Chapter 6, Problem 6.100QP

Electrostatic potential maps for three compounds A. B. and C are shown here

Using the data in the following table, determine which compound corresponds to which electrostatic potential map and fill in the missing data in the table.

Compound Partial charges	Distance between charges (pm)	Dipole moment
±0.16	157
±0.43		3.10
	258	8.89

Interpretation Introduction

Interpretation: Electrostatic potential maps for three compounds A, B, and C should be identified. Also dipole moment, distance between the charges and partial charge of given compound should be found.

Concept Introduction:

• Dipole moment $\left(\text{μ}\right)$ 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.
• $\text{μ=}\text{Q}\text{×}\text{r}$
• $\begin{array}{c}\text{μ}\text{=}\text{Dipole moment}\\ \text{Q}\text{= partial charges}\\ \text{r}\text{= distance between partial charges}\end{array}$
• 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.

Answer to Problem 6.100QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
A	$\pm 0.16$	157	$\text{1}\text{.2}\text{D}$
C	$\pm 0.43$	150	$\text{3}\text{.10}\text{D}$
B	$\pm 0.718$	258	$\text{8}\text{.89}\text{D}$

Explanation of Solution

(a)

Determine the dipole moment charges for the given compound A.

Given,

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.16}{\text{e}}^{-\text{1}}\\ \text{r}\text{=}\text{157}\text{pm}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The partial charges are $+\text{0}\text{.16}{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.16}{\text{e}}^{-\text{1}}$.

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

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.16}{\text{e}}^{-\text{1}}\text{×}\frac{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}{\text{1}{\text{e}}^{-\text{1}}}\\ \text{=}\text{2}\text{.56}{\text{×10}}^{-20}\text{C}\end{array}$

The unit of distance is converted from $\text{pm}$ to meter.

$\begin{array}{c}\text{r}\text{=}\text{157}\text{pm}\frac{\text{1}{\text{×10}}^{-\text{12}}\text{m}}{\text{1}{\text{A}}^{\text{o}}}\\ \text{=}\text{1}\text{.57}{\text{×10}}^{-\text{10}}\text{m}\end{array}$

$\begin{array}{c}\text{dipole}\text{moment,μ}\text{=}\text{Q}\text{×}\text{r}\\ =\left(\text{2}\text{.56}{\text{×10}}^{-20}\text{C}\right)\times \left(\text{1}\text{.57}{\text{×10}}^{-\text{10}}\text{m}\right)\\ =4.019\times {\text{10}}^{-30}\text{C}\text{m}\end{array}$

Dipole moment in electronic charges

$\begin{array}{c}\text{μ}\text{=}4.019\times {\text{10}}^{-30}\text{C}\text{m}\text{×}\frac{\text{1}\text{D}}{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}\\ \text{=}1.2\text{D}\end{array}$

Therefore, the dipole moment for given compound is $1.2\text{D}$.

Using the given data match it with the given electrostatic potential energy maps.

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 green color is dominated comparing to the red color. The molecule seems to have less dipole moment. So the compound with dipole moment $1.2\text{D}$ is compound A.

(b)

Determine the distance between the charges.

Given,

$\begin{array}{c}\text{μ}\text{=3}\text{.10}\text{D}\\ \text{Q}\text{=}\text{0}\text{.43}{\text{e}}^{-}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The unit of dipole moment is converted from Debye to $\text{C}\text{m}$.

$\begin{array}{c}\text{μ}\text{=3}\text{.10}\text{D×}\frac{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}{\text{1}\text{D}}\\ \text{=}1.034\text{×}{\text{10}}^{-29}\text{Cm}\end{array}$

The unit of partial charge is converted to electronic charges.

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.43}{\text{e}}^{-}\frac{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}{\text{1}{\text{e}}^{-}}\\ \text{=}6.89\times {\text{10}}^{-2\text{0}}\text{C}\end{array}$

Therefore,

$\begin{array}{c}\text{Distance between charges, r}\text{=}\frac{\text{μ}}{\text{Q}}\\ \text{=}\frac{1.034\text{×}{\text{10}}^{-29}\text{C}\text{m}}{6.89\times {\text{10}}^{-2\text{0}}\text{C}}\\ \text{=}\text{1}\text{.50}\text{×}{\text{10}}^{-1\text{0}}\text{m}\\ \text{=150 pm}\end{array}$

The distance between charges of given compound is $\text{150 pm}$

Using the given data match it with the given electrostatic potential energy maps.

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 blue color is dominated comparing to the red color. The molecule has less red color comparing to other compounds. So the compound with dipole moment $3.1\text{D}$ is compound C.

(c)

Determine the partial charge for the given compound.

Partial charges are the magnitude of charges

Given,

$\begin{array}{l}\text{μ}\text{=8}\text{.89}\text{D}\\ \text{r}\text{=}\text{258}\text{pm}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The unit of dipole moment is converted from Debye to $\text{C}\text{m}$.

$\begin{array}{c}\text{μ}\text{=8}\text{.89}\text{D}\text{×}\frac{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}{\text{1}\text{D}}\\ \text{=}\text{2}\text{.966}\text{×}{\text{10}}^{-29}\text{Cm}\end{array}$

The unit of distance is converted from $\text{pm}$ to meter.

$\begin{array}{c}\text{r}\text{=}\text{258}\text{pm}\frac{\text{1}{\text{×10}}^{-\text{12}}\text{m}}{\text{1}\text{pm}}\\ \text{=}\text{2}\text{.58}\times {\text{10}}^{-\text{10}}\text{m}\end{array}$

Therefore,

$\begin{array}{c}\text{Partial charge, Q}\text{=}\frac{\text{μ}}{\text{r}}\\ \text{=}\frac{\text{2}\text{.966}\text{×}{\text{10}}^{-29}\text{C}\text{m}}{\text{2}\text{.58}\times {\text{10}}^{-\text{10}}\text{m}}\\ \text{=}\text{1}\text{.150}\text{×}{\text{10}}^{-19}\text{C}\end{array}$

Partial charge in electronic charges

$\begin{array}{c}\text{Q}\text{=}\text{1}\text{.150}\text{×}{\text{10}}^{-19}\text{C}\text{×}\frac{\text{1}{\text{e}}^{-\text{1}}}{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}\\ \text{=}\text{0}\text{.718}{\text{e}}^{-\text{1}}\end{array}$

The partial charges of given compound are $+\text{0}\text{.718}{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.718}{\text{e}}^{-\text{1}}$.

Using the given data match it with the given electrostatic potential energy maps.

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 blue color is dominated comparing to the red color. The molecule has red color comparing too that means it have high dipole. So the compound with dipole moment $8.89\text{D}$ is compound B.

Conclusion

Electrostatic potential maps for three compounds A, B, and C were identified. Also dipole moment of compound A, distance between the charges of compound C and partial charge of compound B were found