The correct order of dipole moments for the given species is

The correct option is BF₃ = NH₄⁺ < NF₃ < NH₃.

Let us analyze the molecular geometry and dipole moment (μ) of each given species step-by-step:

1. BF₃ (Boron Trifluoride) and NH₄⁺ (Ammonium Ion):

BF₃ has a trigonal planar structure with sp² hybridization. The bond dipoles of the three B-F bonds are equal in magnitude and oriented at 120° to each other. Their vector sum cancels out completely, resulting in a zero net dipole moment.

${\mu }_{{\text{BF}}_3}=0\text{D}$

NH₄⁺ has a symmetrical tetrahedral structure with sp³ hybridization. Due to its complete spherical/tetrahedral symmetry, the individual N-H bond moments cancel each other completely, giving a zero net dipole moment.

${\mu }_{{\text{NH}}_4^{+}}=0\text{D}$

Therefore, ${\mu }_{{\text{BF}}_3}={\mu }_{{\text{NH}}_4^{+}}=0$.

2. NF₃ (Nitrogen Trifluoride) vs NH₃ (Ammonia):

Both NF₃ and NH₃ have a trigonal pyramidal geometry with one lone pair on the central nitrogen atom.

In NH₃, nitrogen is more electronegative than hydrogen. Consequently, the N-H bond dipoles point toward the nitrogen atom. The dipole moment due to the lone pair also acts in the same upward direction. Since the bond dipoles and the lone-pair dipole reinforce each other, NH₃ has a relatively high net dipole moment (≈ 1.47 D).

In NF₃, fluorine is more electronegative than nitrogen. Therefore, the N-F bond dipoles point away from nitrogen toward the fluorine atoms (downwards), while the lone-pair dipole acts upwards. The N-F bond dipoles partially oppose the lone-pair dipole, resulting in a very low net dipole moment (≈ 0.23 D).

Hence, ${\mu }_{{\text{NF}}_3}<{\mu }_{{\text{NH}}_3}$.

Conclusion:

Combining all the dipole moments in increasing order:

${\text{BF}}_3={\text{NH}}_4^{+}<{\text{NF}}_3<{\text{NH}}_3$