Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found. Concept Introduction: Dipole moment ( μ ) occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity. μ= Q × r μ = Dipole moment Q = partial charges r = distance between partial charges Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another. To calculate : The dipole moment of given set of compounds and it should be matched with the given respective maps.

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Answer 1

Question

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

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Answer 2

Question

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

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Answer 3

Question

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

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Answer 4

Question

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

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 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

Answer 6

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of given compound were found.

Answer 7

Chapter 6, Problem 6.93QP

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment of given compounds should be found.

Concept Introduction:

Dipole moment $(μ)$ occurs on polar bonds where there is a separation of charges within the molecule. Dipole moment is formed due to the difference in electronegativity of atoms within the molecule and is a measure of polarity.

$μ= Q × r$

$μ = Dipole momentQ = partial chargesr = distance between partial charges$

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

To calculate: The dipole moment of given set of compounds and it should be matched with the given respective maps.

Answer 8

Expert Solution & Answer

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$± 0.19$	213	$1.94 D$
C	$± 0.051$	214	$0.524 D$
A	$± 0.68$	315	$10.3 D$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

(a)

Given,

$Q = 0.19 e −1r = 213 pm$

$μ= Q × r$

The partial charges are $+ 0.19 e−$ and $− 0.19 e−$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.19 e −1 × 1.6022 ×10 −19 C1 e −1= 3.04 ×10 −20 C$

The unit of distance is converted from pm to meter.

$r = 213pm 1 ×10 −12 m1 Ao= 2.13 ×10 −10 m$

$dipole moment,μ = Q × r=(3.04 × 10 −20 C) × (2.13 ×10 −10m)= 6.475 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 1 $μ = 6.475 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 1.94 D$

Therefore, the dipole moment of given compound is $1.94 D$

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is dominated comparing to the blue color. The molecule seems to have some dipole moment and also green cloud separating blue and red clouds. So the compound with dipole moment $1.94 D$ is compound B.

(b)

Given,

$Q = 0.051 e −1r = 214 pm$

$μ= Q × r$

The partial charges are $+ 0.051 e −1$ and $− 0.051 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.051 e −1 × 1.6022 ×10 −19 C1 e −1= 8.171 ×10 −21 C$

The unit of distance is converted from pm to meter.

$r = 214 pm 1 ×10 −10 m1 Ao= 2.14 ×10 −10 m$

$dipole moment, μ = Q × r=(8.171 ×10 −21 C) × (2.14 ×10 −10m)= 1.749 × 10 −30 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 2 $μ = 1.749 × 10 −30 C m × 1 D3.336 × 10 −30 C m= 0.524 D$

Therefore, the dipole moment of given compound is $0.524 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color is equally distributed in the molecule so have low dipole moment. So the compound with dipole moment $0.524 D$ is compound C.

(c)

Given,

$Q = 0.68 e −1r =315 pm$

$μ= Q × r$

The partial charges are $+ 0.68 e −1$ and $− 0.68 e −1$ .

The unit of partial charge is converted from electronic charges to Coulombs.

$Q = 0.68 e −1 × 1.6022 ×10 −19 C1 e −1= 1.09 ×10 −19 C$

The unit of distance is converted from pm to meter.

$r = 315 pm 1 ×10 −10 m1 Ao= 3.15 ×10 −10 m$

$dipole moment,μ = Q × r=(1.09 ×10 −19 C) × (3.15 ×10 −10 m)= 3.433 × 10 −29 C m$

Dipole moment in electronic charges

GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST, Chapter 6, Problem 6.93QP , additional homework tip 3 $μ = 3.433 × 10 −29 C m × 1 D3.336 × 10 −30 C m= 10.3 D$

Therefore, the dipole moment of given compound is $10.3 D$ .

Using the given data match the compound with the respective potential map.

Distribution of electrons in a molecule can be shown using electrostatic potential maps three dimensionally. It is also known as electrostatic energy maps. Distributions of charge help to know the interaction of molecule with one another.

Red color in the electrostatic potential map shows that electrons spends more time in that region whereas the blue color shows that electrons spends less time in that region. Green area represents that electrons spend moderate amount of time.

Here the red color and blue color are equally distributed in the molecule and have visible separation of cloud so have high dipole moment. So the compound with dipole moment $10.3 D$ is compound A.