Chapter 11, Problem 11C.4E

(a)

Interpretation Introduction

Interpretation:

Force constant of the hydrogen-halogen bond has to be determined.

Concept Introduction:

Vibrational frequency of a molecule can be determined using the given formula.

  $\begin{array}{c}Vibrationalfrequency,\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}$

Effective mass of a diatomic molecule is given by,

  $\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}$

Explanation of Solution

• Force constant of the bond in $\underset{\_}{HF}$

Given information is shown below,

  $\text{Vibrational wavenumber, }\overline{\text{ν}}\text{ = 4138}c{m}^{-1}\text{ = }41.38\times {10}^4{m}^{-1}$

Effective mass of $HF$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(1.008m_u\right)\left(18.998m_u\right)}{\left(1.008m_u\right)+\left(18.998m_u\right)}\\ \\ =\text{ 0}{\text{.9572 m}}_u\\ \\ =\text{ 0}\text{.9572}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.589}\times {10}^{-27}\text{ kg}\end{array}$

Force constant of the bond in $HF$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\left(\text{Since }\overline{\text{ν}}\text{ = }\frac{\text{ν}}{\text{c}}\right)\\ \\ \overline{\nu }.c\text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \left(41.38\times {10}^4{m}^{-1}\right)\left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)\text{= }\frac{1}{2\pi }{\left(\frac{k_f}{\left(\text{1}\text{.589}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \left(\frac{k_f}{\left(\text{1}\text{.589}\times {10}^{-27}\text{ kg}\right)}\right)\text{= 6}\text{.0696}\times {\text{10}}^{29}\\ \\ k_f\text{ = 964}{\text{.46 Nm}}^{-1}\end{array}$

Force constant of the bond in $HF$ is $\text{964}{\text{.46 Nm}}^{-1}$.

• Force constant of the bond in $\underset{\_}{HCl}$

Given information is shown below,

  $\text{Vibrational wavenumber, }\overline{\text{ν}}\text{ = 2992}c{m}^{-1}\text{ = }29.92\times {10}^4{m}^{-1}$

Effective mass of $HCl$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(1.008m_u\right)\left(34.969m_u\right)}{\left(1.008m_u\right)+\left(34.969m_u\right)}\\ \\ =\text{ 0}{\text{.9798 m}}_u\\ \\ =\text{ 0}\text{.9798}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.627}\times {10}^{-27}\text{ kg}\end{array}$

Force constant of the bond in $HCl$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\left(\text{Since }\overline{\text{ν}}\text{ = }\frac{\text{ν}}{\text{c}}\right)\\ \\ \overline{\nu }.c\text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \left(29.92\times {10}^4{m}^{-1}\right)\left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)\text{= }\frac{1}{2\pi }{\left(\frac{k_f}{\left(\text{1}\text{.627}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \left(\frac{k_f}{\left(\text{1}\text{.627}\times {10}^{-27}\text{ kg}\right)}\right)\text{= 3}\text{.173}\times {\text{10}}^{29}\\ \\ k_f\text{ = 516}{\text{.25 Nm}}^{-1}\end{array}$

Force constant of the bond in $HCl$ is $\text{516}{\text{.25 Nm}}^{-1}$.

• Force constant of the bond in $\underset{\_}{HBr}$

Given information is shown below,

  $\text{Vibrational wavenumber, }\overline{\text{ν}}\text{ = 2649}c{m}^{-1}\text{ = }26.49\times {10}^4{m}^{-1}$

Effective mass of $HBr$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(1.008m_u\right)\left(79.918m_u\right)}{\left(1.008m_u\right)+\left(79.918m_u\right)}\\ \\ =\text{ 0}{\text{.9954 m}}_u\\ \\ =\text{ 0}\text{.9954}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.653}\times {10}^{-27}\text{ kg}\end{array}$

Force constant of the bond in $HBr$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\left(\text{Since }\overline{\text{ν}}\text{ = }\frac{\text{ν}}{\text{c}}\right)\\ \\ \overline{\nu }.c\text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \left(26.49\times {10}^4{m}^{-1}\right)\left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)\text{= }\frac{1}{2\pi }{\left(\frac{k_f}{\left(\text{1}\text{.653}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \left(\frac{k_f}{\left(\text{1}\text{.653}\times {10}^{-27}\text{ kg}\right)}\right)\text{= 2}\text{.497}\times {\text{10}}^{29}\\ \\ k_f\text{ = 411}{\text{.10 Nm}}^{-1}\end{array}$

Force constant of the bond in $HBr$ is $\text{411}{\text{.10 Nm}}^{-1}$.

• Force constant of the bond in $\underset{\_}{HI}$

Given information is shown below,

  $\text{Vibrational wavenumber, }\overline{\text{ν}}\text{ = 2308}c{m}^{-1}\text{ = }23.08\times {10}^4{m}^{-1}$

Effective mass of $HI$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(1.008m_u\right)\left(126.904m_u\right)}{\left(1.008m_u\right)+\left(126.904m_u\right)}\\ \\ =\text{ 1}{\text{.000 m}}_u\\ \\ =\text{ 1}\text{.000}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 1}\text{.660}\times {10}^{-27}\text{ kg}\end{array}$

Force constant of the bond in $HI$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\left(\text{Since }\overline{\text{ν}}\text{ = }\frac{\text{ν}}{\text{c}}\right)\\ \\ \overline{\nu }.c\text{ = }\frac{1}{2\pi }{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \left(23.08\times {10}^4{m}^{-1}\right)\left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)\text{= }\frac{1}{2\pi }{\left(\frac{k_f}{\left(\text{1}\text{.660}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \left(\frac{k_f}{\left(\text{1}\text{.660}\times {10}^{-27}\text{ kg}\right)}\right)\text{= 1}\text{.888}\times {\text{10}}^{29}\\ \\ k_f\text{ = 313}{\text{.41 Nm}}^{-1}\end{array}$

Force constant of the bond in $HI$ is $\text{313}{\text{.41 Nm}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

Fundamental vibrational wavenumbers of the deuterium-halogen bond has to be determined.

Concept Introduction:

Fundamental vibrational wavenumbers of a molecule can be determined using the given formula.

  $\begin{array}{c}Vibrationalfrequency,\stackrel{\sim }{\nu }\text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}$

Effective mass of a diatomic molecule is given by,

  $\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}$

Explanation of Solution

• Fundamental vibrational wavenumbers of $\underset{\_}{{}^{2}HF}$

Given information is shown below,

  $Forcecons\tan t\text{ = 964}{\text{.46 Nm}}^{-1}$

Effective mass of ${}^{2}HF$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{ 2}\text{.014}1m_u\right)\left(18.998m_u\right)}{\left(\text{ 2}\text{.014}1m_u\right)+\left(18.998m_u\right)}\\ \\ =\text{ 1}{\text{.821 m}}_u\\ \\ =\text{ 0}\text{.9572}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 3}\text{.024}\times {10}^{-27}\text{ kg}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HF$ is calculated as shown,

  $\begin{array}{c}\stackrel{\sim }{\nu }\text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{ = }\frac{1}{2\pi \left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)}{\left(\frac{\left(\text{964}{\text{.46 Nm}}^{-1}\right)}{\left(\text{ 3}\text{.024}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ ={\text{ 299957 m}}^{-1}\\ \\ =\text{ 2999}{\text{.57 cm}}^{-1}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HF$ is $\text{2999}{\text{.57 cm}}^{-1}$.

• Fundamental vibrational wavenumbers of $\underset{\_}{{}^{2}HCl}$

Given information is shown below,

  $Forcecons\tan t\text{ = 516}{\text{.25 Nm}}^{-1}$

Effective mass of ${}^{2}HCl$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{ 2}\text{.014}1m_u\right)\left(34.969m_u\right)}{\left(\text{ 2}\text{.014}1m_u\right)+\left(34.969m_u\right)}\\ \\ =\text{ 1}{\text{.9044 m}}_u\\ \\ =\text{ 1}\text{.9044}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 3}\text{.16}\times {10}^{-27}\text{ kg}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HCl$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{= }\frac{1}{2\pi \left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)}{\left(\frac{\left(\text{516}{\text{.25 Nm}}^{-1}\right)}{\left(\text{3}\text{.16}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \text{= }214682{\text{ m}}^{-1}\\ \\ \text{= }2146.82{\text{ cm}}^{-1}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HCl$ is $2146.82{\text{ cm}}^{-1}$.

• Fundamental vibrational wavenumbers of $\underset{\_}{{}^{2}HBr}$

Given information is shown below,

  $Forcecons\tan t\text{ = 411}{\text{.10 Nm}}^{-1}$

Effective mass of ${}^{2}HBr$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{ 2}\text{.014}1m_u\right)\left(79.918m_u\right)}{\left(\text{ 2}\text{.014}1m_u\right)+\left(79.918m_u\right)}\\ \\ =\text{ 1}{\text{.9646 m}}_u\\ \\ =\text{ 1}\text{.9646}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 3}\text{.262}\times {10}^{-27}\text{ kg}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HBr$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{ = }\frac{1}{2\pi \left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)}{\left(\frac{\text{411}{\text{.10 Nm}}^{-1}}{\left(\text{3}\text{.262}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \text{= }188556{\text{ m}}^{-1}\\ \\ \text{ = }1885.56{\text{ cm}}^{-1}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HBr$ is $1885.56{\text{ cm}}^{-1}$.

• Fundamental vibrational wavenumbers of $\underset{\_}{{}^{2}HI}$

Given information is shown below,

  $Forcecons\tan t=\text{ 313}{\text{.41 Nm}}^{-1}$

Effective mass of ${}^{2}HI$ is given by,

  $\begin{array}{c}\mu =\frac{m_A{m}_B}{\left(m_A+m_B\right)}\\ \\ =\frac{\left(\text{ 2}\text{.014}1m_u\right)\left(126.904m_u\right)}{\left(\text{ 2}\text{.014}1m_u\right)+\left(126.904m_u\right)}\\ \\ =\text{ 1}{\text{.983 m}}_u\\ \\ =\text{ 1}\text{.983}\times 1.66054\times {10}^{-27}kg\\ \\ =\text{ 3}\text{.293}\times {10}^{-27}\text{ kg}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HI$ is calculated as shown,

  $\begin{array}{c}\nu \text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\\ \\ \text{ = }\frac{1}{2\pi \left(\text{2}\text{.998}\times {10}^8\text{ m}{\text{.s}}^{-1}\right)}{\left(\frac{\text{ 313}{\text{.41 Nm}}^{-1}}{\left(\text{3}\text{.293}\times {10}^{-27}\text{ kg}\right)}\right)}^{\text{1/2}}\\ \\ \text{= }163859{\text{ m}}^{-1}\\ \\ \text{ = }1638.59{\text{ cm}}^{-1}\end{array}$

Fundamental vibrational wavenumbers of ${}^{2}HI$ is $1638.59{\text{ cm}}^{-1}$.