Chapter 3, Problem 3.80QP

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{HNO}}_3$ has to be drawn.

Explanation of Solution

The given compound is ${\text{HNO}}_3$.

Step – 1

The skeletal structure and charge has to be drawn.

Step – 2

The total number of valence electrons is determined.

    $\begin{array}{l}{\text{3O atom × 6 valence electrons =18e}}^{\text{-}}\\ {\text{1N atom × 5 valence electrons =5e}}^{\text{-}}\\ {\text{1H atom × 1 valence electrons =1e}}^{\text{-}}\\ \text{----------------------------------------------}\\ {\text{ = 24 e}}^{\text{-}}\end{array}$

Step – 3

Connect the atoms using single bond.

Step – 4

Start placing the lone pairs electrons around the atoms, giving each an octet.  Do not exceed the 24 available electrons.

Step – 5

A final electron count indicates that each atom has eight valence electrons surrounding it.

(b)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{CCl}}_{\text{4}}$ has to be drawn.

Explanation of Solution

The given compound is ${\text{CCl}}_{\text{4}}$.

Step – 1

The skeletal structure and charge has to be drawn.

Step – 2

The total number of valence electrons is determined.

    $\begin{array}{l}{\text{4Cl atom × 7 valence electrons =28e}}^{\text{-}}\\ {\text{1C atom × 4 valence electrons =4e}}^{\text{-}}\\ \text{----------------------------------------------}\\ {\text{ = 32 e}}^{\text{-}}\end{array}$

Step – 3

Connect the atoms using single bond.

Step – 4

Start placing the lone pairs electrons around the atoms, giving each an octet.  Do not exceed the 32 available electrons.

Step – 5

A final electron count indicates that each atom has eight valence electrons surrounding it.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{PBr}}_{\text{3}}$ has to be drawn.

Explanation of Solution

The given compound is ${\text{PBr}}_{\text{3}}$.

Step – 1

The skeletal structure and charge has to be drawn.

Step – 2

The total number of valence electrons is determined.

    $\begin{array}{l}{\text{3Br atom × 7 valence electrons =21e}}^{\text{-}}\\ {\text{1P atom × 5 valence electrons =5e}}^{\text{-}}\\ \text{----------------------------------------------}\\ {\text{ = 26 e}}^{\text{-}}\end{array}$

Step – 3

Connect the atoms using single bond.

Step – 4

Start placing the lone pairs electrons around the atoms, giving each an octet.  Do not exceed the 26 available electrons.

Step – 5

A final electron count indicates that each atom has eight valence electrons surrounding it.

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ has to be drawn.

Explanation of Solution

The given compound is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

Step – 1

The skeletal structure and charge has to be drawn.

Step – 2

The total number of valence electrons is determined.

    $\begin{array}{l}{\text{2C atom × 4 valence electrons =8e}}^{\text{-}}\\ {\text{1O atom × 6 valence electrons =6e}}^{\text{-}}\\ {\text{6H atom × 6 valence electrons =6e}}^{\text{-}}\\ \text{----------------------------------------------}\\ {\text{ = 20 e}}^{\text{-}}\end{array}$

Step – 3

Connect the atoms using single bond.

Step – 4

Start placing the lone pairs electrons around the atoms, giving each an octet.  Do not exceed the 20 available electrons.

Step – 5

A final electron count indicates that each atom has eight valence electrons surrounding it.