Question

The electronegativity difference between $N$ and $F$ is greater than $N$ and $H$, yet the dipole moment of $N{H}_3\left(1.5D\right)$ is greater than that of $N{F}_3\left(0.2D\right)$. This is because :

• A. in $N{H}_3$ as well as $N{F}_3$, the atomic dipole and bond dipole are in opposite direction.
• B. in $N{H}_3$, the atomic dipole and bond dipole are in the opposite direction, where as in $N{F}_3$, these are in same direction.
• C. in $N{H}_3$ as well as in $N{F}_3$, the atomic dipole and bond dipole are in same direction.
• D. in $N{H}_3$, the atomic dipole and bond dipole are in same direction where as in $N{F}_3$ these are in opposite direction.

Answer 1

The correct option is D in $N{H}_3$, the atomic dipole and bond dipole are in same direction where as in $N{F}_3$ these are in opposite direction.

In $N{H}_3$, partial negative charge is towards nitrogen and it is contributed by three hydrogen. These dipoles are in the direction of the atomic dipoles.

In $N{F}_3$, partial negative charge is towards three fluorines contributed by one nitrogen atom, and these dipoles are in opposite directions. The resultant vector is in downward direction which is reduced due to the upward dipole of the lone pair of electrons on the nitrogen atom.

Hence, $N{F}_3$ has less dipole moment compared to $N{H}_3$.