if removing an electron will create molecular ions (He 2 + and Ne 2 + ) more stable than He 2 and Ne 2 .

Question

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Answer 1

Question

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

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Answer 2

Question

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

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Answer 3

Question

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

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Answer 4

Question

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

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

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

Answer 6

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

Answer 7

Chapter 5, Problem 5.104QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Answer 8

Expert Solution & Answer

Answer to Problem 5.104QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED, Chapter 5, Problem 5.104QA

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)2

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

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

(σ1s)2 (σ*1s)1

Bond order =(Bonding electrons – Nonbonding electrons2)= (2-1)2= 0.5

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-10)2= 0

Electronic configuration of Ne₂⁺ is,

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

Bond order =(Bonding electrons – Nonbonding electrons2)= (10-9)2= 0.5

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.

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

if removing an electron will create molecular ions (He 2 + and Ne 2 + ) more stable than He 2 and Ne 2 .

To find: if removing an electron will create molecular ions (He2+ and Ne2+) more stable than He2 and Ne2.

Answer to Problem 5.104QA

Explanation of Solution

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Chapter 5 Solutions

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.