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) SiCl 4 (non polar covalent) (b) CsBr (ionic) (c) F 2 (non polar covalent) (d) CF 4 (polar covalent)

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

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) SiCl 4 (non polar covalent)
(b) CsBr (ionic)
(c) F 2 (non polar covalent)
(d) CF 4 (polar covalent)

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

c) 3. Provide a common name for each of the following alcohols. a) a) OH b) OH c) HO b) c) 4. Provide a common name for each of the following compounds. b) OH a) 5 a) Y OH c) OH

Using the critical constants for water (refer to the table in the lecture slides), calculate the second virial coefficient. Assume that the compression factor (Z) is expressed as an expansion series in terms of pressure.

+3413 pts /4800 Question 38 of 48 > Write the full electron configuration for a Kion. © Macmillan Learning electron configuration: ↓ Resources Solution Penalized → Al Tutor Write the full electron configuration for an Fion. electron configuration: T G 6 & 7 Y H כ Y 00 8 hp 9 J K no L 144 P 112 | t KC 47°F Clear ins prt sc delete ] backspace er

Answer 2

Textbook Question

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) SiCl 4 (non polar covalent)
(b) CsBr (ionic)
(c) F 2 (non polar covalent)
(d) CF 4 (polar covalent)

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

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) SiCl 4 (non polar covalent)
(b) CsBr (ionic)
(c) F 2 (non polar covalent)
(d) CF 4 (polar covalent)

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

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) SiCl 4 (non polar covalent)
(b) CsBr (ionic)
(c) F 2 (non polar covalent)
(d) CF 4 (polar covalent)

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Answer 8

Expert Solution

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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 SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

(a)

Interpretation:

Whether CHEMISTRY-TEXT, Chapter 7, Problem 7.1P 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.

Answer 13

Explanation of Solution

The given compound is SiCl4

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:

ΔE=3.0−1.8=1.2

Since, the value of ΔE 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.

Answer 14

Expert Solution

Interpretation Introduction

(b)

Interpretation:

Whether 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 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Answer 15

Expert Solution

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

(b)

Interpretation:

Whether 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.

Answer 19

Explanation of Solution

The given compound is 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:

ΔE=2.0−0.7=1.3

Since, the value of ΔE 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.

Answer 20

Expert Solution

Interpretation Introduction

(c)

Interpretation:

Whether F2 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 F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Answer 21

Expert Solution

Answer 22

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

(c)

Interpretation:

Whether F2 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.

Answer 25

Explanation of Solution

The given compound is F2

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:

ΔE=4.0−4.0=0

Since, the value of ΔE is 0 which is less 0.5 thus, the bond is non-polar covalent.

Answer 26

Expert Solution

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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 CF4.

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:

ΔE=4.0−2.5=1.5

Since, the value of ΔE is 1.5 which is between 0.5 to 1.7 thus, the bond is polar covalent.

Answer 27

Expert Solution

Answer 28

Expert Solution

Answer 29

Expert Solution

Answer 30

Interpretation Introduction

(d)

Interpretation:

Whether CF4 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.