Chapter 11, Problem 11B.2ST

Interpretation Introduction

Interpretation:

The molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is whether oblate or prolate has to be stated.  The rotational energy term for this molecule has to be calculated.

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 11B.2ST

As the moment of inertia around the principal axis $\left(I_{\parallel }\right)$ is smaller than the moment of inertia around the perpendicular axis so the molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is prolate.  The rotational energy term for the molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is $\underset{\_}{0.454J\left(J+1\right)+4.576K^2}$.

Explanation of Solution

The molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is shown below.

Figure 1

The moment of inertia is given by the formula as shown below.

  $I_{\parallel }=2m_{\text{H}}\left(1-\text{cosθ}\right)R^2$        (1)

  $I_{\perp }=\left(\begin{array}{c}{m}_{\text{H}}\left(1-\text{cosθ}\right)R^2+\frac{m_{\text{H}}}{m}\left(m_{\text{C}}+m_{\text{H}}\right)\left(1+2\text{cosθ}\right)R^2\\ +\frac{m_{\text{Cl}}}{m}\left\{\left(3m_{\text{H}}+m_{\text{C}}\right)R^{\prime }+6m_{\text{H}}R{\left[\frac{1}{3}\left(1+2\text{cosθ}\right)\right]}^{1/2}\right\}{R}^{\prime }\end{array}\right)$        (2)

Where,

• $I$ is the moment of inertia.
• $m_{\text{H}}$ is the mass of hydrogen atom.
• $m_{\text{C}}$ is the mass of carbon atom.
• $m_{\text{Cl}}$ is the mass of the chlorine atom.
• $m$ is the total mass of the molecule.
• $\text{θ}$ is the $\text{HCH}$ bond angle
• $R$ is the $\text{C}-\text{H}$ bond length $\left(R=111\text{pm}\right)$.
• $R^{\prime }$ is the $\text{C}-\text{Cl}$ bond length $\left(R^{\prime }=178\text{pm}\right)$.

The value of $m_{\text{H}}$ is $1.6735\times {10}^{-27}\text{kg}$.  The value of $m_{\text{C}}$ is $1.99\times {10}^{-26}\text{kg}$.  The value of $m_{\text{Cl}}$ is $5.85\times {10}^{-26}\text{kg}$.  The value of $m$ is $8.367\times {10}^{-26}\text{kg}$.  The $\text{HCH}$ bond angle is $\text{110}{\text{.5}}^{\text{ο}}$.  The $\text{C}-\text{H}$ bond length $\left(R\right)$ is $111\text{pm}$.  The $\text{C}-\text{Cl}$ bond length $\left(R^{\prime }\right)$ is $178\text{pm}$.

Substitute the corresponding values in equation (1) as shown below.

  $\begin{array}{c}{I}_{\parallel }=2m_{\text{H}}\left(1-\text{cosθ}\right)R^2\\ =2\left(1.6735\times {10}^{-27}\text{kg}\right)\left(1-\text{cos110}{\text{.5}}^{\text{ο}}\right){\left(111\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)}^2\\ =3.347\times {10}^{-27}\text{kg}\left(1+0.35\right){\left(1.11\times {10}^{-10}\text{m}\right)}^2\\ =5.567\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}\end{array}$

The moment of inertia around the principal axis $\left(I_{\parallel }\right)$ is $5.567\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}$.

Substitute the corresponding values in equation (2) as shown below.

$I_{\perp }=\left(\begin{array}{c}1.6735\times {10}^{-27}\text{kg}\left(1-\text{cos110}{\text{.5}}^{\text{ο}}\right){\left(111\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)}^2\\ +\frac{1.6735\times {10}^{-27}\text{kg}}{8.367\times {10}^{-26}\text{kg}}\left(1.99\times {10}^{-26}\text{kg}+1.6735\times {10}^{-27}\text{kg}\right)\left(1+2\text{cos110}{\text{.5}}^{\text{ο}}\right){\left(111\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)}^2\\ +\frac{5.85\times {10}^{-26}\text{kg}}{8.367\times {10}^{-26}\text{kg}}\left\{\begin{array}{l}\left(3\times 1.6735\times {10}^{-27}\text{kg}+1.99\times {10}^{-26}\text{kg}\right)\left(178\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)\\ +6\times 1.6735\times {10}^{-27}\text{kg}\left(111\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right){\left[\frac{1}{3}\left(1+2\text{cos110}{\text{.5}}^{\text{ο}}\right)\right]}^{1/2}\end{array}\right\}\left(178\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{pm}}\right)\end{array}\right)$ The above equation is further solved as shown below

  $I_{\perp }=\left(\begin{array}{c}1.6735\times {10}^{-27}\text{kg}\left(1+{0.35}^{\text{ο}}\right){\left(1.11\times {10}^{-10}\text{m}\right)}^2\\ +0.02\left(2.1567\times {10}^{-26}\text{kg}\right)\left(1-0.700\right){\left(1.11\times {10}^{-10}\text{m}\right)}^2\\ +0.69\left\{\begin{array}{l}\left(2.49\times {10}^{-26}\text{kg}\right)\left(1.78\times {10}^{-10}\text{m}\right)\\ +\left(1.004\times {10}^{-26}\text{kg}\right)\left(1.11\times {10}^{-10}\text{m}\right){\left[\frac{1}{3}\left(1-0.700\right)\right]}^{1/2}\end{array}\right\}\left(1.78\times {10}^{-10}\text{m}\right)\end{array}\right)$

The above equation is further solved as shown below.

  $\begin{array}{c}{I}_{\perp }=\left(\begin{array}{c}2.783\times {10}^{-47}\text{kg}{\text{m}}^2+0.159\times {10}^{-47}\text{kg}{\text{m}}^2\\ +0.69\left\{4.432\times {10}^{-36}\text{kg}\text{m}+0.352\times {10}^{-36}\text{kg}\text{m}\right\}\left(1.78\times {10}^{-10}\text{m}\right)\end{array}\right)\\ =\left(2.942\times {10}^{-47}\text{kg}{\text{m}}^2+0.69\left(4.784\times {10}^{-36}\text{kg}\text{m}\right)\left(1.78\times {10}^{-10}\text{m}\right)\right)\\ =\left(2.942\times {10}^{-47}\text{kg}{\text{m}}^2+5.875\times {10}^{-46}\text{kg}{\text{m}}^2\right)\\ =6.169\times {10}^{-46}\text{kg}{\text{m}}^2\end{array}$

The moment of inertia around the perpendicular axis $\left(I_{\perp }\right)$ is $6.169\times {10}^{-46}\text{kg}{\text{m}}^2$.

As the moment of inertia around the principal axis $\left(I_{\parallel }\right)$ is smaller than the moment of inertia around the perpendicular axis so the molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is prolate.

The rotational constant $\tilde{\text{A}}$ and $\tilde{\text{B}}$ is calculated by the formula shown below.

  $\tilde{\text{A}}=\frac{\hslash }{4\text{π}c{I}_{\parallel }}$        (3)

  $\tilde{B}=\frac{\hslash }{4\text{π}c{I}_{\perp }}$        (4)

Where

• $I_{\parallel }$ is the moment of inertia around the principal axis.
• $I_{\perp }$ is the moment of inertia around the perpendicular axis.
• $\hslash$ is the constant $\left(\hslash =\frac{h}{2\text{π}}\right)$ where $h$ is the Planck’s constant.
• $\tilde{\text{A}}$ and $\tilde{\text{B}}$ is the rotational constant.
• $c$ is the velocity of light.

The value of $\hslash$ is $1.055\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}$, the value of $c$ is $2.998\times {10}^8{\text{ms}}^{\text{-1}}$, the value of $I_{\parallel }$ is $5.567\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}$ and the value of $I_{\perp }$ is $6.169\times {10}^{-46}\text{kg}{\text{m}}^2$  Substitute the corresponding values in equation (3) as shown below.

  $\begin{array}{c}\tilde{\text{A}}=\frac{1.055\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}}{4\times \left(\text{3}\text{.14}\right)\times \left(2.998\times {10}^8\text{m}{\text{s}}^{\text{-1}}\right)\left(5.567\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}\right)}\\ =\frac{1.055\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}}{12.56\times 16.68\times {10}^{-39}\text{kg}{\text{m}}^3{\text{s}}^{\text{-1}}}\\ =\left(503.57{\text{m}}^{-1}\times \frac{{10}^{-2}{\text{cm}}^{-1}}{1{\text{m}}^{-1}}\right)\\ =5.03{\text{cm}}^{-1}\end{array}$

Substitute the corresponding values in equation (4) as shown below.

  $\begin{array}{c}\tilde{B}=\frac{1.055\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}}{4\times \left(\text{3}\text{.14}\right)\times \left(2.998\times {10}^8\text{m}{\text{s}}^{\text{-1}}\right)\left(6.169\times {10}^{-46}\text{kg}{\text{m}}^2\right)}\\ =\frac{1.055\times {10}^{-34} {\text{m}}^2 {\text{kg s}}^{-1}}{12.56\times 18.49\times {10}^{-38}\text{kg}{\text{m}}^3{\text{s}}^{\text{-1}}}\\ =\left(45.42{\text{m}}^{-1}\times \frac{{10}^{-2}{\text{cm}}^{-1}}{1{\text{m}}^{-1}}\right)\\ =0.454{\text{cm}}^{-1}\end{array}$

The Rotational energy terms are calculated by the formula shown below.

  $\tilde{F}\left(J,K\right)=\tilde{B}J\left(J+1\right)+\left(\tilde{\text{A}}-\tilde{B}\right)K^2$        (5)

Where,

• $K$ is the quantum number and it signifies the component on the principal axis.
• $J$ is the rotational energy quantum number.

Substitute the corresponding values in equation (5) as shown below.

  $\begin{array}{c}\tilde{F}\left(J,K\right)/{\text{cm}}^{-1}=0.454J\left(J+1\right)+\left(5.03-0.454\right)K^2\\ =\underset{\_}{0.454J\left(J+1\right)+4.576K^2}\end{array}$

The rotational energy term for the molecule ${}^{\text{12}}{\text{C}}^{\text{1}}{\text{H}}_{\text{3}}{}^{\text{35}}\text{Cl}$ is $\underset{\_}{0.454J\left(J+1\right)+4.576K^2}$.