Chapter 11, Problem 11B.14P

Interpretation Introduction

Interpretation:

The $\text{CC}$ and $\text{CH}$ bond lengths have to be determined.

Concept introduction:

The moment of inertia of a molecule about an axis passing through the centre of mass is directly proportional to the reduced mass of the molecule and the bond length of the molecule.  The rotational energy of a molecule is usually expressed in terms of its rotational constant.  The rotational constant of a molecule depends upon the moment of inertia of the molecule.

Answer to Problem 11B.14P

The $\text{CC}$ bond length is $\underset{\_}{1.396\times 10^{-10}m}$.  The $\text{CH}$ bond length is $\underset{\_}{1.085\times 10^{-10}m}$.

Explanation of Solution

Rotation about any axis perpendicular to ${\text{C}}_{\text{6}}$ may be represented by the rotation of a pseudo-linear molecule about the $x-$ axis as shown in the figure given below.

Figure 1

The expression for moment of inertia of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ ($I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}$) is given below.

    $I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}=4m_{\text{H}}{R}_{\text{H}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}$        (1)

Where,

• $m_{\text{H}}$ is the mass of hydrogen atom.
• $m_{\text{C}}$ is the mass of carbon atom.
• $R_{\text{H}}$ is the distance of the hydrogen atom from the $x-$ axis.
• $R_{\text{C}}$ is the distance of the carbon atom from the $x-$ axis.

The expression for moment of inertia of ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ ($I_{{\text{C}}_{\text{6}}{\text{D}}_{\text{6}}}$) is given below.

    $I_{{\text{C}}_{\text{6}}{\text{D}}_{\text{6}}}=4m_{\text{D}}{R}_{\text{D}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}$        (2)

Where,

• $m_{\text{D}}$ is the mass of the deuterium atom.
• $m_{\text{C}}$ is the mass of carbon atom.
• $R_{\text{H}}$ is the distance of the deuterium atom from the $x-$ axis.
• $R_{\text{C}}$ is the distance of the carbon atom from the $x-$ axis.

The value of $I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}$ is $1.4759\times {10}^{-45}\text{kg}{\text{m}}^2$.

Substitute the value of $I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}$ in equation (1).

    $4m_{\text{H}}{R}_{\text{H}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}=1.4759\times {10}^{-45}\text{kg}{\text{m}}^2$        (3)

The value of $I_{{\text{C}}_{\text{6}}{\text{D}}_{\text{6}}}$ is $1.7845\times {10}^{-45}\text{kg}{\text{m}}^2$.

Substitute the value of $I_{{\text{C}}_{\text{6}}{\text{D}}_{\text{6}}}$ in equation (2).

    $4m_{\text{D}}{R}_{\text{D}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}=1.7845\times {10}^{-45}\text{kg}{\text{m}}^2$        (4)

Substract equation (3) from equation (4).

    $\begin{array}{c}4m_{\text{D}}{R}_{\text{D}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}-\left(4m_{\text{H}}{R}_{\text{H}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}\right)=1.7845\times {10}^{-45}\text{kg}{\text{m}}^2-1.4759\times {10}^{-45}\text{kg}{\text{m}}^2\\ 4\left(m_{\text{D}}{R}_{\text{D}}^{\text{2}}-m_{\text{H}}{R}_{\text{H}}^{\text{2}}\right)=30.86\times {10}^{-47}\text{kg}{\text{m}}^2\end{array}$

It is assumed that $R_{\text{H}}$ is equal to $R_{\text{D}}$.

Therefore, the above equation can be written as shown below.

    $\begin{array}{c}4\left(m_{\text{D}}{R}_{\text{H}}^{\text{2}}-m_{\text{H}}{R}_{\text{H}}^{\text{2}}\right)=30.86\times {10}^{-47}\text{kg}{\text{m}}^2\\ 4\left(m_{\text{D}}-m_{\text{H}}\right)R_{\text{H}}^{\text{2}}=30.86\times {10}^{-47}\text{kg}{\text{m}}^2\end{array}$        (5)

The value of $m_{\text{D}}$ is $2.01410m_{\text{u}}$.

The conversion of $\text{1}{m}_{\text{u}}$ to $\text{kg}$ is shown below.

    $\text{1}{m}_{\text{u}}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $2.01410m_{\text{u}}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}2.01410m_{\text{u}}=2.01410\times \text{1}\text{.6605}\times {\text{10}}^{-27}\text{kg}\\ =\text{3}\text{.344}\times {\text{10}}^{-27}\text{kg}\end{array}$

The value of $m_{\text{H}}$ is $1.0078m_{\text{u}}$.

The conversion of $\text{1}{m}_{\text{u}}$ to $\text{kg}$ is shown below.

    $\text{1}{m}_{\text{u}}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $1.0078m_{\text{u}}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}1.0078m_{\text{u}}=1.0078\times \text{1}\text{.6605}\times {\text{10}}^{-27}\text{kg}\\ =\text{1}\text{.6735}\times {\text{10}}^{-27}\text{kg}\end{array}$

Substitute the value of $m_{\text{D}}$ and $m_{\text{H}}$ in equation (5).

    $\begin{array}{c}4\left(\text{3}\text{.344}\times {\text{10}}^{-27}\text{kg}-\left(\text{1}\text{.6735}\times {\text{10}}^{-27}\text{kg}\right)\right)R_{\text{H}}^{\text{2}}=30.86\times {10}^{-47}\text{kg}{\text{m}}^2\\ R_{\text{H}}={\left(\frac{30.86\times {10}^{-47}\text{kg}{\text{m}}^2}{6.682\times {\text{10}}^{-27}\text{kg}}\right)}^{1/2}\\ R_{\text{H}}={\left(4.6183\times {\text{10}}^{-20}{\text{m}}^2\right)}^{1/2}\\ =2.149\times {\text{10}}^{-10}\text{m}\end{array}$

The value of $R_{\text{C}}$ can be determined by rearranging equation (1) as shown below.

    $\begin{array}{c}{I}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}=4m_{\text{H}}{R}_{\text{H}}^{\text{2}}+4m_{\text{C}}{R}_{\text{C}}^{\text{2}}\\ 4m_{\text{C}}{R}_{\text{C}}^{\text{2}}=I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}-4m_{\text{H}}{R}_{\text{H}}^{\text{2}}\\ R_{\text{C}}^{\text{2}}=\frac{I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}-4m_{\text{H}}{R}_{\text{H}}^{\text{2}}}{4m_{\text{C}}}\\ R_{\text{C}}={\left(\frac{I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}-4m_{\text{H}}{R}_{\text{H}}^{\text{2}}}{4m_{\text{C}}}\right)}^{1/2}\end{array}$        (6)

The value of $I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}$ is $1.4759\times {10}^{-45}\text{kg}{\text{m}}^2$.

The value of $m_{\text{H}}$ is $\text{1}\text{.6735}\times {\text{10}}^{-27}\text{kg}$.

The value of $R_{\text{H}}$ is $2.149\times {\text{10}}^{-10}\text{m}$.

The value of $m_{\text{C}}$ is $12.011m_{\text{u}}$.

The conversion of $\text{1}{m}_{\text{u}}$ to $\text{kg}$ is shown below.

    $\text{1}{m}_{\text{u}}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $12.011m_{\text{u}}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}12.011m_{\text{u}}=12.011\times \text{1}\text{.6605}\times {\text{10}}^{-27}\text{kg}\\ =\text{19}\text{.944}\times {\text{10}}^{-27}\text{kg}\end{array}$

Substitute the values of $I_{{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}}$, $m_{\text{H}}$, $R_{\text{H}}$, and $m_{\text{C}}$ in equation (6).

    $\begin{array}{c}{R}_{\text{C}}={\left(\frac{1.4759\times {10}^{-45}\text{kg}{\text{m}}^2-\left(4\times \text{1}\text{.6735}\times {\text{10}}^{-27}\text{kg}\times {\left(2.149\times {\text{10}}^{-10}\text{m}\right)}^2\right)}{4\times \text{19}\text{.944}\times {\text{10}}^{-27}\text{kg}}\right)}^{1/2}\\ ={\left(\frac{1.4759\times {10}^{-45}\text{kg}{\text{m}}^2-0.3091\times {10}^{-45}\text{kg}{\text{m}}^2}{79.776\times {\text{10}}^{-27}\text{kg}}\right)}^{1/2}\\ ={\left(1.4626\times {10}^{-20}{\text{m}}^2\right)}^{1/2}\\ =1.209\times {10}^{-10}\text{m}\end{array}$

The relation between $R_{\text{H}}$, $R_{\text{C}}$, $R_{\text{CH}}$, and $R_{\text{CC}}$ is shown in the figure given below.

Figure 2

Where,

• $R_{\text{CC}}$ is the $\text{CC}$ bond length.
• $R_{\text{CH}}$ is the $\text{CH}$ bond length.

From the figure shown above, the relation between $R_{\text{CC}}$ and $R_{\text{C}}$ given below holds true.

    $R_{\text{CC}}=\frac{R_{\text{C}}}{\cos 30°}$        (7)

The value of $R_{\text{C}}$ is $1.209\times {10}^{-10}\text{m}$.

Substitute the value of $R_{\text{C}}$ in equation (7).

    $\begin{array}{c}{R}_{\text{CC}}=\frac{1.209\times {10}^{-10}\text{m}}{0.866}\\ =\underset{\_}{1.396\times 10^{-10}m}\end{array}$

Therefore, the $\text{CC}$ bond length is $\underset{\_}{1.396\times 10^{-10}m}$.

From the figure shown above, the relation between $R_{\text{CH}}$, $R_{\text{C}}$ and $R_{\text{H}}$ given below holds true.

    $R_{\text{CH}}=\frac{R_{\text{H}}-R_{\text{C}}}{\cos 30°}$        (8)

The value of $R_{\text{C}}$ is $1.209\times {10}^{-10}\text{m}$.

The value of $R_{\text{H}}$ is $2.149\times {\text{10}}^{-10}\text{m}$.

Substitute the value of $R_{\text{C}}$ and $R_{\text{H}}$ in equation (8).

    $\begin{array}{c}{R}_{\text{CH}}=\frac{2.149\times {\text{10}}^{-10}\text{m}-1.209\times {10}^{-10}\text{m}}{0.866}\\ =\frac{0.94\times {\text{10}}^{-10}\text{m}}{0.866}\\ =\underset{\_}{1.085\times 10^{-10}m}\end{array}$

Therefore, the $\text{CH}$ bond length is $\underset{\_}{1.085\times 10^{-10}m}$.