Question

The dipole moment of $N{F_3}$ is less than that of $N{H_3}$ because:
a. F is more reactive than H
b. $N{H_3}$ forms associated molecules
c. The resultant of the individual bond polarities is less
d. The resultant of the individual bond polarities is opposed by the polarity of lone pair

Answer 1

Hint – In this question use the concept that electronegativity increases from left to right when we move in a periodic table. So compare the electronegativity of N and F for compound $N{F_3}$ and then compare the electronegativity of N and H for $N{H_3}$. This will help get the right answer.

Complete answer:
As we know that electronegativity is increasing from left to right when we move in the period of periodic table.
As F is right of N in the periodic table so the electronegativity of F is greater than the electronegativity of N.
So each of N – F bonds has its polarity towards F, which results in a net polarity which opposes the polarity of lone pair electrons.
But in the case of $N{H_3}$ the electronegativity of N is lesser than the electronegativity of H.
So each N – H bonds has its polarity towards N, which results in a net polarity which does not oppose the polarity of lone pair electrons.
That’s why the dipole moment of $N{F_3}$ less than that of $N{H_3}$.
So this is the required answer.

Hence option (D) is the correct answer.

Note – The basic definition of dipole moments is very helpful, while dealing with problems of this kind. The mathematical product of the separation of the ends of a dipole (dipole is two opposite charges separated at a distance) and the magnitude of the charges.