Correct Order of Dipole Moments for Given Chemical Species

To determine the correct order of dipole moments ($\mu$) for the given species ($\text{BF}_3$, $\text{NH}_4^+$, $\text{NF}_3$, and $\text{NH}_3$), we analyze the molecular geometry and direction of individual bond dipoles for each species:

1. $\text{BF}_3$ (Boron Trifluoride):

   • $\text{BF}_3$ has a trigonal planar geometry ($sp^2$ hybridized) with bond angles of $120^\circ$.
   • Due to its highly symmetrical planar structure, the vector sum of the three $\text{B}-\text{F}$ bond dipoles cancels out completely.
   • Therefore, the net dipole moment is: $\mu(\text{BF}_3) = 0\text{ D}$

2. $\text{NH}_4^+$ (Ammonium Cation):

   • $\text{NH}_4^+$ has a regular tetrahedral geometry ($sp^3$ hybridized) with bond angles of $109.5^\circ$.
   • Because of its perfectly symmetrical tetrahedral arrangement, the four $\text{N}-\text{H}$ bond dipoles completely cancel each other out.
   • Therefore, the net dipole moment is: $\mu(\text{NH}_4^+) = 0\text{ D}$

3. $\text{NF}_3$ vs. $\text{NH}_3$:

   • Both $\text{NF}_3$ and $\text{NH}_3$ have a trigonal pyramidal geometry ($sp^3$ hybridized with 1 lone pair on the nitrogen atom).
   • In $\text{NF}_3$, Fluorine is more electronegative than Nitrogen ($\chi_{\text{F}} > \chi_{\text{N}}$). The $\text{N}-\text{F}$ bond dipoles point away from Nitrogen toward Fluorine. The dipole moment resulting from the lone pair points in the opposite direction (away from Nitrogen). Since the lone pair dipole and the resultant of the three $\text{N}-\text{F}$ bond dipoles oppose each other, the net dipole moment is relatively small ($\mu \approx 0.235\text{ D}$).
   • In $\text{NH}_3$, Nitrogen is more electronegative than Hydrogen ($\chi_{\text{N}} > \chi_{\text{H}}$). The $\text{N}-\text{H}$ bond dipoles point toward Nitrogen. The lone pair dipole also points in the same general direction away from Nitrogen. Thus, the lone pair dipole and the resultant of the three $\text{N}-\text{H}$ bond dipoles reinforce each other, resulting in a much larger net dipole moment ($\mu \approx 1.47\text{ D}$).

   Therefore: $\mu(\text{NF}_3) < \mu(\text{NH}_3)$

Combining all the results: $\mu(\text{BF}_3) = \mu(\text{NH}_4^+) < \mu(\text{NF}_3) < \mu(\text{NH}_3)$

Thus, the correct order is $\text{BF}_3 = \text{NH}_4^+ < \text{NF}_3 < \text{NH}_3$, which corresponds to Option (A).