Chapter 6.2, Problem 2PPA

Interpretation Introduction

Interpretation: The partial charge of $\text{H}$ and $\text{Br}$ in $\text{HBr}$ should be determined.

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.
• $\text{μ=}\text{Q}\text{×}\text{r}$

$\begin{array}{c}\text{μ}\text{=}\text{Dipole moment}\\ \text{Q}\text{= partial charges}\\ \text{r}\text{= distance between partial charges}\end{array}$

To determine: The partial charge of $\text{H}$ and $\text{Br}$ in $\text{HBr}$

Answer to Problem 2PPA

The partial charge of $\text{H}$ and $\text{Br}$ in $\text{HBr}$ is $\text{+ 0}\text{.12 }{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.12 }{\text{e}}^{-\text{1}}$ respectively.

Explanation of Solution

Partial charges are the magnitude of charges

Given,

$\begin{array}{l}\text{μ}\text{= 0}\text{.82}\text{D}\\ \text{r}\text{=}\text{1}{\text{.41 A}}^0\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

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

$\begin{array}{c}\text{μ}\text{=0}\text{.82}\text{D}\text{×}\frac{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}{\text{1}\text{D}}\\ \text{=}2.74\text{×}{\text{10}}^{-\text{30}}\text{C m}\end{array}$

The unit of distance is converted from pm to meter.

$\begin{array}{c}\text{r}\text{=}\text{1}{\text{.41 A}}^0\frac{\text{1}{\text{×10}}^{-\text{10}}\text{m}}{\text{1}{\text{A}}^0}\\ \text{=}\text{1}\text{.41}\times {\text{10}}^{-\text{10}}\text{ m}\end{array}$

Therefore,

$\begin{array}{c}\text{Partial charge, Q}\text{=}\frac{\text{μ}}{\text{r}}\\ \text{=}\frac{2.74\text{×}{\text{10}}^{-\text{30}}\text{C m}}{\text{1}\text{.41}\times {\text{10}}^{-\text{10}}\text{ m}}\\ \text{=}\text{1}\text{.943}\text{×}{\text{10}}^{-\text{20}}\text{C}\end{array}$

Partial charge in electronic charges

$\begin{array}{c}\text{Q}\text{=}\text{1}\text{.943}\text{×}{\text{10}}^{-\text{20}}\text{C}\text{×}\frac{\text{1}{\text{e}}^{-\text{1}}}{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}\\ \text{=}\text{0}\text{.12 }{\text{e}}^{-\text{1}}\end{array}$

The partial charge of $\text{H}$ and $\text{Br}$ in $\text{HBr}$ is $\text{+ 0}\text{.12 }{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.12 }{\text{e}}^{-\text{1}}$ respectively.

${}^{\text{+0}\text{.12}}\text{H}-{\text{Br}}^{-\text{0}\text{.12}}$

Conclusion

The partial charge of $\text{H}$ and $\text{F}$ in $\text{HBr}$ was determined.