MCQ
Dipole moment of $NF_3$ is smaller than
- ✓$NH_3$
- B$CO_2$
- C$BF_3$
- D$CCl_4$
✓
Answer
Correct option: A.
$NH_3$
a
Both the molecules have a pyramidal shape with a lone pair of electrons on the nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of $\mathrm{NH}_{3}\left(4.90 \times 10^{-30} \mathrm{CM}\right)$ is greater than that of $\mathrm{NF}_{3}\left(0.8 \times 10^{-30} \mathrm{Cm}\right)$
Both the molecules have a pyramidal shape with a lone pair of electrons on the nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of $\mathrm{NH}_{3}\left(4.90 \times 10^{-30} \mathrm{CM}\right)$ is greater than that of $\mathrm{NF}_{3}\left(0.8 \times 10^{-30} \mathrm{Cm}\right)$
This is because in case of $\mathrm{NH}_{3}$ the orbital dipole due to lone pair is in the same direction as the
resultant dipole moment of the $\mathrm{N}-\mathrm{H}$ bonds. Whereas in $\mathrm{NF}_{3}$ the orbital dipole is in the direction opposite to the resultant dipole moment of
the three $N$ - F bonds. The orbital dipole because of the lone pair decreases the effect of the
resultant $N$ - F bond moments, which results in the low dipole moment of $\mathrm{NF}_{3}$.
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