Chapter 11, Problem 11B.7P

Interpretation Introduction

Interpretation:

The bond lengths of $\text{CO}$ and $\text{CS}$ in $\text{OCS}$ molecule have to be stated.

Concept introduction:

A rotational spectrum is a spectrum in which only the rotation of molecule changes.  For a molecule to exhibit pure rotational spectrum, the molecule must possess a permanent electric dipole moment.  The molecules possessing dipole moment is considered as a handle that stirs the electromagnetic field into oscillation.  The energy for a rotational spectrum relates the quantum number and rotational constant to give the expression, $E=\tilde{B}J(J+1)$.  The rotational constant further gives a relation with the moment of inertia of the molecule.

Answer to Problem 11B.7P

The bond length of $\text{CO}$ in $\text{OCS}$ molecule is $\underset{\_}{115.1pm}$ and the bond length of $\text{CS}$ in $\text{OCS}$ molecule is $\underset{\_}{156.8pm}$.

Explanation of Solution

The given data for absorption in the microwave spectrum of ${}^{\text{16}}{\text{O}}^{\text{12}}\text{CS}$ in $\text{GHz}$ is mentioned in the table below.

$J$	1	2	3	4
${}^{\text{32}}\text{S}$	$24.32592$	$36.48882$	$48.65164$	$60.81408$
${}^{\text{34}}\text{S}$	$23.73233$	$47.46240$

The expression for the relationship between frequency of absorption line and $J$ is shown below.

    $\begin{array}{l}\nu =\frac{2\left(J+1\right)h}{8{\pi }^{2}I}\\ \nu I=\frac{2\left(J+1\right)h}{8{\pi }^2}\end{array}$

Where,

• $I$ is the moment inertia of $\text{OCS}$ molecule.
• $h$ is the Planck’s constant.

The value of $h$ is $6.626\times {10}^{-27}\text{ergs}$.

The value of frequency for ${}^{\text{32}}\text{S}$ and $J=1$ is $v=24.325\text{92}\text{GHz}$ or $24.325\text{92}\times {\text{10}}^9\text{Hz}$.

The value of frequency for ${}^{\text{34}}\text{S}$ and $J=1$ is $v=23.73233\text{GHz}$ or $23.73233\times {\text{10}}^9\text{Hz}$.

If $I_1$ and $I_2$ are the moment of inertia for ${}^{\text{32}}\text{S}$ and ${}^{\text{34}}\text{S}$ respectively, then the expression for $\nu I$ is given below.

    $\begin{array}{c}24.325\text{92}\text{GHz}\times I_1=23.73233\text{GHz}\times I_2\\ \frac{I_2}{I_1}=\frac{24.325\text{92}\text{GHz}}{23.73233\text{GHz}}\\ =1.025\end{array}$        (1)

The moment of inertia for ${}^{\text{32}}\text{S}$ is calculated by the expression given below.

    $I=\frac{2\left(J+1\right)h}{8{\pi }^2\nu }$        (2)

Substitute $J$ as $1$, $v=24.325\text{92}\times {\text{10}}^9\text{Hz}$ and $h$ as $6.626\times {10}^{-27}\text{ergs}$ in equation (2).

    $\begin{array}{c}{I}_1=\frac{2\left(1+1\right)\times 6.626\times {10}^{-27}\text{ergs}}{8{\left(3.14\right)}^2\times 24.325\text{92}\times {\text{10}}^9\text{Hz}}\\ =\frac{4\times 6.626\times {10}^{-27}\text{g}{\text{cm}}^2\text{s}}{8\times 9.8596\times 24.325\text{92}\times {\text{10}}^9{\text{s}}^{-1}}\\ =1.3813\times {10}^{-38}\text{g}{\text{cm}}^2\end{array}$

Substitute $I_1=1.3813\times {10}^{-38}\text{g}{\text{cm}}^2$ in equation (1).

    $\begin{array}{c}\frac{I_2}{I_1}=1.025\\ \frac{I_2}{1.3813\times {10}^{-38}\text{g}{\text{cm}}^2}=1.025\\ I_2=1.3813\times {10}^{-38}\text{g}{\text{cm}}^2\times 1.025\\ I_2=1.4158\times {10}^{-38}\text{g}{\text{cm}}^2\end{array}$

The exact atomic mass of ${}^{\text{16}}\text{O}$ is $15.9949m_u$.

The exact atomic mass of ${}^{\text{12}}\text{C}$ is $12.00m_u$.

The exact atomic mass of ${}^{\text{32}}\text{S}$ is $31.9721m_u$.

The exact atomic mass of ${}^{\text{34}}\text{S}$ is $33.9679m_u$.

The inertia of a triatomic linear molecule $\left({}^{16}{\text{O}}^{12}\text{CS}\right)$ is given by the equation as follows.

  $I=m_{\text{O}}{R}_{\text{1}}^2+m_{\text{S}}{R}_{\text{2}}^2-\frac{{\left(m_{\text{O}}{R}_{\text{1}}-m_{\text{S}}{R}_2\right)}^2}{m_{\text{O}}+m_{\text{s}}+m_{\text{C}}}$        (3)

Where,

• $m_{\text{O}}$ is the mass of oxygen.
• $m_{\text{S}}$ is the mass of sulfur.
• $R_{\text{1}}$ is the bond length of $\text{CO}$.
• $R_{\text{2}}$ is the bond length between $\text{CS}$.
• $m_{\text{C}}$ is the mass of carbon.

For ${}^{\text{16}}{\text{O}}^{\text{12}}{\text{C}}^{\text{32}}\text{S}$ molecule, substitute the values of $m_{\text{O}}$, $m_{\text{C}}$, and $m_{\text{S}}$ in equation (3).

    $\begin{array}{c}{I}_1=15.9949m_u\times R_{\text{1}}^2+31.9721m_u\times R_{\text{2}}^2-\frac{{\left(15.9949m_u\times R_{\text{1}}-31.9721m_u\times R_2\right)}^2}{15.9949m_u+31.9721m_u+12.00m_u}\\ =15.9949m_u\times R_{\text{1}}^2+31.9721m_u\times R_{\text{2}}^2-\frac{{\left(15.9949\times R_{\text{1}}-31.9721\times R_2\right)}^2{m}_u^2}{59.967m_u}\\ =15.9949m_u{R}_{\text{1}}^2+31.9721m_u{R}_{\text{2}}^2-4.2663m_u{R}_{\text{1}}^2-17.0463m_u{R}_{\text{2}}^2+17.0557m_u{R}_1{R}_2\\ =11.7286m_u{R}_{\text{1}}^2+14.9258m_u{R}_{\text{2}}^2+17.0557m_u{R}_1{R}_2\end{array}$        (4)

For ${}^{\text{16}}{\text{O}}^{\text{12}}{\text{C}}^{\text{34}}\text{S}$ molecule, substitute the values of $m_{\text{O}}$, $m_{\text{C}}$, and $m_{\text{S}}$ in equation (3).

    $\begin{array}{c}{I}_2=15.9949m_u\times R_{\text{1}}^2+33.9679m_u\times R_{\text{2}}^2-\frac{{\left(15.9949m_u\times R_{\text{1}}-33.9679m_u\times R_2\right)}^2}{15.9949m_u+33.9679m_u+12.00m_u}\\ =15.9949m_u\times R_{\text{1}}^2+33.9679m_u\times R_{\text{2}}^2-\frac{{\left(15.9949\times R_{\text{1}}-33.9679\times R_2\right)}^2{m}_u^2}{59.967m_u}\\ =15.9949m_u{R}_{\text{1}}^2+33.9679m_u{R}_{\text{2}}^2-4.1289m_u{R}_{\text{1}}^2-18.6211m_u{R}_{\text{2}}^2+17.5368m_u{R}_1{R}_2\\ =11.8660m_u{R}_{\text{1}}^2+15.3468m_u{R}_{\text{2}}^2+17.5368m_u{R}_1{R}_2\end{array}$        (5)

Substitute the value of $I_1$ and $m_u$ as $1.66\times {10}^{-24}\text{g}$ in equation (4).

    $\begin{array}{c}1.3813\times {10}^{-38}\text{g}{\text{cm}}^2=1.66\times {10}^{-24}\text{g}\left(11.7286R_{\text{1}}^2+14.9258R_{\text{2}}^2+17.0557R_1{R}_2\right)\\ \frac{1.3813\times {10}^{-38}\text{g}{\text{cm}}^2}{1.66\times {10}^{-24}\text{g}}=\left(11.7286R_{\text{1}}^2+14.9258R_{\text{2}}^2+17.0557R_1{R}_2\right)\\ 11.7286R_{\text{1}}^2+14.9258R_{\text{2}}^2+17.0557R_1{R}_2=8.32\times {10}^{-15}{\text{cm}}^2\end{array}$        (6)

Substitute the value of $I_2$ and $m_u$ as $1.66\times {10}^{-24}\text{g}$ in equation (4).

    $\begin{array}{c}1.415\times {10}^{-38}\text{g}{\text{cm}}^2=1.66\times {10}^{-24}\text{g}\left(11.8660R_{\text{1}}^2+15.3468R_{\text{2}}^2+17.5368R_1{R}_2\right)\\ \frac{1.415\times {10}^{-38}\text{g}{\text{cm}}^2}{1.66\times {10}^{-24}\text{g}}=\left(11.8660R_{\text{1}}^2+15.3468R_{\text{2}}^2+17.5368R_1{R}_2\right)\\ 11.8660R_{\text{1}}^2+15.3468R_{\text{2}}^2+17.5368R_1{R}_2=8.529\times {10}^{-15}{\text{cm}}^2\end{array}$        (7)

The equations (6) and (7) are simplified into one new equation as follows.

    $\begin{array}{l}\left(11.7286R_{\text{1}}^2+14.9258R_{\text{2}}^2+17.0557R_1{R}_2=8.32\times {10}^{-15}{\text{cm}}^2\right)\times 11.8660\\ \underset{\_}{\left(11.8660R_{\text{1}}^2+15.3468R_{\text{2}}^2+17.5368R_1{R}_2=8.529\times {10}^{-15}{\text{cm}}^2\right)\times }11.7286\\ 2.8905R_{\text{2}}^2+3.297R_1{R}_2=1.305\times {10}^{-15}{\text{cm}}^{\text{2}}\end{array}$        (8)

The equations (6) and (7) are again simplified to obtain the second new equation as follows.

    $\begin{array}{l}\left(11.7286R_{\text{1}}^2+14.9258R_{\text{2}}^2+17.0557R_1{R}_2=8.32\times {10}^{-15}{\text{cm}}^2\right)\times 15.3468\\ \underset{\_}{\left(11.8660R_{\text{1}}^2+15.3468R_{\text{2}}^2+17.5368R_1{R}_2=8.529\times {10}^{-15}{\text{cm}}^2\right)\times }14.9258\\ 2.89R_{\text{1}}^2=0.383\times {10}^{-15}{\text{cm}}^{\text{2}}\end{array}$        (9)

The further simplification of equation (9) is give below.

    $\begin{array}{c}2.89R_{\text{1}}^2=0.383\times {10}^{-15}{\text{cm}}^{\text{2}}\\ R_1=\sqrt{\frac{0.383\times {10}^{-15}{\text{cm}}^{\text{2}}}{2.89}}\\ R_1=1.151\times {10}^{-8}\text{cm}\\ R_1=\underset{\_}{115.1pm}\end{array}$

Thus, the bond length of $\text{CO}$ in $\text{OCS}$ is $\underset{\_}{115.1pm}$.

Substitute the value of $R_1$ as $1.151\times {10}^{-8}\text{cm}$ in equation (8).

    $\begin{array}{c}2.8905R_{\text{2}}^2+3.297\times 1.151\times {10}^{-8}\text{cm}\times R_2=1.305\times {10}^{-15}{\text{cm}}^{\text{2}}\\ R_2^2+1.312\times {10}^{-8}\text{cm}\times R_2-0.4516\times {10}^{-15}\text{cm}=\text{0}\end{array}$

The further simplified equation for $R_2$ is given below.

    $\begin{array}{c}{R}_2=\frac{1.312\times {10}^{-8}\text{cm}+\sqrt{{\left(1.312\times {10}^{-8}\text{cm}\right)}^2+4\times 0.4516\times {10}^{-15}\text{cm}}}{2}\\ =\frac{1.312\times {10}^{-8}\text{cm}+\sqrt{\left(1.72\times {10}^{-16}{\text{cm}}^2\right)+1.8064\times {10}^{-15}\text{cm}}}{2}\\ =1.5679\times {10}^{-8}\text{cm}\\ =\underset{\_}{156.8pm}\end{array}$

Therefore, the bond length of $\text{CS}$ in $\text{OCS}$ molecule is $\underset{\_}{156.8pm}$.