Question

Both $B{F}_3$ and $N{F}_3$ are covalent compounds. $B{F}_3$ is non-polar while $N{F}_3$ is polar. The reason is that:

• A. Boron is a metalloid while nitrogen is a non-metal
• B. Boron is a solid and nitrogen is a gas in free state
• C. $B{F}_3$ is planar while $N{F}_3$ is pyramidal in shape
• D. Atomic size of Boron is smaller than that of nitrogen

Answer 1

The correct option is C $B{F}_3$ is planar while $N{F}_3$ is pyramidal in shape

$B{F}_3$ is planar and hence the resultant dipole moment vectors cancel out each other. $\mu =0$.

$N{F}_3$ is pyramidal in shape. There is a net resultant dipole moment. There are two dipole moment vectors in $N{F}_3$. One dipole moment vector is towards the lone pair and the other dipole moment vector is the resultant of the three $N-F$ bonds. These dipole moments do not cancel out each other. Hence, $\mu \ne 0$.

Hence $N{F}_3$ is a polar compound.
\( \mu = 0 \)