Use VSEPR theory to predict the shape of each of the following:
a. ${\mathrm{Cl}}_4$
b. ${\mathrm{NCl}}_3$
c. ${\mathrm{SeBr}}_2$
d. ${\mathrm{CS}}_2$

We are given some compounds to predict the shape using VSEPR theory.

The shape of the carbon tetraiodide molecule, ${\mathrm{CI}}_4$. tetrahedral, was predicted using VSEPR theory.
Around the centre atom, there are 2 bonding pairs and 0 lone pairs.

We are given some compounds to predict the shape using VSEPR theory.

One lone pair and three bonded pairs are found in ${\mathrm{NCl}}_3$
The electron geometry of ${\mathrm{NCl}}_3$is tetrahedral, but the molecular geometry is trigonal pyramidal, according to the VSEPR chart.

We are given some compounds to predict the shape using VSEPR theory.

${\mathrm{SeBr}}_2$ contains two bound atoms and two lone pairs.
As a result, electron group geometry is tetrahedral and molecular geometry is bent, according to VSEPR theory.

We are given some compounds to predict the shape using VSEPR theory.

The ${\mathrm{CS}}_2$ molecule has linear molecular geometry, according to the VSEPR theory.
Because the two sulphur atoms surrounding the central atom, carbon, have two $\mathrm{C}-\mathrm{S}$ double bonds.
In the linear ${\mathrm{CS}}_2$molecular geometry, the $\mathrm{S}-\mathrm{C}-\mathrm{S}$bond angle is $180$degrees.