Explain the shape and bond angle in `BCl_(3)` molecule in terms of Valence Bond Theory.

Explanation

In `BCl_(3)` the central atom is boron. Its atomic number is 5. Therefore its ground state electron configuration is `1s^(2)2s^(2)2p_(x)^(1)2p_(y)^(0)2p_(z)^(0)`. From this configuration it is evident that it exhibits mono-valency. The first excited state configuration of B is `1s^(2)2s^(1)2p_(x)^(1)2p_(y)^(0)2p_(z)^(0)`. Since, there are two half-filled orbitals, the valency of Boron is 3.
In order to explain the formation of `BCl_(3)` molecule `sp^(2)` hybridisation is to be assumed to boron atom in its excited state I. As a result of which three `sp^(2)` hybrid orbitals of 'B' atom overlap Head - Head with `3p_(z)` orbitals of three Cl atoms forming three sigma bonds. The shape of the molecule is plane triangular and the bond angle is `120^@`.