Electronegativity Fajan

Given

HF has a bond length of 0.92 Å (9.2 × 10⁻¹¹ m) and an observed dipole moment of 1.91 D. HCl has a bond length of 1.27 Å (1.27 × 10⁻¹⁰ m) and an observed dipole moment of 1.03 D.

Required

Explain why HF has a larger dipole moment than HCl despite having a shorter bond length d.

Concept

Dipole moment μ = q × d. A larger d alone would increase μ, yet HF (shorter d) has a larger μ than HCl. This means the partial charge q on HF must be significantly larger, overriding the shorter distance.

Formula

μ = q × d, therefore q = μ / d.

Substitution

For HF: q = 1.91 D / 0.92 Å (in relative terms) For HCl: q = 1.03 D / 1.27 Å (in relative terms)

Calculation

Relative charge density comparison: - HF: 1.91 / 0.92 ≈ 2.08 D/Å - HCl: 1.03 / 1.27 ≈ 0.81 D/Å The ratio q(HF)/q(HCl) ≈ 2.08/0.81 ≈ 2.6. Fluorine's much higher electronegativity (4.0 vs chlorine's 3.0) pulls electron density far more strongly, creating a partial charge roughly 2.6 times larger in HF. Note: The bond lengths and dipole moments are given data — they are treated as exact for this comparison. The ratio calculation uses them directly.

Final answer

HF has a larger dipole moment than HCl because fluorine's greater electronegativity (4.0 vs 3.0) creates a much larger partial charge q, which more than compensates for the shorter bond length. The electronegativity difference dominates the μ = q × d product.

Common trap

Students sometimes assume that longer bond length automatically means larger dipole moment (since d appears in μ = qd). This ignores the charge term. When electronegativity difference changes significantly between two molecules, q can dominate over d.

Similar NEET-style question

"Among HF, HCl, HBr, and HI, which has the highest dipole moment? Justify using electronegativity and the dipole moment formula." Answer: HF — highest electronegativity difference (F = 4.0, H = 2.1) produces the largest partial charge, outweighing the effect of bond length. ---