = 180^° the two opposing bond dipoles cancel to zero.
Proof & Explanation: A difference in Pauling electronegativity Δχ = |χ_A - χ_B| shifts shared electron density toward the more electronegative atom, creating partial charges ±δ q and a bond dipole vector μ_bond = δ q · d (1 D = 3.336× 10⁻³⁰ C·m). For a symmetrical triatomic molecule AB₂ with bond angle θ, vector addition gives μₙet = 2μ_bondcos (θ/2). When θ = 180^° (CO₂, BeCl₂), the two equal and opposite bond dipoles cancel (μₙet = 0).
Key Equations & 3D Simulation Parameters
μ_bond = δ q · d, μₙet = ∑ᵢ μᵢ ⇒ μₙet = 2μ_bondcos ≤ft((θ)/(2) ), τ = μ×E
- Pauling electronegativity difference Δχ = |χ_B − χ_A|: 0 … 2.5 (default: 1.24)
- Valence bond angle θ (105° = H₂O, 180° = CO₂): 90 … 180 ° (default: 105 °)
- Interatomic bond length l(A–B): 0.9 … 1.8 Å (default: 0.96 Å)
- External electric field strength E: 0 … 5 kV/cm (default: 2 kV/cm)