Chapter 7, Problem 7E.6P

Interpretation Introduction

Interpretation:

The force constants of the bonds in $\text{HCl}$, $\text{HBr}$, $\text{HI}$, $\text{CO}$, and $\text{NO}$ have to be calculated. The molecules, $\text{HCl}$, $\text{HBr}$, $\text{HI}$, $\text{CO}$, and $\text{NO}$ have to arranged in increasing order of stiffness.

Concept introduction:

A molecule is made up of atoms that are bonded together by covalent bonds.  These bonds undergo a to and fro movement to vibrate.  The vibrational frequency of a bond is given by the expression shown below.

    $v=\frac{1}{2\text{πc}}\sqrt{\frac{k_f}{\mu }}$

Answer to Problem 7E.6P

The table for the force constant of the bond is shown below.

Molecule	${}^1{\text{H}}^{35}\text{Cl}$	${}^1{\text{H}}^{81}\text{Br}$	${}^1{\text{H}}^{127}\text{I}$	${}^{12}{\text{C}}^{16}\text{O}$	${}^{14}{\text{N}}^{16}\text{O}$
Force constant $\left(\text{kg}{\text{s}}^{-2}\right)$	$512.1262$	$408.7319$	$311.9602$	$1902.6209$	$1380.2479$

The arrangement of given molecules in increasing order of their stiffness is shown below.

    ${}^1{\text{H}}^{127}\text{I}{<}^1{\text{H}}^{81}\text{Br}{<}^1{\text{H}}^{35}\text{Cl}{<}^{14}{\text{N}}^{16}\text{O}{<}^{12}{\text{C}}^{16}\text{O}$

Explanation of Solution

The Data table for the wavenumber number of the molecules is shown below.

Molecule	${}^1{\text{H}}^{35}\text{Cl}$	${}^1{\text{H}}^{81}\text{Br}$	${}^1{\text{H}}^{127}\text{I}$	${}^{12}{\text{C}}^{16}\text{O}$	${}^{14}{\text{N}}^{16}\text{O}$
Wavenumber $({\text{cm}}^{-1})$	$2990$	$2650$	$2310$	$2170$	$1904$

The mass of hydrogen atom is $1\text{u}$.

The mass of hydrogen atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{H}}=\left(1\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =1.66054\times {10}^{-27}\text{kg}\end{array}$

The mass of chlorine atom is $35\text{u}$.

The mass of chlorine atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{Cl}}=\left(35\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =58.1189\times {10}^{-27}\text{kg}\end{array}$

The mass of bromine atom is $81\text{u}$.

The mass of bromine atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{Br}}=\left(81\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =134.50374\times {10}^{-27}\text{kg}\end{array}$

The mass of iodine atom is $127\text{u}$.

The mass of iodine atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{I}}=\left(127\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =210.8886\times {10}^{-27}\text{kg}\end{array}$

The mass of carbon atom is $12\text{u}$.

The mass of carbon atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{C}}=\left(12\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =19.9265\times {10}^{-27}\text{kg}\end{array}$

The mass of oxygen atom is $16\text{u}$.

The mass of oxygen atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{O}}=\left(16\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =26.5686\times {10}^{-27}\text{kg}\end{array}$

The mass of nitrogen atom is $14\text{u}$.

The mass of nitrogen atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{N}}=\left(14\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =23.2476\times {10}^{-27}\text{kg}\end{array}$

The reduced mass of a molecule $\left(\mu \right)$ is given by the expression as shown below.

  $\mu =\frac{m_{\text{A}}{m}_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}$                                                                                                 (1)

Where,

• $m_{\text{A}}$ is the mass of atom $\text{A}$.
• $m_{\text{B}}$ is the mass of atom $\text{B}$.

Substitute the values of the mass of hydrogen atom and chlorine atom in the equation (1).

  $\begin{array}{c}{\mu }_{\text{HCl}}=\frac{\left(1.66054\times {10}^{-27}\text{kg}\right)\left(58.1189\times {10}^{-27}\text{kg}\right)}{\left(1.66054\times {10}^{-27}\text{kg}\right)+\left(58.1189\times {10}^{-27}\text{kg}\right)}\\ =1.6144\times {10}^{-27}\text{kg}\end{array}$

Substitute the values of the mass of hydrogen atom and bromine atom in the equation (1).

  $\begin{array}{c}{\mu }_{\text{HBr}}=\frac{\left(1.66054\times {10}^{-27}\text{kg}\right)\left(134.50374\times {10}^{-27}\text{kg}\right)}{\left(1.66054\times {10}^{-27}\text{kg}\right)+\left(134.50374\times {10}^{-27}\text{kg}\right)}\\ =1.6403\times {10}^{-27}\text{kg}\end{array}$

Substitute the values of the mass of hydrogen atom and iodine atom in the equation (1).

  $\begin{array}{c}{\mu }_{\text{HI}}=\frac{\left(1.66054\times {10}^{-27}\text{kg}\right)\left(210.8886\times {10}^{-27}\text{kg}\right)}{\left(1.66054\times {10}^{-27}\text{kg}\right)+\left(210.8886\times {10}^{-27}\text{kg}\right)}\\ =1.6476\times {10}^{-27}\text{kg}\end{array}$

Substitute the values of the mass of carbon atom and oxygen atom in the equation (1).

  $\begin{array}{c}{\mu }_{\text{CO}}=\frac{\left(19.9265\times {10}^{-27}\text{kg}\right)\left(26.5686\times {10}^{-27}\text{kg}\right)}{\left(19.9265\times {10}^{-27}\text{kg}\right)+\left(26.5686\times {10}^{-27}\text{kg}\right)}\\ =1.1387\times {10}^{-26}\text{kg}\end{array}$

Substitute the values of the mass of nitrogen atom and oxygen atom in the equation (1).

  $\begin{array}{c}{\mu }_{\text{NO}}=\frac{\left(19.9265\times {10}^{-27}\text{kg}\right)\left(23.2476\times {10}^{-27}\text{kg}\right)}{\left(19.9265\times {10}^{-27}\text{kg}\right)+\left(23.2476\times {10}^{-27}\text{kg}\right)}\\ =1.073\times {10}^{-26}\text{kg}\end{array}$

The wavenumber $\left(\overline{v}\right)$ for the corresponding vibrational frequency of a bond is given by the expression shown below.

    $\overline{v}=\frac{1}{2\text{πc}}\sqrt{\frac{k_f}{\mu }}$

Where,

• $k_f$ is the force constant of the bond.
• $\mu$ is the reduced mass of the molecule.
• $c$ isthe speed of light with value of $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

Rearrange the above equation for the value of $k_f$.

    $k_f=\mu {\left(2\text{πc}\overline{v}\right)}^2$                                                                                                (2)

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{HCl}$ bond and reduced mass of $\text{HCl}$ in the equation (2).

    $\begin{array}{c}{k}_f\left(\text{HCl}\right)=\left(1.6144\times {10}^{-27}\text{kg}\right){\left(2\text{π}\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2990{\text{cm}}^{-1}\right)\right)}^2\\ =\left(1.6144\times {10}^{-27}\right){\left(\left(1.8837\times {10}^{11}\right)\left(2990\right)\right)}^2\text{kg}{\text{s}}^{-2}\\ =512.1262\text{kg}{\text{s}}^{-2}\end{array}$

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{HBr}$ bond and reduced mass of $\text{HBr}$ in the equation (2).

    $\begin{array}{c}{k}_f\left(\text{HBr}\right)=\left(1.6403\times {10}^{-27}\text{kg}\right){\left(2\text{π}\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2650{\text{cm}}^{-1}\right)\right)}^2\\ =\left(1.6403\times {10}^{-27}\text{kg}\right){\left(\left(1.8837\times {10}^{11}\right)\left(2650\right)\right)}^2\text{kg}{\text{s}}^{-2}\\ =408.7319\text{kg}{\text{s}}^{-2}\end{array}$

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{HI}$ bond and reduced mass of $\text{HI}$ in the equation (2).

    $\begin{array}{c}{k}_f\left(\text{HI}\right)=\left(1.6476\times {10}^{-27}\text{kg}\right){\left(2\text{π}\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2310{\text{cm}}^{-1}\right)\right)}^2\\ =\left(1.6476\times {10}^{-27}\text{kg}\right){\left(\left(1.8837\times {10}^{11}\right)\left(2310\right)\right)}^2\text{kg}{\text{s}}^{-2}\\ =311.9602\text{kg}{\text{s}}^{-2}\end{array}$

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{CO}$ bond and reduced mass of $\text{CO}$ in the equation (2).

    $\begin{array}{c}{k}_f\left(\text{CO}\right)=\left(1.1387\times {10}^{-26}\text{kg}\right){\left(2\text{π}\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2170{\text{cm}}^{-1}\right)\right)}^2\\ =\left(1.1387\times {10}^{-26}\text{kg}\right){\left(\left(1.8837\times {10}^{11}\right)\left(2170\right)\right)}^2\text{kg}{\text{s}}^{-2}\\ =1902.6209\text{kg}{\text{s}}^{-2}\end{array}$

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{NO}$ bond and reduced mass of $\text{NO}$ in the equation (2).

    $\begin{array}{c}{k}_f\left(\text{NO}\right)=\left(1.073\times {10}^{-26}\text{kg}\right){\left(2\text{π}\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(1904{\text{cm}}^{-1}\right)\right)}^2\\ =\left(1.073\times {10}^{-26}\text{kg}\right){\left(\left(1.8837\times {10}^{11}\right)\left(1904\right)\right)}^2\text{kg}{\text{s}}^{-2}\\ =1380.2479\text{kg}{\text{s}}^{-2}\end{array}$

Therefore, the table for the force constant of the bond is shown below.

Molecule	${}^1{\text{H}}^{35}\text{Cl}$	${}^1{\text{H}}^{81}\text{Br}$	${}^1{\text{H}}^{127}\text{I}$	${}^{12}{\text{C}}^{16}\text{O}$	${}^{14}{\text{N}}^{16}\text{O}$
Force constant $\left(\text{kg}{\text{s}}^{-2}\right)$	$512.1262$	$408.7319$	$311.9602$	$1902.6209$	$1380.2479$

The stiffness of the bond increases with increase in force constant of the bond.  Therefore, the arrangement of given molecules in increasing order of their stiffness is shown below.

    ${}^1{\text{H}}^{127}\text{I}{<}^1{\text{H}}^{81}\text{Br}{<}^1{\text{H}}^{35}\text{Cl}{<}^{14}{\text{N}}^{16}\text{O}{<}^{12}{\text{C}}^{16}\text{O}$