Chapter 6, Problem 91A

Interpretation Introduction

Interpretation:

Ionization energy in joules for $\text{Li, Be, B}$ and $\text{C}$ is to be calculated and then converted to electron volt.

Concept introduction:

Ionization energy is required to remove an electron from a gaseous atom. Ionization energy of atoms of elements down group decreases as atomic number increases as new shells are added and distance between nucleus and valence shell increases. It becomes easier to remove valence electrons away. Along a period, the Ionization energy increases as the electrons are added to same shell and are pulled by high nuclear charge.

Answer to Problem 91A

Ionization energy for $\text{Li, Be, B}$ and $\text{C}$ in joules and electron volt is as follows, respectively:

$8.635\times {10}^{-19}\text{J}$, $5.396\text{eV}$

$1.494\times {10}^{-18}\text{J}$, $9.33\text{eV}$

$1.328\times {10}^{-18}\text{J}$, $8.3\text{eV}$ and

$1.810\times {10}^{-18}\text{J}$, $11.312\text{eV}$.

Explanation of Solution

Given information:

Ionization energies in Kilojoules per moles for $\text{Li, Be, B}$ and $\text{C}$ is $\text{520}\text{kJ/mol, 900}\text{kJ/mol, 800}\text{kJ/mol}\text{\&1090}\text{kJ/mol}$ respectively.

Ionization energy for $\text{Li, Be, B}$ and $\text{C}$ in joules can be calculated as follows:

$\begin{array}{c}\text{Ionization energy of Li}\left(\text{J}\right)=520\text{kJ/mol}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\left(\frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}\right)\\ =8.635\times {10}^{-19}\text{J}\end{array}$

$\begin{array}{c}\text{Ionization energy of B}\left(\text{J}\right)=900\text{kJ/mol}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\left(\frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}\right)\\ =1.494\times {10}^{-18}\text{J}\end{array}$

$\begin{array}{c}\text{Ionization energy of Be}\left(\text{J}\right)=800\text{kJ/mol}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\left(\frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}\right)\\ =1.328\times {10}^{-18}\text{J}\end{array}$

$\begin{array}{c}\text{Ionization energy of C}\left(\text{J}\right)=1090\text{kJ/mol}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\left(\frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}\right)\\ =1.810\times {10}^{-18}\text{J}\end{array}$

The unit $\text{eV}$ is related to $\text{J}$ by conversion factor that is as follows:

$1\text{eV}=\text{1}\text{.60}\times {10}^{-19}\text{J}$

The conversion of $8.635\times {10}^{-19}\text{J}$ into $\text{eV}$ is as follows:

$\begin{array}{c}8.635\times {10}^{-19}\text{J}=8.635\times {10}^{-19}\text{J}\left(\frac{\text{1}\text{eV}}{\text{1}\text{.60}\times {10}^{-19}\text{J}}\right)\\ =5.396\text{eV}\end{array}$

The conversion of $1.494\times {10}^{-18}\text{J}$ into $\text{eV}$ is as follows:

$\begin{array}{c}1.494\times {10}^{-18}\text{J}=1.494\times {10}^{-18}\text{J}\left(\frac{\text{1}\text{eV}}{\text{1}\text{.60}\times {10}^{-19}\text{J}}\right)\\ =9.33\text{eV}\end{array}$

The conversion of $1.328\times {10}^{-18}\text{J}$ into $\text{eV}$ is as follows:

$\begin{array}{c}1.328\times {10}^{-18}\text{J}=1.328\times {10}^{-18}\text{J}\left(\frac{\text{1}\text{eV}}{\text{1}\text{.60}\times {10}^{-19}\text{J}}\right)\\ =8.3\text{eV}\end{array}$

The conversion of $1.810\times {10}^{-18}\text{J}$ into $\text{eV}$ is as follows:

$\begin{array}{c}1.810\times {10}^{-18}\text{J}=1.810\times {10}^{-18}\text{J}\left(\frac{\text{1}\text{eV}}{\text{1}\text{.60}\times {10}^{-19}\text{J}}\right)\\ =11.312\text{eV}\end{array}$

Conclusion

Ionization energy for $\text{Li, Be, B}$ and $\text{C}$ in joules and electron volt is as follows, respectively:

$8.635\times {10}^{-19}\text{J}$, $5.396\text{eV}$

$1.494\times {10}^{-18}\text{J}$, $9.33\text{eV}$

$1.328\times {10}^{-18}\text{J}$, $8.3\text{eV}$ and

$1.810\times {10}^{-18}\text{J}$, $11.312\text{eV}$.