Question

Why the shape of ${\mathrm{BeCl}}_2$molecule is linear?

Answer 1

Step 1: Formula used for calculation of hybridization:

• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right)$
• Where V is the number of valence electrons, M is the monovalent electron, C is the positive charge and A represents the negative charge.

Step 2: Calculation for Hybridisation of ${\mathrm{BeCl}}_2$:

• The total number of valence electrons on Beryllium in ${\mathrm{BeCl}}_2$ is $4$ and there are $2$ monovalent atoms.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(2+2\right)=\frac{1}{2}\times 4 \\ \mathrm{Steric}\mathrm{number}=2 \end{gathered}$$
• The steric number is $2$, thus the hybridization of ${\mathrm{BeCl}}_2$ is $\mathrm{sp}$.

Step 3: Shape of Beryllium chloride:

• Beryllium chloride is a molecule with formula ${\mathrm{BeCl}}_2$.
• During the formation of ${\mathrm{BeCl}}_2$_, the beryllium atom is covalently bonded with two chlorine atoms.
• In this molecule, there are two bond pairs while no lone pair is found in the molecule.
• The hybridization of ${\mathrm{BeCl}}_2$is $\mathrm{sp}$ hybridization.
• Thus, as per VSEPR theory, the shape of ${\mathrm{BeCl}}_2$is linear.