Chapter 11, Problem 11C.4BE

Interpretation Introduction

Interpretation:

The fundamental vibrational wavenumbers of deuterium halides have to be stated.

Concept introduction:

The internal mode of vibration of molecule leads to a periodic change in the internal internuclear separation of a molecule.  Due to this change, the molecules are considered as harmonic oscillators.  The strength of a bond in a bond is measured by force constant. The force constant of a molecule is related to the effective mass and vibrational frequency.  The vibrational energy of a molecule is given by the equation as shown below.

  $E_{\nu }=\left(\nu +\frac{1}{2}\right)\hslash \omega$.

Answer to Problem 11C.4BE

The fundamental vibrational wavenumbers of $\text{H}\text{F}\text{,}\text{H}\text{C}\text{l,}\text{H}\text{B}\text{r and }\text{H}\text{I}$ are $\underset{\_}{2997.568c{m}^{-1}},\underset{\_}{2141.949c{m}^{-1}},\underset{\_}{1883.851c{m}^{-1}}$ and $\underset{\_}{1638.390c{m}^{-1}}$ respectively.

Explanation of Solution

The wavenumber of a molecule is given by the equation shown below.

  $\tilde{\nu }=\frac{1}{2\text{π}c}{\left(\frac{k_{\text{f}}}{m_{\text{eff}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$        (1)

Where,

• $k_{\text{f}}$ is the force constant.
• $m_{\text{eff}}$ is the effective mass.
• $c$ is the speed of light

It is assumed that the force constants of deuterium halides are same as that of hydrogen halides.

The effective mass of $\text{H}\text{F}$ is calculated as shown below.

  $m_{\text{eff}}=\frac{m_a{m}_b}{m_a+m_b}$        (2)

Where,

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

For $\text{H}\text{F}$, substitute the value of $m_a$ as $\text{2 g/mol}$ and $m_b$ as $19\text{ g/mol}$ and in equation (2).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{2 g/mol}\times 19\text{ g/mol}}{2+19\text{ g/mol}}\\ =\frac{38}{21}\text{g/mol}\\ =1.81\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 1.81\text{ g/mol}=\frac{\text{1}\text{.81 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =3.01\times {10}^{-27}\text{ kg}\end{array}$

Substitute $k_{\text{f}}$ as $958.71\text{}{\text{N m}}^{-1}$, $m_{\text{eff}}$ as $3.01\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (1).

  $\begin{array}{c}\tilde{\nu }=\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}{\left(\frac{958.71{\text{ N m}}^{-1}}{3.01\times {10}^{-27}\text{kg}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}\times 5.644\times {10}^{14}{\text{ s}}^{-1}\\ =\underset{\_}{2997.568c{m}^{-1}}\end{array}$

Therefore, the fundamental vibrational wavenumber for $\text{H}\text{F}$ is $\underset{\_}{2997.568c{m}^{-1}}$.

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

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{2 g/mol}\times 35\text{ g/mol}}{2+35\text{ g/mol}}\\ =\frac{70}{37}\text{g/mol}\\ =1.892\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 1.892\text{ g/mol}=\frac{\text{1}\text{.892 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =3.142\times {10}^{-27}\text{ kg}\end{array}$

Substitute $k_{\text{f}}$ as $511.06\text{}{\text{N m}}^{-1}$, $m_{\text{eff}}$ as $3.142\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (2).

  $\begin{array}{c}\tilde{\nu }=\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}{\left(\frac{511.06{\text{ N m}}^{-1}}{3.142\times {10}^{-27}\text{kg}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}\times 4.034\times {10}^{14}{\text{ s}}^{-1}\\ =\underset{\_}{2141.949c{m}^{-1}}\end{array}$

Therefore, the fundamental vibrational wavenumber for $\text{H}\text{C}\text{l}$ is $\underset{\_}{2141.949c{m}^{-1}}$.

For $\text{H}\text{B}\text{r}$, substitute the value of $m_a$ as $\text{2 g/mol}$ and $m_b$ as $81\text{ g/mol}$ and in equation (2).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{2 g/mol}\times 81\text{ g/mol}}{2+81\text{ g/mol}}\\ =\frac{162}{83}\text{g/mol}\\ =1.952\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 1.952\text{ g/mol}=\frac{\text{1}\text{.952 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =3.242\times {10}^{-27}\text{ kg}\end{array}$

Substitute $k_{\text{f}}$ as $407.90\text{}{\text{N m}}^{-1}$, $m_{\text{eff}}$ as $3.242\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (1).

  $\begin{array}{c}\tilde{\nu }=\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}{\left(\frac{407.90{\text{ N m}}^{-1}}{3.242\times {10}^{-27}\text{kg}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}\times 3.548\times {10}^{14}{\text{ s}}^{-1}\\ =\underset{\_}{1883.851c{m}^{-1}}\end{array}$

Therefore, the fundamental vibrational wavenumber for $\text{H}\text{B}\text{r}$ is $\underset{\_}{1883.851c{m}^{-1}}$.

For $\text{H}\text{I}$, substitute the value of $m_a$ as $\text{2 g/mol}$ and $m_b$ as $127\text{ g/mol}$ and in equation (2).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{2 g/mol}\times 127\text{ g/mol}}{2+127\text{ g/mol}}\\ =\frac{254}{129}\text{g/mol}\\ =1.9689\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 1.9689\text{ g/mol}=\frac{\text{1}\text{.9689 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =3.269\times {10}^{-27}\text{ kg}\end{array}$

Substitute $k_{\text{f}}$ as $311.098\text{}{\text{N m}}^{-1}$, $m_{\text{eff}}$ as $3.269\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (1).

  $\begin{array}{c}\tilde{\nu }=\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}{\left(\frac{311.098{\text{ N m}}^{-1}}{3.269\times {10}^{-27}\text{kg}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{2\times 3.14\times 2.998\times {10}^{10}{\text{ cm s}}^{-\text{1}}}\times 3.085\times {10}^{14}{\text{ s}}^{-1}\\ =\underset{\_}{1638.390c{m}^{-1}}\end{array}$

Therefore, the fundamental vibrational wavenumber for $\text{H}\text{I}$ is $\underset{\_}{1638.390c{m}^{-1}}$.