Chapter 18, Problem 18A.4P

Interpretation Introduction

Interpretation:

The data given has to be assessed in terms of Harpoon mechanism.

Explanation of Solution

Given information:

Ionization energy values of alkali metals is given as,

    $\begin{array}{l}\text{Na}\text{K}\text{Rb}\text{Cs}\\ \text{5}\text{.1}\text{eV}4.3\text{eV}\text{4}\text{.2}\text{eV}3.9\text{eV}\end{array}$

Electron affinity values of halogens are given as,

    $\begin{array}{l}{\text{Cl}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{I}}_{\text{2}}\\ \text{1}\text{.3}\text{eV}1.2\text{eV}\text{1}\text{.7}\text{eV}\end{array}$

Experimental reactive cross-section given as,

    $\begin{array}{l}\sigma */n{m}^2{\text{Cl}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{I}}_{\text{2}}\\ \text{Na}\text{1}\text{.24}1.160.97\\ \text{K}\text{1}\text{.54}1.511.27\\ \text{Rb}\text{1}\text{.90}1.971.67\\ \text{Cs}\text{1}\text{.96}2.041.95\end{array}$

According to the Harpoon mechanism, the charge transfer is determined as,

    $R^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})$

Where,

  $\Delta E_0=I-E_{ea}$

Total reaction cross section can be given as,

    $\sigma *=\pi R^{*}{}^2$

Considering sodium and chlorine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{5}\text{.1}\text{eV}\text{-}\text{1}\text{.3}\text{eV}\\ \text{=}\text{3}\text{.8}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.8eV}(\text{nm})\\ =0.379\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.379nm)}}^{\text{2}}\\ =\text{0}\text{.45}{\text{nm}}^{\text{2}}\end{array}$

Considering potassium and chlorine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.3}\text{eV}\text{-}\text{1}\text{.3}\text{eV}\\ \text{=}\text{3}\text{.0}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.0eV}(\text{nm})\\ =0.48\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.48nm)}}^{\text{2}}\\ =\text{0}\text{.72}{\text{nm}}^{\text{2}}\end{array}$

Considering rubidium and chlorine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.2}\text{eV}\text{-}\text{1}\text{.3}\text{eV}\\ \text{=}\text{2}\text{.9}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.9eV}(\text{nm})\\ =0.496\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.496nm)}}^{\text{2}}\\ =\text{0}\text{.77}{\text{nm}}^{\text{2}}\end{array}$

Considering cesium and chlorine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{3}\text{.9}\text{eV}\text{-}\text{1}\text{.3}\text{eV}\\ \text{=}\text{2}\text{.6}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.6eV}(\text{nm})\\ =0.553\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.553nm)}}^{\text{2}}\\ =\text{0}\text{.96}{\text{nm}}^{\text{2}}\end{array}$

Considering sodium and bromine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{5}\text{.1}\text{eV}\text{-}\text{1}\text{.2}\text{eV}\\ \text{=}\text{3}\text{.9}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.9eV}(\text{nm})\\ =0.369\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.369nm)}}^{\text{2}}\\ =\text{0}\text{.42}{\text{nm}}^{\text{2}}\end{array}$

Considering potassium and bromine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.3}\text{eV}\text{-}\text{1}\text{.2}\text{eV}\\ \text{=}\text{3}\text{.1}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.1eV}(\text{nm})\\ =0.46\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.46nm)}}^{\text{2}}\\ =\text{0}\text{.66}{\text{nm}}^{\text{2}}\end{array}$

Considering rubidium and bromine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.2}\text{eV}\text{-}\text{1}\text{.2}\text{eV}\\ \text{=}\text{3}\text{.0}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.0eV}(\text{nm})\\ =0.48\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.48nm)}}^{\text{2}}\\ =\text{0}\text{.72}{\text{nm}}^{\text{2}}\end{array}$

Considering cesium and bromine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{3}\text{.9}\text{eV}\text{-}\text{1}\text{.2}\text{eV}\\ \text{=}\text{2}\text{.7eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.7eV}(\text{nm})\\ =0.53\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.53nm)}}^{\text{2}}\\ =\text{0}\text{.88}{\text{nm}}^{\text{2}}\end{array}$

Considering sodium and iodine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{5}\text{.1}\text{eV}\text{-}\text{1}\text{.7}\text{eV}\\ \text{=}\text{3}\text{.4}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 3.4eV}(\text{nm})\\ =0.423\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.423nm)}}^{\text{2}}\\ =\text{0}\text{.56}{\text{nm}}^{\text{2}}\end{array}$

Considering potassium and iodine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.3}\text{eV}\text{-}\text{1}\text{.7}\text{eV}\\ \text{=}\text{2}\text{.6}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.6eV}(\text{nm})\\ =0.553\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.553nm)}}^{\text{2}}\\ =\text{0}\text{.96}{\text{nm}}^{\text{2}}\end{array}$

Considering rubidium and iodine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{4}\text{.2}\text{eV}\text{-}\text{1}\text{.7}\text{eV}\\ \text{=}\text{2}\text{.5}\text{eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.5eV}(\text{nm})\\ =0.57\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.57nm)}}^{\text{2}}\\ =\text{1}\text{.02}{\text{nm}}^{\text{2}}\end{array}$

Considering cesium and iodine atom,

    $\begin{array}{l}\Delta E_0=I-E_{ea}\\ =\text{3}\text{.9}\text{eV}\text{-}\text{1}\text{.7}\text{eV}\\ \text{=}\text{2}\text{.2eV}\end{array}$

Value of $R^{*}$ can be calculated as shown below,

    $\begin{array}{l}{R}^{*}=\frac{\text{14}\text{.4}\text{eV}}{\text{10}\Delta E_{\text{0}}}(\text{nm})\\ =\frac{\text{14}\text{.4}\text{eV}}{\text{10}\times 2.2eV}(\text{nm})\\ =0.65\text{nm}\end{array}$

Total cross-section can be calculated as,

    $\begin{array}{l}\sigma *=\pi R^{*}{}^2\\ =3.14\times {(\text{0}\text{.65nm)}}^{\text{2}}\\ =\text{1}\text{.32}{\text{nm}}^{\text{2}}\end{array}$

Table can be constructed from the obtained values as shown below,

    $\begin{array}{l}\sigma */n{m}^2{\text{Cl}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{I}}_{\text{2}}\\ \text{Na}\text{0}\text{.45}0.420.56\\ \text{K}\text{0}\text{.72}0.660.96\\ \text{Rb}\text{0}\text{.77}0.721.02\\ \text{Cs}\text{0}\text{.96}0.881.32\end{array}$

The calculated values are not same as that of the experimental values that is given in the problem statement.  But the trends in the values are similar to that of the experimental values.