Solution:
The $B-C l$ bond is normally polar as a result of the distinction in the electronegativities of $\mathrm{Cl}$ and $B$. However the particle of $\mathrm{BCl}_{3}$ is non-polar. That is on the grounds that the state of $\mathrm{BCl}_{3}$ is three-sided planar. The particle is hilter kilter. So the particular $B-C I$ bond dipole minutes drop one another, making a zero dipole second.
