Chapter 11, Problem 11C.4AE

Interpretation Introduction

Interpretation:

The force constants of the hydrogen-halogen bonds 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.4AE

The force constants for $\text{H}\text{F}\text{, }\text{H}\text{C}\text{l, }\text{H}\text{B}\text{r}$ and $\text{H}\text{I}$ are $\underset{\_}{958.71N{m}^{-1}},\underset{\_}{511.06N{m}^{-1}},\underset{\_}{407.90N{m}^{-1}}$ and $\underset{\_}{311.098N{m}^{-1}}$ 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

Rearrange the above equation for force constant as shown below.

  $\begin{array}{c}\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.}\\ 4{\tilde{\nu }}^2{\text{π}}^2{c}^2=\frac{k_{\text{f}}}{m_{\text{eff}}}\end{array}$

Thus, the above equation reduces as shown below.

  $k_{\text{f}}=4{\text{π}}^2{c}^2{\tilde{\nu }}^2{m}_{\text{eff}}$        (2)

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

  $m_{\text{eff}}=\frac{m_a{m}_b}{m_a+m_b}$        (3)

Where,

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

For $\text{H}\text{F}$, substitute the value of $m_a$ as $\text{1 g/mol}$ and $m_b$ as $19\text{ g/mol}$ in equation (3).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{1 g/mol}\times 19\text{ g/mol}}{1+19\text{ g/mol}}\\ =\frac{19}{20}\text{g/mol}\\ =0.95\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 0.95\text{ g/mol}=\frac{\text{0}\text{.95 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.577\times {10}^{-27}\text{ kg}\end{array}$

Substitute the value of $\tilde{\nu }$ as $4141.3{\text{ cm}}^{-1}$, $m_{\text{eff}}$ as $1.577\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (2).

  $\begin{array}{c}{k}_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(2.998\times {10}^{10}\text{ cm}{\text{s}}^{-1}\right)}^2\times {\left(4141.3{\text{ cm}}^{-1}\right)}^2\times \left(1.577\times {10}^{-27}\text{kg}\right)\\ =4\times 9.8596\times 8.988\times {10}^{20}{\text{ cm}}^2{\text{s}}^{-2}\times 17150365.69{\text{ cm}}^{-2}\times 1.577\times {10}^{-27}\text{kg}\\ =\underset{\_}{958.71N{m}^{-1}}\end{array}$

Therefore, the force constant for $\text{H}\text{F}$ is $\underset{\_}{958.71N{m}^{-1}}$.

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

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

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 0.9722\text{ g/mol}=\frac{\text{0}\text{.9722 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.614\times {10}^{-27}\text{ kg}\end{array}$

Substitute the value of $\tilde{\nu }$ as $2988.9{\text{ cm}}^{-1}$, $m_{\text{eff}}$ as $1.614\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (2).

  $\begin{array}{c}{k}_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(2.998\times {10}^{10}\text{ cm}{\text{s}}^{-1}\right)}^2\times {\left(2988.9{\text{ cm}}^{-1}\right)}^2\times \left(1.614\times {10}^{-27}\text{kg}\right)\\ =\underset{\_}{511.06N{m}^{-1}}\end{array}$

Therefore, the force constant for $\text{H}\text{C}\text{l}$ is $\underset{\_}{511.06N{m}^{-1}}$.

For $\text{H}\text{B}\text{r}$, substitute the value of $m_a$ as $\text{1 g/mol}$ and $m_b$ as $81\text{ g/mol}$ in equation (3).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{1 g/mol}\times 81\text{ g/mol}}{1+81\text{ g/mol}}\\ =\frac{81}{82}\text{g/mol}\\ =0.9878\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 0.9878\text{ g/mol}=\frac{\text{0}\text{.9878 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.64\times {10}^{-27}\text{ kg}\end{array}$

Substitute $\tilde{\nu }$ as $2649.7{\text{ cm}}^{-1}$, $m_{\text{eff}}$ as $1.639\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (2).

  $\begin{array}{c}{k}_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(2.998\times {10}^{10}\text{ cm}{\text{s}}^{-1}\right)}^2\times {\left(2649.7{\text{ cm}}^{-1}\right)}^2\times \left(1.64\times {10}^{-27}\text{kg}\right)\\ =4\times 9.8596\times \left(8.988\times {10}^{20}{\text{ cm}}^2{\text{s}}^{-2}\right)\times \left(7.02\times {10}^6{\text{ cm}}^{-2}\right)\times \left(1.64\times {10}^{-27}\text{kg}\right)\\ =\underset{\_}{407.90N{m}^{-1}}\end{array}$

Therefore, the force constant for $\text{H}\text{B}\text{r}$ is $\underset{\_}{407.90N{m}^{-1}}$.

For $\text{H}\text{I}$, substitute the value of $m_a$ as $\text{1 g/mol}$ and $m_b$ as $81\text{ g/mol}$ in equation (3).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{1 g/mol}\times 127\text{ g/mol}}{1+127\text{ g/mol}}\\ =\frac{127}{128}\text{g/mol}\\ =0.9922\text{ g/mol}\end{array}$

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

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 0.9922\text{ g/mol}=\frac{\text{0}\text{.9922 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.647\times {10}^{-27}\text{ kg}\end{array}$

Substitute $\tilde{\nu }$ as $2309.5{\text{ cm}}^{-1}$, $m_{\text{eff}}$ as $1.639\times {10}^{-27}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (2).

  $\begin{array}{c}{k}_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(2.998\times {10}^{10}\text{ cm}{\text{s}}^{-1}\right)}^2\times {\left(2308.5{\text{ cm}}^{-1}\right)}^2\times \left(1.647\times {10}^{-27}\text{kg}\right)\\ =4\times 9.8596\times \left(8.988\times {10}^{20}{\text{ cm}}^2{\text{s}}^{-2}\right)\times \left(5329172.25{\text{ cm}}^{-2}\right)\times \left(1.647\times {10}^{-27}\text{kg}\right)\\ =\underset{\_}{311.098N{m}^{-1}}\end{array}$

Therefore, the force constant for $\text{H}\text{I}$ is $\underset{\_}{311.098N{m}^{-1}}$.