Question

The correct decreasing order of dipole moments for the given molecules is:
$(B{F}_3,N{F}_3$ and $N{H}_3)$

• A. $B{F}_3>N{F}_3>N{H}_3$
• B. $N{F}_2>B{F}_3>N{H}_3$
• C. $N{H}_3>N{F}_3>B{F}_3$
• D. $N{H}_3>B{F}_3>N{F}_3$

Answer 1

The correct option is C $N{H}_3>N{F}_3>B{F}_3$

In $N{H}_3$ and $N{F}_3$, the central nitrogen has three $\sigma$ bonds and one lone pair. Hence, both are $s{p}^3$ hybridized and posses trigonal pyramidal shape. In $B{F}_3$, the central boron has only three $\sigma$ bonds and so it is $s{p}^2$ hybridized and has trigonal planer shape.
In $B{F}_3$, the dipole moment vectors are arranged at ${120}^{\circ }$. Therefore, the resultant of the two dipole moment vectors cancel out with the third dipole moment vector and hence $B{F}_3$ is non polar in nautre. In $N{H}_3$ and $N{F}_3$, the dipole moment vectors do not cancel out each other. Hence, both the molecules are polar in nature. In $N{H}_3$, the dipole moment vector of the bond and the lone pairs are in the same direction, while in $N{F}_3$, the dipole moment vector of the lone pair and the bond pairs are opposite in direction. Hence, the dipole moment of $N{H}_3$ is larger than $N{F}_3$.