Chapter 11, Problem 11.5IA

Interpretation Introduction

Interpretation:

The bond lengths of ${\text{HgCl}}_2,{\text{HgBr}}_{\text{2}}$ and ${\text{HgI}}_{\text{2}}$ 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 11.5IA

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The value of bond lengths of ${\text{HgCl}}_2,{\text{HgBr}}_{\text{2}}$ and ${\text{HgI}}_{\text{2}}$ are $\underset{\_}{3.486\times 10^{-10}m},\underset{\_}{4.0077\times 10^{-10}m}$ and $\underset{\_}{4.57\times 10^{-10}m}$.

Explanation of Solution

The Placzek-Teller relation is shown below.

  $\delta ={\left(\frac{32\tilde{B}kT}{hc}\right)}^{1/2}$        (1)

Where,

• $\tilde{B}$ is the rotational constant.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.
• $h$ is the Planck’s constant.
• $c$ is the speed of light.

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

  $\tilde{B}=\frac{h}{{\text{8π}}^{2}Ic}$        (2)

Substitute equation (2) in equation (1).

  $\begin{array}{c}\delta ={\left(\frac{32kT}{hc}\times \left(\frac{h}{8{\text{π}}^{2}Ic}\right)\right)}^{1/2}\\ ={\left(\frac{4kT}{{\text{π}}^{2}I{c}^2}\right)}^{1/2}\end{array}$

Rearrange the above equation for value of inertia of molecule.

  $\begin{array}{c}\delta ={\left(\frac{4kT}{{\text{π}}^{2}I{c}^2}\right)}^{1/2}\\ {\delta }^2=\left(\frac{4kT}{{\text{π}}^{2}I{c}^2}\right)\end{array}$

The equation for moment of inertia is shown below.

  $I=\frac{4kT}{{\text{π}}^2{\delta }^2{c}^2}$        (3)

For ${\text{HgCl}}_2$.

Convert $282°\text{C}$ in $\text{K}$ as shown below.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =282+273\\ =555\text{K}\end{array}$

Substitute the value of $\delta$ as $23.8{\text{ cm}}^{-1}$, $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ $T$ as $555\text{K}$ and $k$ as $1.38\times {10}^{-23}{\text{JK}}^{-1}$ in equation (3)

  $\begin{array}{c}I=\frac{4\times 1.38\times {10}^{-23}{\text{J K}}^{-1}\times 555\text{K}}{{\left(3.14\right)}^2\times {\left(23.8{\text{cm}}^{-1}\right)}^2\times {\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)}^2}\\ =\frac{3.06\times {10}^{-20}}{9.8596\times 566.44\times 8.988\times {10}^{20}}\text{kg}{\text{m}}^2\\ =6.089\times {10}^{-45}\text{kg}{\text{m}}^2\end{array}$

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

  $I=\mu R^2$        (4)

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.

  $\mu =\frac{m_a{m}_b}{m_a+m_b}$        (5)

Where,

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

For ${\text{HgCl}}_2$, substitute the value of $m_a$ as $200.59\text{g}/\text{mol}$ and $m_b$ as $35.5\text{g}/\text{mol}$ and in equation (5).

  $\begin{array}{c}\mu =\frac{\text{200}\text{.59 }\text{g}/\text{mol}\times 35.5\text{g}/\text{mol}}{200.59\text{g}/\text{mol}+35.5\text{g}/\text{mol}}\\ =\frac{7120.945}{236.09}\text{g}/\text{mol}\\ =30.161\text{g}/\text{mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 30.161\text{ g/mol}=\frac{\text{30}\text{.161 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =5.008\times {10}^{-26}\text{ kg}\end{array}$

Substitute $\mu$ as $5.008\times {10}^{-26}\text{ kg}$ and $I$ as $6.089\times {10}^{-45}{\text{ kg m}}^2$ in equation (4) as shown below.

  $\begin{array}{c}6.089\times {10}^{-45}{\text{kg m}}^2=5.008\times {10}^{-26}\text{ kg}\times R^2\\ R^2=1.21\times {10}^{-19}{\text{ m}}^2\\ R=\underset{\_}{3.486\times 10^{-10}m}\end{array}$

Therefore, the value of bond length of ${\text{HgCl}}_2$ is $\underset{\_}{3.486\times 10^{-10}m}$.

For ${\text{HgBr}}_2$.

Convert $282°\text{C}$ in $\text{K}$ as shown below.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =292+273\\ =565\text{K}\end{array}$

Substitute the value of $\delta$ as $15.2{\text{ cm}}^{-1}$, $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ $T$ as $565\text{K}$ and $k$ as $1.38\times {10}^{-23}{\text{JK}}^{-1}$ in equation (3)

  $\begin{array}{c}I=\frac{4\times 1.38\times {10}^{-23}{\text{JK}}^{-1}\times 565\text{K}}{{\left(3.14\right)}^2\times {\left(15.2{\text{cm}}^{-1}\right)}^2\times {\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)}^2}\\ =\frac{3.118\times {10}^{-20}}{9.8596\times 231.04\times 8.988\times {10}^{20}}\text{kg}{\text{m}}^2\\ =1.524\times {10}^{-44}\text{kg}{\text{m}}^2\end{array}$

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

  $I=\mu R^2$        (4)

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.

  $\mu =\frac{m_a{m}_b}{m_a+m_b}$        (5)

Where,

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

For ${\text{HgBr}}_2$, substitute the value of $m_a$ as $200.59\text{g}/\text{mol}$ and $m_b$ as $79.9\text{g}/\text{mol}$ and in equation (5).

  $\begin{array}{c}\mu =\frac{\text{200}\text{.59 }\text{g}/\text{mol}\times 79.9\text{g}/\text{mol}}{200.59\text{g}/\text{mol}+79.9\text{g}/\text{mol}}\\ =\frac{16027.141}{280.49}\text{g}/\text{mol}\\ =57.139\text{g}/\text{mol}\end{array}$

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

  $\begin{array}{c}1\text{g}/\text{mol}=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 57.139\text{g}/\text{mol}=\frac{\text{57}\text{.139 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =9.488\times {10}^{-26}\text{ kg}\end{array}$

Substitute $\mu$ as $9.488\times {10}^{-26}\text{ kg}$ and $I$ as $1.524\times {10}^{-44}{\text{ kg m}}^2$ in equation (4) as shown below.

  $\begin{array}{c}1.524\times {10}^{-44}{\text{kg m}}^2=9.488\times {10}^{-26}\text{ kg}\times R^2\\ R^2=1.606\times {10}^{-19}{\text{ m}}^2\\ R=\underset{\_}{4.0077\times 10^{-10}m}\end{array}$

Therefore, the value of bond length of ${\text{HgBr}}_2$ is $\underset{\_}{4.0077\times 10^{-10}m}$.

For ${\text{HgI}}_2$.

Convert $282°\text{C}$ in $\text{K}$ as shown below.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =292+273\\ =565\text{K}\end{array}$

Substitute the value of $\delta$ as $11.4{\text{ cm}}^{-1}$, $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ $T$ as $565\text{K}$ and $k$ as $1.38\times {10}^{-23}{\text{JK}}^{-1}$ in equation (3)

  $\begin{array}{c}I=\frac{4\times 1.38\times {10}^{-23}{\text{JK}}^{-1}\times 565\text{K}}{{\left(3.14\right)}^2\times {\left(11.4{\text{cm}}^{-1}\right)}^2\times {\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)}^2}\\ =\frac{3.118\times {10}^{-20}}{9.8596\times 129.96\times 8.988\times {10}^{20}}\text{kg}{\text{m}}^2\\ =2.704\times {10}^{-44}\text{kg}{\text{m}}^2\end{array}$

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

  $I=\mu R^2$        (4)

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.

  $\mu =\frac{m_a{m}_b}{m_a+m_b}$        (5)

Where,

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

For ${\text{HgI}}_2$, substitute the value of $m_a$ as $200.59\text{g}/\text{mol}$ and $m_b$ as $126.9\text{g}/\text{mol}$ and in equation (5).

  $\begin{array}{c}\mu =\frac{\text{200}\text{.59 }\text{g}/\text{mol}\times 126.9\text{g}/\text{mol}}{200.59\text{g}/\text{mol}+126.9\text{g}/\text{mol}}\\ =\frac{25454.871}{327.49}\text{g}/\text{mol}\\ =77.72\text{g}/\text{mol}\end{array}$

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

  $\begin{array}{c}1\text{g}/\text{mol}=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 77.72\text{g}/\text{mol}=\frac{\text{77}\text{.72 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.29\times {10}^{-25}\text{ kg}\end{array}$

Substitute $\mu$ as $1.29\times {10}^{-25}\text{ kg}$ and $I$ as $2.704\times {10}^{-44}{\text{ kg m}}^2$ in equation (4) as shown below.

  $\begin{array}{c}2.704\times {10}^{-44}{\text{kg m}}^2=1.29\times {10}^{-25}\text{ kg}\times R^2\\ R^2=2.096\times {10}^{-19}{\text{ m}}^2\\ R=\underset{\_}{4.57\times 10^{-10}m}\end{array}$

Therefore, the value of bond length of ${\text{HgI}}_2$ is $\underset{\_}{4.57\times 10^{-10}m}$.