**Formation and structure of Boron Trichloride (BCI _{3)} molecule:** The electronic configuration of boron (5) is 1s

^{2}2s

^{2}2p

_{x}

^{1}2p

_{y}

^{0}2p

_{z}

^{0}. When one electron is transferred from 2s orbital to 2p orbital, it has three unpaired electrons, which explains the valency of three for boron. The sp

^{2}hybridization takes place on boron atom.

**At normal state, B (5): 1s ^{2} 2s^{2} 2p_{x}^{1} 2p_{y}^{0} 2p_{z}^{0}**

**At excited state. B* (5): 1s ^{2} 2s^{1} 2p_{x}^{1} 2p_{y}^{0} 2p_{z}^{0}**

**After sp ^{2} hybridization, B (5): 1s^{2} ψ_{1}^{1} ψ_{2}^{1} ψ_{3}^{1}**

Where ψ_{1}, ψ_{2} and ψ_{3} are hybrid orbitals. They overlap with 3p_{z} orbitals (Containing one electron each) of three chlorine atoms as follows and form three B-CI bonds. Then BCl_{3} molecule is formed.

So, Boron Trichloride (BCl_{3)} is a planar molecule, where all bond angles are 120^{0}. Although each B-Cl bonds is polar, the vector summation of three dipoles is zero and Bell has a zero dipole moment.