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 $\left(\text{μ}\right)$ 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.

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

Answer to Problem 6.93QP

Compound	Partial charge	Distance between charges (pm)	Dipole moment
B	$\pm 0.19$	213	$\text{1}\text{.94}\text{D}$
C	$\pm 0.051$	214	$\text{0}\text{.524}\text{D}$
A	$\pm 0.68$	315	$\text{10}\text{.3}\text{D}$

Explanation of Solution

(a)

Given,

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.19}{\text{e}}^{-\text{1}}\\ \text{r}\text{=}\text{213}\text{pm}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The partial charges are $+0.19{\text{e}}^{-}$ and $-0.19{\text{e}}^{-}$.

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

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.19}{\text{e}}^{-\text{1}}\text{×}\frac{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}{\text{1}{\text{e}}^{-\text{1}}}\\ \text{=}\text{3}\text{.04}{\text{×10}}^{-20}\text{C}\end{array}$

The unit of distance is converted from pm to meter.

$\begin{array}{c}\text{r}\text{=}\text{213pm}\frac{\text{1}{\text{×10}}^{-\text{12}}\text{m}}{\text{1}{\text{A}}^{\text{o}}}\\ \text{=}\text{2}\text{.13}{\text{×10}}^{-\text{10}}\text{m}\end{array}$

$\begin{array}{c}\text{dipole}\text{moment,μ}\text{=}\text{Q}\text{×}\text{r}\\ =\left(\text{3}\text{.04}\text{×}{\text{10}}^{-20}\text{C}\right)\times \left(2.13\times {\text{10}}^{-\text{10}}\text{m}\right)\\ =6.475\times {\text{10}}^{-30}\text{C}\text{m}\end{array}$

Dipole moment in electronic charges

$\begin{array}{c}\text{μ}\text{=}6.475\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}\text{×}\frac{\text{1}\text{D}}{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}\\ \text{=}1.94\text{D}\end{array}$

Therefore, the dipole moment of given compound is $1.94\text{D}$

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 $1.94\text{D}$ is compound B.

(b)

Given,

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.051}{\text{e}}^{-\text{1}}\\ \text{r}\text{=}\text{214}\text{pm}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The partial charges are $+\text{0}\text{.051}{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.051}{\text{e}}^{-\text{1}}$.

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

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.051}{\text{e}}^{-\text{1}}\text{×}\frac{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}{\text{1}{\text{e}}^{-\text{1}}}\\ \text{=}\text{8}\text{.171}{\text{×10}}^{-21}\text{C}\end{array}$

The unit of distance is converted from pm to meter.

$\begin{array}{c}\text{r}\text{=}\text{214}\text{pm}\frac{\text{1}{\text{×10}}^{-\text{10}}\text{m}}{\text{1}{\text{A}}^{\text{o}}}\\ \text{=}\text{2}\text{.14}\times {\text{10}}^{-\text{10}}\text{m}\end{array}$

$\begin{array}{c}\text{dipole}\text{moment,}\text{μ}\text{=}\text{Q}\text{×}\text{r}\\ =\left(\text{8}\text{.171}{\text{×10}}^{-21}\text{C}\right)\times \left(\text{2}\text{.14}\times {\text{10}}^{-\text{10}}\text{m}\right)\\ =1.749\times {\text{10}}^{-30}\text{C}\text{m}\end{array}$

Dipole moment in electronic charges

$\begin{array}{c}\text{μ}\text{=}1.749\times {\text{10}}^{-30}\text{C}\text{m}\text{×}\frac{\text{1}\text{D}}{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}\\ \text{=}\text{0}\text{.524}\text{D}\end{array}$

Therefore, the dipole moment of given compound is $\text{0}\text{.524}\text{D}$.

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 $\text{0}\text{.524}\text{D}$ is compound C.

(c)

Given,

$\begin{array}{l}\text{Q}\text{=}\text{0}\text{.68}{\text{e}}^{-\text{1}}\\ \text{r}\text{=315 pm}\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

The partial charges are $+\text{0}\text{.68}{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.68}{\text{e}}^{-\text{1}}$.

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

$\begin{array}{c}\text{Q}\text{=}\text{0}\text{.68}{\text{e}}^{-\text{1}}\text{×}\frac{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}{\text{1}{\text{e}}^{-\text{1}}}\\ \text{=}1.09{\text{×10}}^{-19}\text{C}\end{array}$

The unit of distance is converted from pm to meter.

$\begin{array}{c}\text{r}\text{=}\text{315 pm}\frac{\text{1}{\text{×10}}^{-\text{10}}\text{m}}{\text{1}{\text{A}}^{\text{o}}}\\ \text{=}\text{3}\text{.15}\times {\text{10}}^{-\text{10}}\text{m}\end{array}$

$\begin{array}{c}\text{dipole}\text{moment,μ}\text{=}\text{Q}\text{×}\text{r}\\ =\left(1.09{\text{×10}}^{-19}\text{C}\right)\times \left(\text{3}\text{.15}\times {\text{10}}^{-\text{10}}\text{m}\right)\\ =3.433\times {\text{10}}^{-29}\text{C}\text{m}\end{array}$

Dipole moment in electronic charges

$\begin{array}{c}\text{μ}\text{=}3.433\times {\text{10}}^{-29}\text{C}\text{m}\text{×}\frac{\text{1}\text{D}}{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}\\ \text{=}\text{10}\text{.3}\text{D}\end{array}$

Therefore, the dipole moment of given compound is $\text{10}\text{.3}\text{D}$.

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 $\text{10}\text{.3}\text{D}$ is compound A.

Conclusion

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