The dipole moment of a molecule depends on the polarity of its individual bonds and the molecular geometry. It is a vector quantity, so bond dipoles can add up or cancel out.

BF₃ has a trigonal planar geometry due to sp² hybridization of boron. The three B-F bond dipoles are equal in magnitude and oriented at 120° to each other. Due to this symmetrical arrangement, the net dipole moment is zero.

μ(BF₃) = 0 D

NH₃ has a trigonal pyramidal geometry due to sp³ hybridization of nitrogen, with one lone pair. The N-H bond dipoles are directed towards nitrogen (N is more electronegative than H). The lone pair also contributes a significant dipole moment in the same direction as the resultant of the N-H bond dipoles. Therefore, the bond dipoles and the lone pair dipole add up, resulting in a high net dipole moment.

μ(NH₃) ≈ 1.47 D

NF₃ also has a trigonal pyramidal geometry due to sp³ hybridization of nitrogen, with one lone pair. However, in NF₃, fluorine is more electronegative than nitrogen, so the N-F bond dipoles are directed towards fluorine, away from the nitrogen atom. The lone pair dipole is directed away from the nitrogen atom. The bond dipoles and the lone pair dipole are in opposite directions, leading to partial cancellation. This results in a much smaller net dipole moment compared to NH₃.

μ(NF₃) ≈ 0.23 D

H₂O has a bent (V-shaped) geometry due to sp³ hybridization of oxygen, with two lone pairs. Oxygen is more electronegative than hydrogen, so the O-H bond dipoles are directed towards oxygen. The two lone pairs also contribute significantly to the dipole moment. Both the bond dipoles and the lone pair dipoles add up constructively, resulting in a very high net dipole moment.

μ(H₂O) ≈ 1.85 D

Based on the analysis:

• μ(BF₃) = 0 D
• μ(NF₃) ≈ 0.23 D
• μ(NH₃) ≈ 1.47 D
• μ(H₂O) ≈ 1.85 D

The correct order of increasing dipole moment is: ${\mathrm{BF}}_3<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$

• A) ${\mathrm{NH}}_3<{\mathrm{BF}}_3<{\mathrm{NF}}_3<{\mathrm{H}}_2\mathrm{O}$: Incorrect, BF₃ has zero dipole moment and is the lowest.
• B) ${\mathrm{BF}}_3<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$: Correct, this order matches our derived values.
• C) ${\mathrm{BF}}_3<{\mathrm{NH}}_3<{\mathrm{NF}}_3<{\mathrm{H}}_2\mathrm{O}$: Incorrect, NF₃ has a smaller dipole moment than NH₃.
• D) ${\mathrm{H}}_2\mathrm{O}<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{BF}}_3$: Incorrect, this is almost the reverse order and BF₃ is the lowest.

Correct Answer: (B)