Question

Discuss the shape of the following molecules using the VSEPR model:
$BeC{l}_2,BC{l}_3,SiC{l}_4,As{F}_5,H_{2}S,P{H}_3$

Answer 1

$BeC{l}_2$:

$Cl:Be:Cl$

The central atom has no lone pair and there are two bond pairs. i.e., $BeC{l}_2$ is of the type $A{B}_2$. Hence, it has a linear shape.

$BC{l}_3$:

The central atom has no lone pair and there are three bond pairs. Hence, it is of the type $A{B}_3$. Hence, it is trigonal planar.

$SiC{l}_4$:

The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl4 is tetrahedral being the $A{B}_4$ type molecule.

$As{F}_5$:

The central atom has no lone pair and there are five bond pairs. Hence, $As{F}_5$ is of the type $A{B}_5$. Therefore, the shape is trigonal bipyramidal.

$H_{2}S$:

The central atom has one lone pair and there are two bond pairs. Hence, $H_{2}S$ is of the type $A{B}_{2}E$. The shape is Bent.

$P{H}_3$ :

The central atom has one lone pair and there are three bond pairs. Hence, $P{H}_3$ is of the $A{B}_{3}E$ type. Therefore, the shape is trigonal pyramidal.