Question

What is the correct dipole moment of $N{H}_3$ and $N{F}_3$ respectively?
(A)- $4.90\times {10}^{-30}$ C m and $0.80\times {10}^{-30}$ C m
(B)- $0.80\times {10}^{-30}$ C m and $4.90\times {10}^{-30}$ C m
(C)- $4.90\times {10}^{-30}$ C m and $4.90\times {10}^{-30}$ C m
(D)- $0.80\times {10}^{-30}$ C m and $0.80\times {10}^{-30}$ C m

Answer 1

Hint: Dipole moment is a measure of polarity of a bond. It is the product of the charges and the distance between partial charges. It is a vector quantity and its direction is always given from less electronegative atom to more electronegative atom.
It is generally expressed in debye (D) and 1 D = $3.33564\times {10}^{-30}$ C m.
Dipole moment of polar molecules containing lone pairs is the vector sum of dipole of lone pair and net dipole moments of bonds.


Complete answer:
Both $N{H}_3$ and $N{F}_3$ have trigonal pyramidal shape.

The dipole moment of lone pairs in $N{H}_3$ and $N{F}_3$ is away from nitrogen.
Dipole moment of $N{H}_3$
We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of $N-H$bond will be form H to N. The net dipole moment of three $N-H$ bond will add up to 1.4 D. As we know that 1 D = $3.33564\times {10}^{-34}$ C m.
Then, 1.4 D will be equal to $1.4\times 3.33564\times {10}^{-30}$ C m, i.e. $4.90\times {10}^{-30}$ C m.

Dipole moment of $N{F}_3$
Electronegativity of F is more than that of N, thus the direction of dipole moment of $N-F$ bond will be from F to N. As we can see that the direction of $N-F$ bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of $N{F}_3$ has been found to be 0.24 D.
Multiplying 0.24 D with $3.33564\times {10}^{-30}$ C m, we get the dipole moment of $0.80\times {10}^{-30}$C m.
So, the correct answer is “Option A”.

Note: The dipole moment of lone pairs in $N{H}_3$ and $N{F}_3$ is away from nitrogen.
Dipole moment of $N{H}_3$
We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of $N-H$bond will be form H to N. The net dipole moment of three $N-H$ bond will add up to 1.4 D. As we know that 1 D = $3.33564\times {10}^{-34}$ C m.
Then, 1.4 D will be equal to $1.4\times 3.33564\times {10}^{-30}$ C m, i.e. $4.90\times {10}^{-30}$ C m.

Dipole moment of $N{F}_3$
Electronegativity of F is more than that of N, thus the direction of dipole moment of $N-F$ bond will be from F to N. As we can see that the direction of $N-F$ bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of $N{F}_3$ has been found to be 0.24 D.
Multiplying 0.24 D with $3.33564\times {10}^{-30}$ C m, we get the dipole moment of $0.80\times {10}^{-30}$C m.