Chapter 6.2, Problem 6.2.3SR

Using data from Table 6.2, calculate the magnitude of the partial charges in HI.

(a) 0.39

(b) 1.8

(c) 0.057

(d) 0.60

(e) 0.15

Interpretation Introduction

Interpretation:

The partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ should be identified from the given options.

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 $\text{I}$ in $\text{HI}$

Answer to Problem 6.2.3SR

The magnitude of partial charges is found to be option (c) $0.057$

Explanation of Solution

Reason for correct option

Given,

$\begin{array}{l}\text{μ}\text{=0}\text{.44}\text{D}\\ \text{r}\text{=}1.82{\text{ A}}^0\end{array}$

$\text{μ=}\text{Q}\text{×}\text{r}$

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

$\begin{array}{c}\text{μ}\text{=}\text{0}\text{.44}\text{D}\text{×}\frac{\text{3}\text{.336}\text{×}{\text{10}}^{-\text{30}}\text{C}\text{m}}{\text{1}\text{D}}\\ \text{=}\text{1}\text{.468}\text{×}{\text{10}}^{-\text{30}}\text{C m}\end{array}$

The unit of distance is converted from pm to meter.

$\begin{array}{c}\text{r}\text{=}1.82{\text{ A}}^0\frac{\text{1}{\text{×10}}^{-\text{10}}\text{m}}{\text{1}{\text{A}}^0}\\ \text{=}1.8\text{2}\times {\text{10}}^{-\text{10}}\text{m}\end{array}$

Therefore,

$\begin{array}{c}\text{Partial charge, Q}\text{=}\frac{\text{μ}}{\text{r}}\\ \text{=}\frac{\text{1}\text{.47}\text{×}{\text{10}}^{-\text{30}}\text{C m}}{1.8\text{2}\times {\text{10}}^{-\text{10}}\text{m}}\\ \text{=}\text{8}\text{.08}\text{×}{\text{10}}^{-\text{21}}\text{C}\end{array}$

Partial charge in electronic charges

$\begin{array}{c}\text{Q}\text{=}\text{8}\text{.08}\text{×}{\text{10}}^{-\text{21}}\text{C}\text{×}\frac{\text{1}{\text{e}}^{-\text{1}}}{\text{1}\text{.6022}{\text{×10}}^{-\text{19}}\text{C}}\\ \text{=}\text{0}\text{.057}{\text{e}}^{-\text{1}}\end{array}$

The partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ is $\text{+ 0}\text{.057}{\text{e}}^{-\text{1}}$ and $-\text{0}\text{.057}{\text{e}}^{-\text{1}}$ respectively.

Hence the correct option for partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ is option c.

Reasons for incorrect options

Writes all the partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ given in the options except option c.

(a) $\text{0}\text{.39}$

(b) $1.8$

(d) $0.60$

(e) $0.60$

The partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ is found as $0.057$.

So, this partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ is not matching with any of the above partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ in remaining options.

Hence the incorrect options are a, b, d and e.

Conclusion

The correct option for partial charge of $\text{H}$ and $\text{I}$ in $\text{HI}$ is identified from the given options.  The correct $\text{H}$ and $\text{I}$ in $\text{HI}$ is determined by using the formula $\text{μ=}\text{Q}\text{×}\text{r}$.