Question

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. ${\mathrm{CS}}_2$

Short answer

Part(a) ${\mathrm{CI}}_4$has tetrahedral shape.

Part(b) ${\mathrm{NCl}}_3$has trigonal pyramidal shape

Part(c) ${\mathrm{SeBr}}_2$has bent shape.

Part(d) ${\mathrm{CS}}_2$ has linear shape.

Step-by-step solution

Part(a) Step 1 : Given information

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

Part(a) Step 2 : Simplify

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.

Part(b) Step 1 : Given information

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

Part(b) Step 2 : Simplify

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.

Part(c) Step 1 : Given information

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

Part(c) Step 2 : Simplify

${\mathrm{SeBr}}_2$ 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.

Part(d) Step 1 : Given information

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

Part(d) Step 2 : Simplify

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.