Chapter 11, Problem 11B.5E

(a)

Interpretation Introduction

Interpretation:

Rotational constants for $S{O}_3$ molecule have to be calculated.

Concept Introduction:

Rotational constant B can be calculated using the following formula.

$B=\frac{h}{8{\pi }^{2}cI}$

Where,

    $\begin{array}{l}B=rotationalcons\tan t\\ h=plank\text{'}scons\tan t\\ c=velocityoflight\\ I=momentofinertia\end{array}$

The total moment of inertia can be calculated by following equation.

$I=\Sigma m_i{r}_i^2$

Where,

    $\begin{array}{l}I=momentofinertia\\ m_i=massof{i}^{th}atom\\ r_i=dis\tan cefromtherotatingaxis\end{array}$

Answer to Problem 11B.5E

The rotational constant A and B are $17.8$ and $26.92$ respectively.

Explanation of Solution

The $S{O}_3$ molecule is planar.  So it has two axes of rotation.  Along each axes it has different moment of inertias.  One rotating axis is perpendicular to the plane of molecule and the other two degenerate axis lies in the plane of molecule.  Let the moment of inertia along the axis perpendicular to the plane of the molecule be $I_A$ and the moment of inertia along the axes that lie in the plane of the molecule be $I_B.$

$I_A$ can be calculated through following steps,

$I_A=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2$

Since mass of all oxygen atoms and their distance from the sulfur atom is same, moment of inertia can be written as follows,

Given,

    $\begin{array}{l}m=massofoxygenatom=16\times 1.6\times {10}^{-27}kg\\ r=dis\tan ceofoxygenfromsulfur=143\times {10}^{-12}m\end{array}$

$\begin{array}{c}{I}_A=m{r}_{}^2+m{r}_{}^2+m{r}_{}^2\\ =3m{r}_{}^2\\ =3\times \left(16\times 1.6\times {10}^{-27}kg\right)\times {\left(143\times {10}^{-12}m\right)}^2\\ =1.57\times 10^{-45}kg{m}^2\end{array}$

Let the rotational constant corresponds to $I_A$ be A and it can be calculated as follows,

Given,

    $\begin{array}{l}\text{h}=\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{ m}}^{\text{2}}\text{ kg/s}\\ \text{c}={\text{3×10}}^{\text{8}}\text{m/s}\\ {\text{I}}_{\text{A}}=\text{1}{\text{.57×10}}^{\text{-45}}\text{kg}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}A=\frac{h}{8{\pi }^{2}c{I}_A}\\ \\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{ m}}^{\text{2}}\text{ kg/s}}{\text{8×}{\left(\text{3}\text{.14}\right)}^{\text{2}}\text{×}\left({\text{3×10}}^{\text{8}}\text{m/s}\right)\left(\text{1}{\text{.57×10}}^{\text{-45}}\text{kg}{\text{m}}^{\text{2}}\right)}\\ \\ =17.8\end{array}$

$I_B$ can be calculated through following steps,

$I_B=m_1{r}_1^2+m_2{r}_2^2$

Since mass of all oxygen atoms and their distance from the sulfur atom is same, moment of inertia can be written as follows,

Given,

    $\begin{array}{l}m=massofoxygenatom=16\times 1.6\times {10}^{-27}kg\\ r=dis\tan ceofoxygenfromsulfur=143\times {10}^{-12}m\end{array}$

$\begin{array}{c}{I}_B=m{r}_{}^2+m{r}_{}^2\\ =2m{r}_{}^2\\ =2\times \left(16\times 1.6\times {10}^{-27}kg\right)\times {\left(143\times {10}^{-12}m\right)}^2\\ =1.04\times 10^{-45}kg{m}^2\end{array}$

Let the rotational constant corresponds to $I_A$ be A and it can be calculated as follows,

Given,

    $\begin{array}{l}\text{h}\text{=6}{\text{.626 × 10}}^{\text{-34}}{\text{ m}}^{\text{2}}\text{ kg/s}\\ \text{c}\text{=}{\text{3×10}}^{\text{8}}\text{m/s}\\ {\text{I}}_B\text{=}\text{1}{\text{.04×10}}^{\text{-45}}\text{kg}{\text{m}}^{\text{2}}\end{array}$

$\begin{array}{c}B=\frac{h}{8{\pi }^{2}c{I}_B}\\ \\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{ m}}^{\text{2}}\text{ kg/s}}{\text{8×}{\left(\text{3}\text{.14}\right)}^{\text{2}}\text{×}\left({\text{3×10}}^{\text{8}}\text{m/s}\right)\left(\text{1}{\text{.04×10}}^{\text{-45}}\text{kg}{\text{m}}^{\text{2}}\right)}\\ \\ =26.92\end{array}$

The rotational constant A and B are $17.8$ and $26.92$ respectively.

(b)

Interpretation Introduction

Interpretation:

The use of microwave spectroscopy for distinguishing the relative abundance of $SO{}_{3}andSO{}_3$ has to be explained.

Concept Introduction:

Rotational constant B can be calculated using the following formula.

$B=\frac{h}{8{\pi }^{2}cI}$

Where,

    $\begin{array}{l}B=rotationalcons\tan t\\ h=plank\text{'}scons\tan t\\ c=velocityoflight\\ I=momentofinertia\end{array}$

The total moment of inertia can be calculated by following equation.

$I=\Sigma m_i{r}_i^2$

Where,

    $\begin{array}{l}I=momentofinertia\\ m_i=massof{i}^{th}atom\\ r_i=dis\tan cefromtherotatingaxis\end{array}$

Explanation of Solution

Microwave spectroscopy cannot be used to distinguish the relative abundance of $SO{}_{3}andSO{}_3$.  Because the center of mass lies on sulfur atom so the axis pass through sulfur atom.  Hence the rotational constant obtained for both the molecule will be same.  But the isotopes of oxygen in the same compound can be distinguished due to their contribution in moment of inertia.