Chapter 12, Problem 59QAP

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;  $\begin{array}{l}\text{a}.{\text{ NBr}}_{\text{3}} \text{c}.{\text{ CBr}}_{\text{4}}\\ \text{b}.\text{ HF} \text{d}.{\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}\end{array}$

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 59QAP

Lewis structure of the molecule is shown in Figure 1.

Explanation of Solution

The Lewis structure of ${\text{NBr}}_{\text{3}}$ indicates that there are total $26$ valence electrons in ${\text{NBr}}_{\text{3}}$.

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

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

Where,

• $\text{n}$ is the number of atoms.

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

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

One valence electron of each bromine atom combines with three valence electrons of nitrogen, which results in the completion of octet in $\text{N}$ and $\text{Br}$ atoms. 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 59QAP

Lewis structure of the molecule is shown in Figure 2.

Explanation of Solution

The Lewis structure of ${\text{CBr}}_4$ indicates that there are total $32$ valence electrons in ${\text{CBr}}_4$.

Total valence electrons in ${\text{CBr}}_4$ is calculated as shown below.

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

Where,

• $\text{n}$ is the number of atoms.

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

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

One valence electron of each bromine atom combines with four valence electrons of carbon, which results in the completion of octet in $\text{C}$ and $\text{Br}$ atoms. 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 59QAP

Lewis structure of the molecule is shown in Figure 3.

Explanation of Solution

The Lewis structure of $\text{HF}$ indicates that there are total $8$ valence electrons in $\text{HF}$.

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

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

Where,

• $\text{n}$ is the number of atoms.

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

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{F}\right)\\ =1\left(1\right)+1\left(7\right)\\ =8\text{valence}\text{electrons}\end{array}$

The example of a Lewis structure is shown below.

Figure 3

One valence electron of hydrogen atom combines with one valence electron of fluorine, which results in the completion of octet in $\text{F}$ and duplet in $\text{H}$ atoms.

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 59QAP

Lewis structure of the molecule is shown in Figure 4.

Explanation of Solution

The Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ indicates that there are total $10$ valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.

Total valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ 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.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)\\ =2\left(4\right)+2\left(1\right)\\ =10\text{valence}\text{electrons}\end{array}$

The Lewis structure is shown below.

Figure 4.