Chapter 9, Problem 75AP

Predict the bond angles for the following molecules: $\left(\text{a}\right){\text{ BeCl}}_{\text{2}}\text{, }\left(\text{b}\right){\text{ BCl}}_{\text{3}}\text{, }\left(\text{c}\right){\text{ CCl}}_{\text{4}}\text{, }\left(\text{d}\right){\text{ CH}}_{\text{3}}\text{Cl, }\left(\text{e}\right){\text{ Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (arrangement of atoms: $\text{ClHgHgCl), }\left(\text{f}\right){\text{ SnCl}}_{\text{2}}\text{, }\left(\text{g}\right){\text{ H}}_{\text{2}}{\text{O}}_{\text{2}}\text{, }\left(\text{h}\right){\text{ SnH}}_{\text{4}}$.

Interpretation Introduction

Interpretation:

The bond angles of the given molecules are to be determined.

Concept introduction:

Electron domain is the number of bonds and lone pair of electrons around the central atom.

According to VSEPR, the electron domains repel each other and arrange themselves as far apart as possible in space.

Molecular geometry represents the arrangement of bonded atoms. If no lone pairs are present, then the molecular geometry is the same as the electron domain geometry.

The repulsion in lone pair–lone pair is greater than that in lone pair–bond pair, which in turn is greater than that in bond pair–bond pair. The repulsion due to a double bond is more than that due to single bonds.

Answer to Problem 75AP

Solution:

a) $\text{180}°$

b) $\text{120}°$

c) $\text{109}\text{.5}°$

d) $\text{109}\text{.5}°$

e) $\text{180}°$

f) Less than $\text{120}°$

g) Less than $\text{109}\text{.5}°$

h) $\text{109}\text{.5}°$

Explanation of Solution

a) ${\text{BeCl}}_{\text{2}}$

The Lewis structure of ${\text{BeCl}}_{\text{2}}$

is as follows:

In this structure, the $\text{Be}$

atom is bonded to two chlorine atoms by single bonds, and there is no lone pair on $\text{Be}$. Thus, there are two electron domains around the $\text{Be}$

atom. According to VSEPR, two electron domains are arranged on opposite sides to give a linear shape, with a bond angle of $\text{180}°$.

b) ${\text{BCl}}_{\text{3}}$

The Lewis structure of ${\text{BCl}}_{\text{3}}$ is as follows:

In this structure, the $\text{B}$

atom is bonded to three chlorine atoms by single bonds, and there is no lone pair on $\text{B}$. Thus, there are three electron domains around the $\text{B}$

atom. According to VSEPR, three electron domains are arranged in a trigonal planar shape with a bond angle of $\text{120}°$.

c) ${\text{CCl}}_{\text{4}}$

The Lewis structure of ${\text{CCl}}_{\text{4}}$

is as follows:

In this structure, the $\text{C}$

atom is bonded to four chlorine atoms by single bonds, and there is no lone pair on $\text{C}$. Thus, there are four electron domains around the $\text{C}$

atom. According to VSEPR, four electron domains are arranged in a tetrahedral shape with a bond angle of $\text{109}\text{.5}°$.

d) ${\text{CH}}_{\text{3}}\text{Cl}$

The Lewis structure of ${\text{CH}}_{\text{3}}\text{Cl}$

is as follows:

In this structure, the $\text{C}$

atom is bonded to three hydrogen atoms and one chlorine atom by single bonds, and there is no lone pair on $\text{C}$. Thus, there are four electron domains around the $\text{C}$

atom. According to VSEPR, four electron domains are arranged in a tetrahedral shape with a bond angle of $\text{109}\text{.5}°$.

e) ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$

The Lewis structure of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$

is as follows:

In this structure, each $\text{Hg}$

atom is bonded to one chorine atom, and to each other. Thus, there are two electron domains around each $\text{Hg}$

atom. Thus, the shape is linear with a bond angle of $\text{180}°$.

f) ${\text{SnCl}}_{\text{2}}$

The Lewis structure of ${\text{SnCl}}_{\text{2}}$

is as follows:

In this structure, the $\text{Sn}$

atom is bonded to two chlorine atoms by single bonds, and there is a lone pair on $\text{Sn}$. Thus, there are three electron domains around the $\text{Sn}$

atom. According to VSEPR, three electron domains are arranged in a trigonal planar shape with a bond angle of $\text{120}°$. However, due to the presence of a lone pair on the $\text{Sn}$

atom, the shape of ${\text{SnCl}}_{\text{2}}$

is bent with a bond angle less than $\text{120}°$.

g) ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

The Lewis structure of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

is as follows:

In this structure, each $\text{O}$

atom is bonded to one hydrogen atom, and to each other. There are two lone pairs on each oxygen atom. Thus, there are four electron domains around each oxygen atom. According to VSEPR, four electron domains have a tetrahedral shape. Due to the presence of two lone pairs, the shape is bent with a bond angle less than $\text{109}\text{.5}°$.

h) ${\text{SnH}}_{\text{4}}$

The Lewis structure is as follows:

In this structure, the $\text{Sn}$

atom is bonded to four chlorine atoms by single bonds, and there is no lone pair on $\text{Sn}$. Thus, there are four electron domains around the $\text{Sn}$

atom. According to VSEPR, four electron domains are arranged in a tetrahedral shape with a bond angle of $\text{109}\text{.5}°$.