Chapter 11, Problem 11C.11P

Interpretation Introduction

Interpretation:

The bond lengths of $\text{CC}$ and $\text{CH}$ bonds has to be calculated using the spacing between the rotational lines of the $\text{P}$ and $\text{R}$ branches.

Concept introduction:

Rotational spectroscopy measures the electronic energy between the rotational states.  The rotational constant is inversely proportional to the moment of inertia.  The rotational constant is given by an expression shown below.

  $B=\frac{h}{8{\pi }^2{I}_{B}c}$

Answer to Problem 11C.11P

The $\text{CC}$ and $\text{CH}$ bond length is $\underset{\_}{1.214\times 10^{-10}m}$ and $\underset{\_}{1.04\times 10^{-10}m}$ respectively.

Explanation of Solution

The $\text{CC}$ and $\text{CH}$ bond length is approximately same in the molecule ${}^{\text{12}}{\text{C}}_{\text{2}}{}^{\text{1}}{\text{H}}_{\text{2}}$ (which is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$) and ${}^{\text{12}}{\text{C}}_{\text{2}}{}^2{\text{H}}_{\text{2}}$ (which is ${\text{C}}_{\text{2}}{\text{D}}_{\text{2}}$).  The spacing between the two rotational lines is equal to $\text{2B}$ where $\text{B}$ is the rotational constant.  The rotational constant for two molecules is calculated as shown below.

  $\begin{array}{c}\text{2B}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=2.352{\text{cm}}^{-1}\\ \text{B}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=\frac{\left(2.352{\text{cm}}^{-1}\times \frac{{10}^2{\text{m}}^{-1}}{1{\text{cm}}^{-1}}\right)}{2}\\ \text{B}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=117.6{\text{m}}^{-1}\end{array}$   and   $\begin{array}{c}\text{2B}\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)=1.696{\text{cm}}^{-1}\\ \text{B}\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)=\frac{\left(1.696{\text{cm}}^{-1}\times \frac{{10}^2{\text{m}}^{-1}}{1{\text{cm}}^{-1}}\right)}{2}\\ \text{B}\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)=84.8{\text{m}}^{-1}\end{array}$

The moment of inertia is calculated by the formula shown below.

  $I=\frac{h}{8{\pi }^{2}Bc}$        (1)

Where

• $I$ is the moment of inertia.
• $h$ is the planks constant.
• $\text{B}$ is the rotational constant.
• $c$ is the velocity of light.

The value of $h$ is $6.626\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}$, the value of $c$ is $2.998\times {10}^8{\text{ms}}^{\text{-1}}$, the value of two rotational constants $\text{B}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$ and $\text{B}\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)$ is $117.6{\text{m}}^{-1}$ and $84.8{\text{m}}^{-1}$ respectively.

Substitute the values of corresponding variables in equation (1) to calculate $I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$ and $I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)$ as shown below.

  $\begin{array}{c}I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=\frac{6.626\times {10}^{-34} {\text{m}}^2{\text{kg s}}^{-1}}{8\times {\left(3.14\right)}^2\times 117.6{\text{m}}^{-1}\times 2.998\times {10}^8{\text{ms}}^{\text{-1}}}\\ =2.382\times {10}^{-46} \text{kg}{\text{m}}^2 \end{array}$

Similarly,

  $\begin{array}{c}I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)=\frac{6.626\times {10}^{-34} {\text{m}}^2{\text{kg s}}^{-1}}{8\times {\left(3.14\right)}^2\times 84.8{\text{m}}^{-1}\times 2.998\times {10}^8{\text{ms}}^{\text{-1}}}\\ =3.30\times {10}^{-46} \text{kg}{\text{m}}^2 \end{array}$

The expression for moment of inertia in terms of masses and and distances is shown below.

  $2m_c{a}^2+2m_H{b}^2=I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$        (2)

  $2m_c{a}^2+2m_D{b}^2=I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)$        (3)

Where,

• $a$ is the distance of $\text{C}$-atom from the center of mass.
• $b$ is the distance of $\text{H}\text{or}\text{D}$ atom from the center of mass.

The value of $a$ is half of the $\text{CC}$ bond length

The value of $m_{\text{C}}$ is $12.000\text{u}$ $m_{\text{H}}$ is $1.0078\text{u}$ and $m_{\text{D}}$ is $2.0140\text{u}$ where $\text{u}$ is the atomic mass unit and $1\text{u}=1.66054\times {10}^{-27}\text{kg}$.

The moment of inertia $I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$ and $I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)$ is $2.382\times {10}^{-46} \text{kg}{\text{m}}^2$ and $3.30\times {10}^{-46} \text{kg}{\text{m}}^2$ respectively.

Subtract equation (2) from equation (3) as shown below.

  $\begin{array}{c}\left(2m_c{a}^2+2m_D{b}^2\right)-\left(2m_c{a}^2+2m_H{b}^2\right)=I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)-I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)\\ 2m_c{a}^2+2m_D{b}^2-2m_c{a}^2-2m_H{b}^2=I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)-I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)\\ 2m_D{b}^2-2m_H{b}^2=I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)-I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)\end{array}$        (4)

Substitute the values of corresponding variables in equation (4)

  $\begin{array}{c}2m_D{b}^2-2m_H{b}^2=I\left({\text{C}}_{\text{2}}{\text{D}}_{\text{2}}\right)-I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)\\ 2b^2\left(2.0140\text{u}-1.0078\text{u}\right)=3.30\times {10}^{-46} \text{kg}{\text{m}}^2 -2.382\times {10}^{-46} \text{kg}{\text{m}}^2\\ 2b^2=\frac{9.18\times {10}^{-47} \text{kg}{\text{m}}^2}{1.0062\times 1.66054\times {10}^{-27}\text{kg}}\\ b=1.65\times {10}^{-10}\text{m}\end{array}$

Substitute the value of $b$ in equation (2) to calculate the value of $a$ as shown below.

  $\begin{array}{c}I\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=2m_c{a}^2+2m_H{b}^2\\ 2.382\times {10}^{-46} \text{kg}{\text{m}}^2=\left(\begin{array}{l}2m_c{a}^2+2\times \left(1.0078\times 1.66054\times {10}^{-27}\text{kg}\right)\times \\ {\left(1.65\times {10}^{-10}\text{m}\right)}^2\end{array}\right)\\ \frac{1.470\times {10}^{-46} \text{kg}{\text{m}}^2}{2\times \left(12.000\text{u}\times 1.66054\times {10}^{-27}\text{kg}\right)}=a^2\\ a=0.607\times {10}^{-10}\text{m}\end{array}$

The bond length of $\text{CC}$ is calculated as shown below.

  $\begin{array}{c}\text{Bond}\text{length}\text{of}\text{CC}=\text{2a}\\ \text{=2}\times 0.607\times {10}^{-10}\text{m}\\ =\underset{\_}{1.214\times 10^{-10}m}\end{array}$

The bond length of $\text{CH}$ is calculated as shown below.

  $\begin{array}{c}\text{Bond}\text{length}\text{of}\text{CH}=b-a\\ =1.65\times {10}^{-10}\text{m}-0.607\times {10}^{-10}\text{m}\\ =\underset{\_}{1.04\times 10^{-10}m}\end{array}$

Therefore, the bond length of $\text{CC}$ is $\underset{\_}{1.214\times 10^{-10}m}$ and the bond length of $\text{CH}$ is $\underset{\_}{1.04\times 10^{-10}m}$.