Explain the formation of bonds using $sp^2$ hybrid orbitals with an example.

(N/A) In $sp^2$ hybridization,one $s$ orbital and two $p$ orbitals of the same shell mix to form three equivalent $sp^2$ hybrid orbitals.
Example: $BCl_3$ molecule.
The electronic configuration of Boron $(Z=5)$ in the ground state is $[He] 2s^2 2p^1$. In the excited state,one $2s$ electron is promoted to the $2p$ orbital,resulting in $[He] 2s^1 2p_x^1 2p_y^1$.
These three orbitals $(2s, 2p_x, 2p_y)$ undergo $sp^2$ hybridization to form three $sp^2$ hybrid orbitals. These orbitals are oriented in a trigonal planar geometry at an angle of $120^{\circ}$.
Each of these three $sp^2$ hybrid orbitals overlaps with the half-filled $3p$ orbital of a Chlorine atom to form three $B-Cl$ $\sigma$ bonds. Thus,$BCl_3$ has a trigonal planar geometry with $Cl-B-Cl$ bond angles of $120^{\circ}$.