Explain bond formation in $BCl_{3}$. It is a symmetrical trigonal molecule.

(N/A) In the ground state,the electron configuration of Boron $(Z=5)$ is $B: [He] 2s^{2} 2p^{1}$.
In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $B^{*}: [He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1}$.
Boron now has three half-filled orbitals $(2s, 2p_{x}, 2p_{y})$,which undergo $sp^{2}$ hybridization to form three equivalent $sp^{2}$ hybrid orbitals.
These three $sp^{2}$ hybrid orbitals are arranged in a trigonal planar geometry with a bond angle of $120^{\circ}$.
Each of these three $sp^{2}$ hybrid orbitals overlaps with a half-filled $3p$ orbital of a Chlorine atom to form three $B-Cl$ $\sigma$ bonds.
Thus,$BCl_{3}$ has a trigonal planar shape with $Cl-B-Cl$ bond angles of $120^{\circ}$.