Answer
See work below for details.
Work Step by Step
The bonds in the boron chloride are stronger and shorter than a typical single bond, which we can explain using valence bond theory and hybridization. The boron atom uses $sp^{2}$ hybridized orbitals to form sigma bonds with the three halogen atoms.
Because boron’s three valence electrons are used to form the sigma bonds, the third p orbital of boron is an empty orbital that is perpendicular to the trigonal plane of the molecule. Each halogen atom has a filled p orbital, also perpendicular to the trigonal plane of the molecule. The empty p orbital on the boron atom can overlap with the full p orbitals on the halogens, forming a coordinate-covalent-type second bond. In $BCl_{3}$ , the boron and chlorine are joined by bonds resembling a double bond. Like normal double bonds, the boron–chlorine bond is shorter and stronger than a single bond.
