The Bond order of N 2 + , O 2 + , C 2 + , and Br 2 2- from the MO theory and predict the bond orders.

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Answer 1

Question

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

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Answer 2

Question

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

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Answer 3

Question

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

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Answer 4

Question

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

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Answer 5

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

Answer 6

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

Answer 7

Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Answer 8

Expert Solution & Answer

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2-are below

CHEM:ATOM FOC 2E CL (TEXT), Chapter 5, Problem 5.103QA

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)2(σ2pz)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (9-4)2= 2.5

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (8-3)2= 2.5

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px)2(π2py)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (7-4)2= 1.5

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

(σ4pz)2(π4px)2(π4py)2(π*4px)2(π*4py)2(σ*4pz)2

Bond order =(Bonding electrons – Nonbonding electrons2)= (6-6)2= 0

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.

Answer 12

Ch. 5 - Prob. 5.1VP Ch. 5 - Prob. 5.2VP Ch. 5 - Prob. 5.3VP Ch. 5 - Prob. 5.4VP Ch. 5 - Prob. 5.5VP Ch. 5 - Prob. 5.6VP Ch. 5 - Prob. 5.7VP Ch. 5 - Prob. 5.8VP Ch. 5 - Prob. 5.9VP Ch. 5 - Prob. 5.10VP

The Bond order of N 2 + , O 2 + , C 2 + , and Br 2 2- from the MO theory and predict the bond orders.

To find: The Bond order of N2+, O2+, C2+, and Br22- from the MO theory and predict the bond orders.

Answer to Problem 5.103QA

Explanation of Solution

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Chapter 5 Solutions

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.