The dipole moment of a molecule depends on both the polarity of individual bonds and the overall molecular geometry. In BF_3, boron has three bonding pairs and no lone pairs (sp^2 hybridised), forming a trigonal planar molecule. The three B-F bond dipoles are arranged at 120 degrees to each other in a plane, and they cancel vectorially to give a net dipole moment of zero. In NF_3, nitrogen has three bonding pairs and one lone pair (sp^3 hybridised), giving a trigonal pyramidal geometry. Unlike BF_3, the three N-F bond dipoles in NF_3 do not cancel because the molecular geometry is pyramidal and the lone pair on N adds its own dipole contribution (directed away from the N-F bonds). The net effect is a non-zero dipole moment for NF_3. It is important to note that in NF_3, the direction of the net dipole actually points from N towards F (because F is more electronegative and pulls electron density), resulting in a small but non-zero dipole moment of about 0.234 D. Option 'N-F bond is less polar than B-F bond' is incorrect in explaining the zero versus non-zero dipole; even if B-F is more polar, complete cancellation in the planar BF_3 geometry gives zero. Option 'BF_3 has more electrons' is irrelevant to dipole moment. Option 'NF_3 is planar while BF_3 is pyramidal' reverses the geometries. This directly tests NCERT molecular geometry and dipole moment concepts. Plausibility: the experimental dipole moments confirm BF_3 = 0 D and NF_3 = 0.234 D, consistent with the geometrical analysis.