Question

The correct order of a dipole moment is

• A. ${\mathrm{CH}}_4<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$
• B. ${\mathrm{NH}}_3<{\mathrm{NF}}_3<{\mathrm{CH}}_4<{\mathrm{H}}_2\mathrm{O}$
• C. ${\mathrm{NF}}_3<{\mathrm{CH}}_4<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$
• D. ${\mathrm{H}}_2\mathrm{O}<{\mathrm{NH}}_3<{\mathrm{NF}}_3<{\mathrm{CH}}_4$

Answer 1

The correct option is A

${\mathrm{CH}}_4<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$

The explanation for the correct option:

Option (A): ${\mathrm{CH}}_4<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$

1. The bond dipole moment is a measurement of the polarity of a chemical bond within a molecule that is based on the electric dipole moment concept. It occurs whenever positive and negative charges are separated.
2. The bond's dipole moment and polarity increase as the electronegativity difference between the two atoms grows.
3. Because they are vectors in nature, molecules with symmetrical and linear geometries have zero dipole moment because the dipoles of distinct bonds cancel out.
4. Hence ${\mathrm{CH}}_4$ has symmetrical tetrahedral geometry it has zero dipole moment.
5. Water has the highest dipole moment due to the high electronegativity difference.
6. Nitrogen is more electronegative than hydrogen because it has a single pair of electrons. So, in the instance of ${\mathrm{NH}}_3$, the orbital moment due to the lone pair of electrons on nitrogen is in the same direction as the three $\mathrm{N}-\mathrm{H}$ dipoles. As a result, the molecule has a net dipole moment.
7. Because fluorine is more electronegative than nitrogen, it attracts the $\mathrm{N}-\mathrm{F}$ bonds' common pair of electrons. As a result, the orbital moment of a lone pair of electrons and the consequent dipole moment of three $\mathrm{N}-\mathrm{F}$ bonds point in opposite directions. This reduces ${\mathrm{NF}}_3$ dipole moment.
8. Hence, it is the correct option.

The explanation for the incorrect option:

Option (B) ${\mathrm{NH}}_3<{\mathrm{NF}}_3<{\mathrm{CH}}_4<{\mathrm{H}}_2\mathrm{O}$

1. Symmetrical and linear geometries have zero dipole moment because the dipoles of separate bonds cancel out.
2. ${\mathrm{CH}}_4$ is tetrahedral in shape, and as a result of this structure, each bond pair is at an equal distance apart, i.e. they are symmetrically placed, and so each dipole moment of the bond is balanced and the net dipole moment is zero.
3. Hence, it is the incorrect option

Option (C): ${\mathrm{NF}}_3<{\mathrm{CH}}_4<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$

1. In ${\mathrm{NF}}_3$ structure Fluorine is more electronegative than Nitrogen due to which the dipole moment is in opposite directions thus reducing its dipole moment compared to ${\mathrm{NH}}_3$.
2. Hence it is an incorrect option

Option (D): ${\mathrm{H}}_2\mathrm{O}<{\mathrm{NH}}_3<{\mathrm{NF}}_3<{\mathrm{CH}}_4$

1. Dipole moment is defined as the measure of polarity in a chemical bond i.e. Higher the electronegative difference between bonds, the higher will be its dipole moment.
2. Water has the largest electronegative difference between its atoms, hence it has the largest dipole moment.
3. Hence, it is an incorrect option.

Therefore, the correct order of a dipole moment is dipole moment${\mathrm{CH}}_4<{\mathrm{NF}}_3<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{O}$, option (A) is the correct answer.