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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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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8:57 PM Sun Jan 26 Content ← Explanation Page X Content X ALEKS Jade Nicol - Le A https://www-av C www-awa.aleks.com O States of Matter Understanding consequences of important physical properties of liquids ? QUESTION Liquid A is known to have a lower viscosity and lower surface tension than Liquid B. Use these facts to predict the result of each experiment in the table below, if you can. experiment Liquid A and Liquid B are each pumped through tubes with an inside diameter of 27.0 mm, and the pressures PA and PB needed to produce a steady flow of 2.4 mL/s are measured. 25.0 mL of Liquid A are poured into a beaker, and 25.0 mL of Liquid B are poured into an identical beaker. Stirrers in each beaker are connected to motors, and the forces FA and FB needed to stir each liquid at a constant rate are measured. predicted outcome OPA will be greater than PB OPA will be less than PB OPA will be equal to PB It's impossible to predict whether PA or PB will be greater without more information.…

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Answer 2

Question

Chapter 5, Problem 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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 5.97QA

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

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.97QA

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 5.97QA

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.