Add up the bond dipoles.

Bond dipole arrows on a molecule add tip-to-tail into a net dipole; changing the bond angle or swapping atoms shows when symmetry cancels it, and lone-pair moments are drawn for NH3 vs NF3.

Bend the AB2 molecule until the net dipole is zero (180°); then pick which of NH3/NF3 has the larger μ and confirm with the vector sum.

A molecule with polar bonds must have a dipole moment; NF3 has a larger μ than NH3 because F is more electronegative.

• A bond between two identical atoms (H₂, O₂) is non-polar; when the atoms differ in electronegativity, the shared pair shifts towards the more electronegative atom and the bond becomes polar (as in HF).
• Dipole moment μ = charge × distance between the charge centres; it is a vector, drawn from the positive to the negative end in the convention chemists use (crossed arrow).
• The unit is the Debye: 1 D = 3.33564 × 10⁻³⁰ C m.
• A molecule's net dipole moment is the vector sum of its bond dipoles, so symmetry can cancel it: BeF₂, BF₃, CO₂, CH₄ and CCl₄ have zero dipole moment.
• Water is bent, so its two O–H bond dipoles add to a net value (about 1.85 D).
• NH₃ has a larger dipole moment than NF₃: in NH₃ the lone-pair moment adds to the N–H bond moments, while in NF₃ it opposes the N–F bond moments.
• Ionic bonds carry some covalent character too, and Fajans' rules say when it is larger: a small cation paired with a large anion, and a cation with a higher charge.
• Of two cations with the same size and charge, a transition-metal type cation with an (n−1)dⁿ ns⁰ configuration (for example Cu⁺, 3d¹⁰) polarises the anion more than one with a noble gas ns² np⁶ configuration.

Teaching this idea? Put the simulation on your own page or LMS with this code.