Answer
On going from $\mathrm{BeCl}_{2}$ to $\left(\mathrm{BeCl}_{2}\right)_{2}$ and $\left(\mathrm{BeCl}_{2}\right)_{n}$, the hybridization of $\mathrm{Be}$ changes from $sp$ to $s p^{2}$ to $s p^{3}$; correspondingly, the geometry around the $\mathrm{Be}$ atom changes from linear to trigonal planar to tetrahedral.
Work Step by Step
See answer.