Chapter 11, Problem 11B.4P

Interpretation Introduction

Interpretation:

The moment of inertia and bond length of $\text{H}\text{C}\text{l}$ and positions of corresponding lines in $\text{H}\text{C}\text{l}$ have to be calculated.

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.4P

The moment of inertia and bond length of $\text{H}\text{C}\text{l}$ is $\underset{\_}{2.73\times 10^{-47}kg{m}^2}$ and $\underset{\_}{1.30\times 10^{-10}m}$ respectively.

The separation of lines in $\text{H}\text{C}\text{l}$ is $\underset{\_}{42.74c{m}^{-1}}$, ${\underset{\_}{53.41cm}}^{-1}$, ${\underset{\_}{63.98cm}}^{-1}$, ${\underset{\_}{74.57cm}}^{-1}$, ${\underset{\_}{85.10cm}}^{-1}$, ${\underset{\_}{95.53cm}}^{-1}$, ${\underset{\_}{105.98cm}}^{-1}$, ${\underset{\_}{116.38cm}}^{-1}$.

Explanation of Solution

From the wavenumbers at various transitions, the separation is calculated as follows.

  $\begin{array}{c}2{\tilde{\text{B}}}_1\left({\text{in cm}}^{-1}\right)=104.13\text{}{\text{ cm}}^{-1}-83.32{\text{ cm}}^{-1}\\ =20.81\text{}{\text{ cm}}^{-1}\end{array}$

For the next transition,

  $\begin{array}{c}2{\tilde{\text{B}}}_2\left({\text{in cm}}^{-1}\right)=124.73\text{}{\text{ cm}}^{-1}-104.13{\text{ cm}}^{-1}\\ =20.6\text{}{\text{ cm}}^{-1}\end{array}$

For the next transition,

  $\begin{array}{c}2{\tilde{\text{B}}}_3\left({\text{in cm}}^{-1}\right)=145.37{\text{ cm}}^{-1}-124.73\text{}{\text{ cm}}^{-1}\\ =20.61{\text{ cm}}^{-1}\end{array}$

For the next transition,

$\begin{array}{c}2{\tilde{\text{B}}}_4\left({\text{in cm}}^{-1}\right)=165.89{\text{ cm}}^{-1}-145.37{\text{ cm}}^{-1}\\ =20.52{\text{ cm}}^{-1}\end{array}$

For the next transition,

  $\begin{array}{c}2{\tilde{\text{B}}}_5\left({\text{in cm}}^{-1}\right)=186.23{\text{ cm}}^{-1}-165.89{\text{ cm}}^{-1}\\ =20.34{\text{ cm}}^{-1}\end{array}$

For the next transition,

  $\begin{array}{c}2{\tilde{\text{B}}}_6\left({\text{in cm}}^{-1}\right)=206.60{\text{ cm}}^{-1}-186.23{\text{ cm}}^{-1}\\ =20.37{\text{ cm}}^{-1}\end{array}$

For the next transition,

  $\begin{array}{c}2{\tilde{\text{B}}}_7\left({\text{in cm}}^{-1}\right)=226.86{\text{ cm}}^{-1}-206.60{\text{ cm}}^{-1}\\ =20.26{\text{ cm}}^{-1}\end{array}$

The mean of the separation values is shown below.

  $\begin{array}{c}2{\tilde{B}}_{mean}=\frac{20.81+20.6+20.61\text{}+20.52+20.34+20.37+20.26\text{}}{7}{\text{cm}}^{-1}\\ =20.50\text{}{\text{cm}}^{-1}\end{array}$

The rotational constant is calculated as shown below.

  $\begin{array}{c}2\tilde{B}=20.50{\text{ cm}}^{-1}\\ \tilde{B}=\frac{20.50}{2}{\text{ cm}}^{-1}\\ =10.25\text{}{\text{cm}}^{-1}\end{array}$

The value of rotational constant helps in determining the moment of inertia of a molecule as shown below.

  $\tilde{B}=\frac{\hslash }{4\text{π}Ic}$        (1)

Where,

• $\hslash$ is the reduced Planck’s constant.
• $I$ is the moment of inertia
• $c$ is the speed of light.

Substitute the value of $\tilde{B}$ as $10.25{\text{ cm}}^{-1}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$, $\hslash$ as $1.05457\times {10}^{-34}\text{ J s}$ in equation (1).

  $\begin{array}{c}10.25{\text{ cm}}^{-1}=\frac{1.05457\times {10}^{-34}\text{ Js}}{4\times 3.14\times I\times 2.998\times {10}^{10}{\text{ cm s}}^{-1}}\\ I=\frac{1.05457\times {10}^{-34}\text{ J s}}{3.86\times {10}^{12}{\text{ s}}^{-1}}\\ I=\underset{\_}{2.73\times 10^{-47}kg{m}^2}\end{array}$

The moment of inertia of $\text{H}\text{C}\text{l}$ is $\underset{\_}{2.73\times 10^{-47}kg{m}^2}$

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

  $I=\mu R^2$        (2)

Where,

• $\mu$ is the reduced mass
• $R$ is the bond length of the molecule.

From the masses of the molecules, the reduced mass is calculated as shown below.

  $\mu =\frac{m_a{m}_b}{m_a+m_b}$        (3)

Where,

• $m_a$ is the mass of first atom
• $m_b$ is the mass of second atom

For $\text{H}\text{C}\text{l}$, substitute the value of $m_a$ as $1\text{ g/mol}$ and $m_b$ as $35$ and in equation (3).

  $\begin{array}{c}\mu =\frac{\text{1 g/mol}\times 35\text{ g/mol}}{1+35\text{ g/mol}}\\ =\frac{35}{36}\text{g/mol}\\ =0.9722\text{ g/mol}\end{array}$

Convert the reduced mass in $\text{kg}$ as shown below.

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 0.9722\text{ g/mol}=\frac{\text{0}\text{.9722 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.61\times {10}^{-27}\text{ kg}\end{array}$

Substitute $\mu$ as $1.61\times {10}^{-27}\text{ kg}$ and $I$ as $2.73\times {10}^{-47}{\text{ kg m}}^2$ in equation (2) as shown below.

  $\begin{array}{c}2.73\times {10}^{-47}{\text{ kg m}}^2=1.61\times {10}^{-27}\text{ kg}\times R^2\\ R^2=1.69\times {10}^{-20}{\text{ m}}^2\\ R=\underset{\_}{1.30\times 10^{-10}m}\end{array}$

Therefore, the value of bond length of $\text{H}\text{C}\text{l}$ is $\underset{\_}{1.30\times 10^{-10}m}$.

On replacing the $\text{H}$ of $\text{H}\text{C}\text{l}$ with $\text{H}\text{C}\text{l}$, the separation of line differs as the rotational constant changes with the change is effective mass.

The change is rotational constant is inversely related to the effective mass as shown below.

  $\begin{array}{c}\tilde{B}\propto \frac{1}{I}\\ I\propto m_{\text{eff}}\end{array}$

Therefore, the relation between rotational constant and effective mass is shown below.

  $\tilde{B}\propto \frac{1}{m_{\text{eff}}}$

Therefore, the ratio rotational constant of $\text{H}\text{C}\text{l}$ and $\text{H}\text{C}\text{l}$ is calculated as shown below.

  $\frac{{\tilde{B}}_{\text{H}\text{C}\text{l}}}{B_{HCl}}=\frac{{\left(m_{\text{eff}}\right)}_{{}_{\text{H}\text{C}\text{l}}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$        (4)

The reduced mass of $\text{H}\text{C}\text{l}$ is shown below.

For $\text{H}\text{C}\text{l}$, substitute the value of $m_a$ as $2\text{ g/mol}$ and $m_b$ as $35$ and in equation (3).

  $\begin{array}{c}\mu =\frac{\text{2 g/mol}\times 35\text{ g/mol}}{2+35\text{ g/mol}}\\ =\frac{90}{37}\text{g/mol}\\ =1.89\text{ g/mol}\end{array}$

Convert the reduced mass in $\text{kg}$ as shown below.

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 1.89\text{ g/mol}=\frac{\text{1}\text{.89 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =3.138\times {10}^{-27}\text{ kg}\end{array}$

Substitute the value of ${\left(m_{\text{eff}}\right)}_{{}_{\text{H}\text{C}\text{l}}}$ as $3.138\times {10}^{-27}\text{ kg}$ and ${\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}$ as $1.61\times {10}^{-27}\text{ kg}$ in equation (4).

  $\begin{array}{c}\frac{{\tilde{B}}_{\text{H}\text{C}\text{l}}}{B_{HCl}}=\frac{1.61\times {10}^{-27}\text{ kg}}{3.138\times {10}^{-27}\text{ kg}}\\ =0.5130\end{array}$

The separation of lines of $\text{H}\text{C}\text{l}$ is calculated as shown below.

  $S_{\text{H}\text{C}\text{l}}=\frac{{\left(m_{\text{eff}}\right)}_{{}_{\text{H}\text{C}\text{l}}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}\times S_{\text{H}\text{C}\text{l}}{\text{ cm}}^{-1}$        (5)

Where,

• $S_{\text{H}\text{C}\text{l}}$ is the separation of $\text{H}\text{C}\text{l}$
• $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ is the ratio of effective masses
• $S_{\text{H}\text{C}\text{l}}$ is the separation of $\text{H}\text{C}\text{l}$.

Substitute $S_{\text{H}\text{C}\text{l}}$ as $83.32{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_1=0.5130\times 83.32{\text{ cm}}^{-1}\\ =\underset{\_}{42.74c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $104.13{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_2=0.5130\times 104.13{\text{ cm}}^{-1}\\ =\underset{\_}{53.41c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $124.73{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_3=0.5130\times 124.73{\text{ cm}}^{-1}\\ =\underset{\_}{63.98c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $145.37{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_4=0.5130\times 145.37{\text{ cm}}^{-1}\\ =\underset{\_}{74.57c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $165.89{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_5=0.5130\times 165.89{\text{ cm}}^{-1}\\ =\underset{\_}{85.10c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $186.23{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_6=0.5130\times 186.23{\text{ cm}}^{-1}\\ =\underset{\_}{95.53c{m}^{-1}}\end{array}$

Substitute $S_{\text{H}\text{C}\text{l}}$ as $206.60{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_7=0.5130\times 206.60{\text{ cm}}^{-1}\\ =\underset{\_}{105.98c{m}^{-1}}\end{array}$

Substitute the value of $S_{\text{H}\text{C}\text{l}}$ as $226.86{\text{ cm}}^{-1}$ and $\frac{{\left(m_{\text{eff}}\right)}_{{}_{HCl}}}{{\left(m_{\text{eff}}\right)}_{\text{H}\text{C}\text{l}}}$ as $0.5130$ in equation (5).

  $\begin{array}{c}{\left(S_{\text{H}\text{C}\text{l}}\right)}_8=0.5130\times 226.86{\text{ cm}}^{-1}\\ =\underset{\_}{116.38c{m}^{-1}}\end{array}$

Therefore, the separation of lines in $\text{H}\text{C}\text{l}$ are $\underset{\_}{42.74c{m}^{-1}}$, ${\underset{\_}{53.41cm}}^{-1}$, ${\underset{\_}{63.98cm}}^{-1}$, ${\underset{\_}{74.57cm}}^{-1}$, ${\underset{\_}{85.10cm}}^{-1}$, ${\underset{\_}{95.53cm}}^{-1}$, ${\underset{\_}{105.98cm}}^{-1}$, ${\underset{\_}{116.38cm}}^{-1}$.