Chapter 13, Problem 9E

Interpretation Introduction

Interpretation:

The Lewis diagram for ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4,{\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ and ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ are to be drawn.

Concept introduction:

The Lewis structure shows the connectivity between atoms by identifying the lone pairs of electrons in a compound. Lewis structures are also known as Lewis dot structures. The valence electrons around an atom are shown by dots. Bonds between atoms are shown by lines and the lone pair of electrons is shown by a pair of dots.

Answer to Problem 9E

The Lewis diagram for ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4,{\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ and ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ are shown below.

Explanation of Solution

The number of valence electrons in the carbon atom is $4$. The number of valence electrons in the hydrogen atom is $1$. Chlorine, fluorine, bromine and iodine atoms belong to the same group and have $7$ valence electrons.

In ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4$, the number of carbon atoms is $2$ which gives the valence electrons from carbon atom to be $\left(4\times 2\right)8$. The number of hydrogen atoms in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4$ is $2$. This gives the valence electrons from hydrogen atoms to be $\left(2\times 1\right)2$. The number of chlorine atoms in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4$ is $4$. This gives the valence electrons from chlorine atoms to be $\left(4\times 7\right)28$.

Therefore, the total number of valence electrons in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4$ is $\left(8+2+28\right)38$.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains $2$ electrons. This uses total $14$ electrons in making seven bonds. The remaining electrons are distributed around the chlorine atoms in order to complete their octet. Therefore, the Lewis diagram for ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4$ is drawn as shown in Figure 1.

Figure 1

In ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$, the number of carbon atoms is two which gives the valence electrons from carbon atom to be $\left(4\times 2\right)8$. The number of hydrogen atom in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ is $2$. This gives the valence electrons from hydrogen atoms to be $\left(2\times 1\right)2$. The number of fluorine atoms in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ is $2$. This gives the valence electrons from fluorine atoms to be $\left(2\times 7\right)14$. The number of chlorine atoms in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ is $2$. This gives the valence electrons from chlorine atoms to be $\left(2\times 7\right)14$.

Therefore, the total number of valence electrons in ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ is $\left(8+2+14+14\right)38$.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains $2$ electrons. This uses total $14$ electrons in making seven bonds. The remaining electrons are distributed around the chlorine and fluorine atoms in order to complete their octet. Therefore, the Lewis diagram for ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ is drawn as shown in Figure 2.

Figure 2

In ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$, the number of carbon atoms is three which give the valence electrons from carbon atoms to be $\left(4\times 3\right)12$. The number of hydrogen atoms in ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ is $4$. This gives the valence electrons from hydrogen atoms to be $\left(4\times 1\right)4$. The number of bromine atoms in ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ is $3$. This gives the valence electrons from bromine atoms to be $\left(7\times 3\right)21$. The number of iodine atoms in ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ is $1$. This gives the valence electrons from iodine atom to be $\left(1\times 7\right)7$.

Therefore, the total number of valence electrons in ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ is $\left(12+4+7+21\right)44$.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains $2$ electrons. This uses total $20$ electrons in making ten bonds. The remaining electrons are distributed around the bromine and iodine atoms in order to complete their octet. Therefore, the Lewis diagram for ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ is drawn as shown in Figure 3.

Figure 3

Conclusion

The Lewis diagram for ${\text{C}}_2{\text{H}}_2{\text{Cl}}_4,{\text{C}}_2{\text{H}}_2{\text{Cl}}_2{\text{F}}_2$ and ${\text{C}}_3{\text{H}}_4{\text{Br}}_3\text{I}$ are shown in Figure 1, 2 and 3 respectively.