Chapter 14, Problem 14A.5AE

Interpretation Introduction

Interpretation:

The polarizability and the dipole moment of fluorobenzene are to be calculated.

Concept introduction:

The dipole forces are attractive forces between two permanent dipoles.  The interaction is between the positive end of one polar molecule and the negative end of another polar molecule.  The induced dipole interaction occurs when an ion or a dipole induces a temporary dipole in an atom or a molecule with no dipole.  The induced dipole forces are the attraction forces between molecules having temporary dipoles.

Answer to Problem 14A.5AE

The dipole moment of fluorobenzene is $\underset{\_}{1.65D}$.

The polarizability of fluorobenzene is $\underset{\_}{1.0075\times 10^{-39}{J}^{-1}{C}^2{m}^2}$.

Explanation of Solution

The molar polarizations of fluorobenzene vapour are shown below.

(1) $70.62{\text{cm}}^3{\text{mol}}^{-1}=70.62\times {10}^{-6}{\text{m}}^3{\text{mol}}^{-1}$ at $351.0\text{K}$.

(2) $62.47{\text{cm}}^3{\text{mol}}^{-1}=62.47\times {10}^{-6}{\text{m}}^3{\text{mol}}^{-1}$ at $423.2\text{K}$.

The polarization $\left(P_m\right)$, polarizability $\left(\alpha \right)$ and dipole moment $\left(\mu \right)$ are related to each other by the expression shown below.

    $P_m=\left(\frac{N_A}{3{\epsilon }_{\text{ο}}}\right)\times \left(\alpha +\frac{{\mu }^2}{3kT}\right)$        (1)

Therefore, the polarizability is calculated by the formula shown below.

    $\begin{array}{c}\alpha +\frac{{\mu }^2}{3kT}=\frac{3{\epsilon }_{\text{ο}}{P}_m}{N_A}\\ \alpha =\frac{3{\epsilon }_{\text{ο}}{P}_m}{N_A}-\frac{{\mu }^2}{3kT}\end{array}$        (2)

In the given case, the polarizability remains constant and the polarization varies with respect to temperature. Therefore, the dipole moment is calculated by the formula shown below.

    $\begin{array}{c}\frac{{\mu }^2}{3k}\times \left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)=\left(\frac{3{\epsilon }_{\text{ο}}}{N_A}\right)\times \left(P-P^{\text{'}}\right)\\ {\mu }^2=\frac{\left(\frac{9{\epsilon }_{\text{ο}}k}{N_A}\right)\times \left(P-P^{\text{'}}\right)}{\left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)}\\ {\mu }^2=\frac{\left(9{\epsilon }_{\text{ο}}k\right)\times \left(P-P^{\text{'}}\right)}{N_A\left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)}\end{array}$        (3)

The value of ${\epsilon }_o$ is $8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$.

The value of Boltzmann constant is $1.38\times {10}^{-23}\text{J}/\text{K}$.

The value of Avogadro’s number is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

Now, Substitute the values of ${\epsilon }_o$, $k$, $N_A$, molar polarizations and temperatures in equation (3) to calculate the dipole moment.

    $\begin{array}{c}{\mu }^2=\frac{\begin{array}{l}\left(9\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times 1.38\times {10}^{-23}\text{J}/\text{K}\right)\\ \times \left(70.62-62.47\right){10}^{-6}{\text{m}}^3{\text{mol}}^{-1}\end{array}}{6.022\times {10}^{23}{\text{mol}}^{-1}\left(\frac{1}{351.0\text{K}}-\frac{1}{423.2\text{K}}\right)}\\ =\frac{1.099\times {10}^{-33}\times 8.15\times {10}^{-6}}{6.022\times {10}^{23}\times 4.8606\times {10}^{-4}}{\text{C}}^2{\text{m}}^2\\ =\frac{8.956\times {10}^{-39}}{2.9270\times {10}^{20}}{\text{C}}^2{\text{m}}^2\\ =3.06\times {10}^{-59}{\text{C}}^2{\text{m}}^2\end{array}$

Simplify the above equation.

    $\begin{array}{c}{\mu }^2=3.06\times {10}^{-59}{\text{C}}^2{\text{m}}^2\\ \mu =\sqrt{3.06\times {10}^{-59}{\text{C}}^2{\text{m}}^2}\\ =\text{5}{\text{.531×10}}^{-30}\text{C}\text{m}\end{array}$

The conversion of C m to Debye is done as shown below.

    $1\text{C}\text{m}=2.997\times {10}^{29}\text{D}$

Therefore, the conversion of $\text{5}{\text{.531×10}}^{-30}\text{C}\text{m}$ to Debye is done as shown below.

    $\begin{array}{c}\text{5}{\text{.531×10}}^{-30}\text{C}\text{m}=\text{5}{\text{.531×10}}^{-30}\times 2.997\times {10}^{29}\text{D}\\ =\underset{\_}{1.65D}\end{array}$

Therefore, the dipole moment of fluorobenzene is $\underset{\_}{1.65D}$.

Substitute the value of ${\mu }^2=3.06\times {10}^{-59}{\text{C}}^2{\text{m}}^2$ in equation (2) to calculate the polarizability.

    $\begin{array}{c}\alpha =\frac{3\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times 70.62\times {10}^{-6}{\text{m}}^3{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ -\frac{3.06\times {10}^{-59}{\text{C}}^2{\text{m}}^2}{3\times 1.38\times {10}^{-23}\text{J}/\text{K}\times 351.0\text{K}}\\ =\frac{1.875\times {10}^{-15}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}{6.022\times {10}^{23}}-\frac{3.06\times {10}^{-59}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}{1.453\times {10}^{-20}}\\ =\left(3.1135\times {10}^{-39}-2.106\times {10}^{-39}\right){\text{J}}^{-1}{\text{C}}^2{\text{m}}^2\\ =\underset{\_}{1.0075\times 10^{-39}{J}^{-1}{C}^2{m}^2}\end{array}$

Therefore, the polarizability of fluorobenzene is $\underset{\_}{1.0075\times 10^{-39}{J}^{-1}{C}^2{m}^2}$.