Chapter 7, Problem 7.1P

Use the electro negativity values in Figure 7.4 top redict whether the bonds in the following compounds are non-polar covalent, polar covalent, or ionic. Which is incorrectly labeled?
(a) ${\text{SiCl}}_{\text{4}}$ (non polar covalent)
(b) CsBr (ionic)
(c) ${\text{F}}_2$ (non polar covalent)
(d) ${\text{CF}}_{\text{4}}$ (polar covalent)

Interpretation Introduction

(a)

Interpretation:

Whether is non-polar covalent or not needs to be determined.

Concept introduction:

The nature of the compound depends on the difference in the electronegativity of two atoms. The given compound is considered as a non-polar covalent compound if the electronegativity difference between two atoms is less than 0.5. If the electronegtaivity difference is between 0.5 to 1.7 then the bond is polar covalent in nature. If the difference is more than 1.7, the bond between two atoms will be ionic in nature.

Explanation of Solution

The given compound is ${\text{SiCl}}_{\text{4}}$

According to the given figure in the question, the electronegativity value of Si and Cl is 1.8 and 3.0 respectively.

The difference between the two atoms can be calculated as follows:

$\Delta E=3.0-1.8=1.2$

Since, the value of $\Delta E$ is 1.2 which is more than 0.5 thus, it is not non-polar covalent in nature. The calculated value is between 0.5 to 1.7 thus, the bond is polar covalent in nature.

Interpretation Introduction

(b)

Interpretation:

Whether $\text{CsBr}$ is ionic or not needs to be determined.

Concept introduction:

The nature of the compound depends on the difference in the electronegativity of two atoms. The given compound is considered as a non-polar covalent compound if the electronegativity difference between two atoms is less than 0.5. If the electronegtaivity difference is between 0.5 to 1.7 then the bond is polar covalent in nature. If the difference is more than 1.7, the bond between two atoms will be ionic in nature.

Explanation of Solution

The given compound is $\text{CsBr}$

According to the given figure in the question, the electronegativity value of Cs and Br is 0.7 and 2.8 respectively.

The difference between the two atoms can be calculated as follows:

$\Delta E=2.0-0.7=1.3$

Since, the value of $\Delta E$ is 1.3 which is less than 1.7 thus, the bond is not ionic. The calculated value is between 0.5 to 1.7 thus, the bond is polar covalent in nature.

Interpretation Introduction

(c)

Interpretation:

Whether ${\text{F}}_2$ is non-polar covalent or not needs to be determined.

Concept introduction:

The nature of the compound depends on the difference in the electronegativity of two atoms. The given compound is considered as a non-polar covalent compound if the electronegativity difference between two atoms is less than 0.5. If the electronegtaivity difference is between 0.5 to 1.7 then the bond is polar covalent in nature. If the difference is more than 1.7, the bond between two atoms will be ionic in nature.

Explanation of Solution

The given compound is ${\text{F}}_2$

According to the given figure in the question, the electronegativity value of F is 4.0.

The difference between the two atoms can be calculated as follows:

$\Delta E=4.0-4.0=0$

Since, the value of $\Delta E$ is 0 which is less 0.5 thus, the bond is non-polar covalent.

Interpretation Introduction

(d)

Interpretation:

Whether ${\text{CF}}_4$ is polar covalent or not needs to be determined.

Concept introduction:

The nature of the compound depends on the difference in the electronegativity of two atoms. The given compound is considered as a non-polar covalent compound if the electronegativity difference between two atoms is less than 0.5. If the electronegtaivity difference is between 0.5 to 1.7 then the bond is polar covalent in nature. If the difference is more than 1.7, the bond between two atoms will be ionic in nature.

Explanation of Solution

The given compound is ${\text{CF}}_4$.

According to the given figure in the question, the electronegativity value of C and F is 2.5 and 4.0 respectively.

The difference between the two atoms can be calculated as follows:

$\Delta E=4.0-2.5=1.5$

Since, the value of $\Delta E$ is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.