Chapter 12, Problem 61QAP

Write a Lewis structure for each of the following simple molecules. Show all bonding valence electron pairs as lines and all nonbonding valence electron pairs as dots.

msp;  $\begin{array}{l}\text{a}.{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}} \text{d}.{\text{ C}}_{\text{4}}{\text{H}}_{\text{1}0}\\ \text{b}.{\text{ NF}}_{\text{3}} \text{d}.{\text{ SiCl}}_{\text{4}}\end{array}$

Interpretation Introduction

(a)

Interpretation:

The Lewis structure of the molecule is to be drawn.

Concept Introduction:

The representation of an element along with its valence electrons is referred to as Lewis symbol or Electron Dot Symbol. The electrons which are present in valence shell are represented by dots around the symbol of an element. The Lewis structure exhibits the connection between atoms.

Answer to Problem 61QAP

Lewis structure of the molecule is shown in Figure 1.

Explanation of Solution

Total valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{C}$ and $\text{H}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)\\ =2\left(4\right)+6\left(1\right)\\ =14\text{valence}\text{electrons}\end{array}$

The Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ indicates that there are total $14$ valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

The three electrons of carbon atoms combine with one electron each of hydrogen and one electron of carbon combines with another carbon atom.

The Lewis structure is shown below.

Figure 1.

Interpretation Introduction

(b)

Interpretation:

The Lewis structure of the molecule is to be drawn.

Concept Introduction:

The representation of an element along with its valence electrons is referred to as Lewis symbol or Electron Dot Symbol. The electrons which are present in valence shell are represented by dots around the symbol of an element. The Lewis structure exhibits the connection between atoms.

Answer to Problem 61QAP

Lewis structure of the molecule is shown in Figure 2.

Explanation of Solution

Total valence electrons in ${\text{NF}}_3$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{N}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{F}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{N}$ and $\text{F}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{N}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{F}\right)\\ =1\left(5\right)+3\left(7\right)\\ =26\text{valence}\text{electrons}\end{array}$

The Lewis structure of ${\text{NF}}_3$ indicates that there are total $26$ valence electrons in ${\text{NF}}_3$.

The three electrons of nitrogen atoms combine with one electron each of fluorine.

The Lewis structure is shown below.

Figure 2.

Interpretation Introduction

(c)

Interpretation:

The Lewis structure of the molecule is to be drawn.

Concept Introduction:

The representation of an element along with its valence electrons is referred to as Lewis symbol or Electron Dot Symbol. The electrons which are present in valence shell are represented by dots around the symbol of an element. The Lewis structure exhibits the connection between atoms.

Answer to Problem 61QAP

Lewis structure of the molecule is shown in Figure 3.

Explanation of Solution

Total valence electrons in ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{C}$ and $\text{H}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)\\ =4\left(4\right)+10\left(1\right)\\ =26\text{valence}\text{electrons}\end{array}$

The Lewis structure of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ indicates that there are total $26$ valence electrons in ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$.The one electron of carbon atoms combines with one electron of the carbon atom and the rest combines with hydrogen.

The Lewis structure is shown below.

Figure 3.

Interpretation Introduction

(d)

Interpretation:

The Lewis structure of the molecule is to be drawn.

Concept Introduction:

The representation of an element along with its valence electrons is referred to as Lewis symbol or Electron Dot Symbol. The electrons which are present in valence shell are represented by dots around the symbol of an element. The Lewis structure exhibits the connection between atoms.

Answer to Problem 61QAP

Lewis structure of the molecule is shown in Figure 4.

Explanation of Solution

Total valence electrons in ${\text{SiCl}}_{\text{4}}$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{Si}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{Cl}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{C}$ and $\text{H}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{Si}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{Cl}\right)\\ =1\left(4\right)+4\left(7\right)\\ =32\text{valence}\text{electrons}\end{array}$

The Lewis structure of ${\text{SiCl}}_{\text{4}}$ indicates that there are total $32$ valence electrons in ${\text{SiCl}}_{\text{4}}$. The four electrons of silicon atoms combine with one electron each of chlorine.

The Lewis structure is shown below.

Figure 4.