Question 3

Dipole moment ($\mu$) depends on molecular symmetry and the vector sum of bond moments.

1. $BF_3$: Trigonal planar geometry ($sp^2$ hybridized) with no lone pairs. The vector sum of three $B-F$ bond moments is zero. Thus, $\mu = 0$.
2. $NH_4^+$: Tetrahedral geometry ($sp^3$ hybridized) and perfectly symmetric. The vector sum of four $N-H$ bond moments is zero. Thus, $\mu = 0$.
3. $NH_3$ and $NF_3$: Both have pyramidal geometry with one lone pair. In $NH_3$, the $N-H$ bond moments and the lone pair moment are in the same direction, reinforcing each other. In $NF_3$, the $N-F$ bond moments are in the opposite direction to the lone pair moment, partially cancelling it. Thus, $\mu(NH_3) > \mu(NF_3) > 0$. Combining these, the order is $BF_3 = NH_4^+ (0) < NF_3 < NH_3$.