Explain the bond formation by $sp$ orbitals.

(N/A) In $BeCl_{2}$,the electronic configuration of $Be$ is $[He] 2s^{2}$.
It forms an excited state $Be^{*}: [He] 2s^{1} 2p^{1}$.
The divalency of $Be$ in the excited state occurs with one electron in the $2s$ orbital and one in the $2p$ orbital.
One $2s$ and one $2p$ orbital of the excited $Be$ atom undergo hybridization to form two $sp$ hybrid orbitals.
These two $sp$ hybrid orbitals are oriented in opposite directions,forming an angle of $180^{\circ}$.
The $Cl$ atom has the configuration $[Ne] 3s^{2} 3p_{x}^{2} 3p_{y}^{2} 3p_{z}^{1}$.
Each $sp$ hybrid orbital of $Be$ overlaps with the $3p_{z}$ orbital of a chlorine $(Cl)$ atom along the $z$-axis to form two $Be-Cl$ $\sigma$ bonds.
This axial overlapping results in a linear shape for the $BeCl_{2}$ molecule.