Arrange the following species in increasing order of their dipole moments: ${\text{NH}}_3,{\text{NF}}_3,{\text{BF}}_3,{\text{H}}_2\text{O}$

${\text{BF}}_3$: Trigonal planar geometry with identical bonds. The individual bond dipoles cancel out, so $\mu =0$.${\text{NF}}_3$ vs ${\text{NH}}_3$: In ${\text{NH}}_3$, the orbital dipole due to the lone pair is in the same direction as the resultant dipole of the $\text{N-H}$ bonds. In ${\text{NF}}_3$, the lone pair dipole is in the opposite direction to the resultant of the $\text{N-F}$ bonds. Thus, ${\mu }_{{\text{NH}}_3}>{\mu }_{{\text{NF}}_3}$.${\text{H}}_2\text{O}$: Highly polar with two lone pairs and strong $\text{O-H}$ bond dipoles, giving it the highest dipole moment among these.Order: ${\text{BF}}_3(0)<{\text{NF}}_3<{\text{NH}}_3<{\text{H}}_2\text{O}$.