Question

Boron trifluoride $(B{F}_3)$ is a nonpolar molecule, whereas ammonia $(N{H}_3)$ is a polar molecule. The difference in polarity is related to the fact that:
A.$$B{F}_3$$ has no hydrogen bonding and $N{H}_3$ does
B.$$B{F}_3$$ is a triangular planar and $N{H}_3$ is pyramidal
C.$$B{F}_3$$ is a Lewis base and $N{H}_3$ is a lewis acid
D.The $$B-F$$ bond is less polar than the $$N-H$$ bond
E.Boron is more electronegative than nitrogen

Answer 1

Hint: The dipole moments of each $$B-F$$ bond cancel each other and $N{H}_3$ s lone pair of electrons sit at one corner of tetrahedral arrangement of electrons with other three electrons forming bond with H.

Complete step by step answer:
The shape of $N{H}_3$ is trigonal pyramidal. This is because N requires 4 electron pairs to fulfil its octet whereas $$B{F}_3$$ is trigonal planar in shape, because it only requires 3 electron pairs to fulfil its octet. This is because there are only three electrons available for bonding which are one unpaired electron per fluoride and three valence electrons total on the boron. In the case of $N{H}_3$ , nitrogen has 5 valence electrons, so when it forms three bonds with the fluorides, a lone pair occupies another space, causing the bonds to bend onwards and resulting in a trigonal pyramidal shape.
 Hence, Boron trifluoride $(B{F}_3)$ has a trigonal planar structure, therefore the dipole moments of each $$B-F$$ bond cancel each other. Ammonia’s $(N{H}_3)$ lone pair of electrons sit at one corner of a tetrahedral arrangement of electrons with the other three electrons forming bonds with hydrogen at the other corners. This gives the molecule a net dipole moment and thus making $N{H}_3$ polar and because $$B{F}_3$$ doesn’t meet the requirement of polar molecules, it is considered non-polar.
Therefore, the correct answer is option (B).

Note: Molecules having polar covalent bond may be polar and nonpolar molecules. Molecules which have dipole moment are called polar molecules & which do not have dipole moment are called nonpolar molecules. $N{H}_3$ is also more polar due to the reason that the H being the least electronegative element and N being the most electronegative element. So, the electronegative N attracts the electron pairs it is sharing with three H closer to it leaving it delta negative and H as delta positive.