Question

Is ${\mathrm{SiCl}}_4$ square planar or tetrahedral?

Answer 1

VSPER theory:

• The VSEPR theory is known as the theory, which helps in predicting the shape and geometry of the molecules by using the electron pairs that surround the central atoms of the molecule.
• This theory is based on the basis that there is a repulsion between the pairs of valence electrons in all atoms, hence the atoms will always tend to arrange themselves in a manner, in which this electron pair repulsion is minimalized.

VSPER formula to predict the structure of any molecule:

• Total number of electrons around the central atom = $$\begin{gathered} \frac{1}{2}(\mathrm{number}\mathrm{of}\mathrm{valence}\mathrm{electrons}\mathrm{of}\mathrm{central}\mathrm{atom}+\mathrm{number}\mathrm{of}\mathrm{atoms}\mathrm{linked}\mathrm{to}\mathrm{central}\mathrm{atom}\mathrm{by}\mathrm{single}\mathrm{bonds} \\ +\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{negative}\mathrm{ions}-\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{positive}\mathrm{ions}) \end{gathered}$$
• The number of Bond pairs = Total number of atoms linked to the central atom by single bonds.
• Number of lone pairs = Total number of electrons – No of shared pair

Shape of ${\mathrm{SiCl}}_4$:

• The shape of any molecule can be determined by VSPER theory in which molecular geometry is based on the number of electronic bond pairs in the valence shell between atoms of a molecule or ion.
• Hence, by using the VSPER formula for the molecule, ${\mathrm{SiCl}}_4$, Total number of electrons = $\frac{1}{2}\left(4+4\right)=4$
• Also, here the number of Bond pairs $=4$(as 4 atoms of Chlorine are attached to the central Silicon atom)
• And, the number of lone pairs on Si $=4-4=0.$
• Hence, by calculation we get that, the ${\mathrm{SiCl}}_4$molecule has $4$ Bond pairs and 0 lone pairs.
• This suggests that the ${\mathrm{SiCl}}_4$ is an${\mathrm{AX}}_4$ -type molecule, with four $\mathrm{Si}-\mathrm{Cl}$ bonds.
• According to the theory, these bonds must be pointing towards the corners to form a regular tetrahedral structure which can be shown as;

Hence, this concludes, that ${\mathrm{SiCl}}_4$has a tetrahedral shape.