Chapter 8, Problem 8.104QP

Interpretation Introduction

Interpretation: Second ionization energy of lithium has to be calculated for the reaction $L{i}_{\left(g\right)}\underset{}{\overset{}{\to }}L{i}^{3+}{}_{\left(g\right)}+3\stackrel{-}{e}$.

Concept Introduction:

First ionization energy: The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

Third ionization energy: The minimum energy needed to remove an electron from a dipositive gaseous ion to form a tripositive ion in the ground state is known as third ionization energy.

The energies of the electron in hydrogen like ion can be calculated by,

$\begin{array}{c}{E}_n=-\left(2.18\times {10}^{-18}J\right)Z^2\left(\frac{1}{n^2}\right)\\ where,\\ \\ n=\text{Principal quantum number}\\ \\ \text{Z}\text{=}\text{Atomic number of the element}\end{array}$

Answer to Problem 8.104QP

The second ionization energy of lithium is $\text{7}\text{.28}\times {10}^3\text{kJ/mol}\text{.}$

Explanation of Solution

The reaction is given as:

$L{i}_{\left(g\right)}\underset{}{\overset{}{\to }}L{i}^{3+}{}_{\left(g\right)}+3\stackrel{-}{e}$

The energy needed for the process is $\text{1}\text{.96 }\times {\text{10}}^4\text{kJ/mol}$

The first ionization energy of lithium is $\text{520 kJ/mol}$

The total energy needed to remove three electrons from lithium is the sum of first, second and third ionization energies of lithium

$TotalEnergy={\text{IE}}_{\text{1}}{\text{ + IE}}_{\text{2}}{\text{ + IE}}_{\text{3}}$

Ionization energy can be also defined as the difference between final and initial state.

Here, the atomic number of lithium is 3. The third electron is removed from the 1s orbital so $n=1$

The ionization energy for hydrogen like ion can be calculated as:

$E_n=-\left(2.18\times {10}^{-18}J\right)Z^2\left(\frac{1}{n^2}\right)$

The third ionization energy is calculated as follows:

$\begin{array}{c}{\text{IE}}_{\text{3}}=\Delta \text{E}\text{=}{\text{E}}_{\infty }-E_1\\ \\ =\left[-\left(2.18\times {10}^{-18}J\right){\left(3\right)}^2\left(\frac{1}{{\infty }^2}\right)\right]-\left[-\left(2.18\times {10}^{-18}J\right){\left(3\right)}^2\left(\frac{1}{1^2}\right)\right]\\ \\ =1.96\times {10}^{-17}J\\ \\ =\left(1.96\times {10}^{-17}J\right)\times \left(\frac{1kJ}{1000J}\right)\times \left(\frac{6.022\times {10}^{23}}{1mol}\right)\\ \\ =1.18\times {10}^4\text{kJ/mol}\end{array}$

The third ionization energy of lithium is $1.18\times {10}^4\text{kJ/mol}$

The second ionization energy of lithium is calculated as follows:

$TotalEnergy={\text{IE}}_{\text{1}}{\text{ + IE}}_{\text{2}}{\text{ + IE}}_{\text{3}}$

Substitute the value as follows:

$\begin{array}{c}\left(\text{1}\text{.96 }\times {\text{10}}^4\text{kJ/mol}\right)=\left(\text{520 kJ/mol}\right){\text{ + IE}}_{\text{2}}\text{ + }\left(1.18\times {10}^4\text{kJ/mol}\right)\\ \\ {\text{IE}}_{\text{2}}=\left(\text{1}\text{.96 }\times {\text{10}}^4\text{kJ/mol}\right)-\left[\left(1.18\times {10}^4\text{kJ/mol}\right)+\left(\text{520 kJ/mol}\right)\right]\\ \\ =\text{7}\text{.28}\times {10}^3\text{kJ/mol}\end{array}$

The second ionization energy of lithium is $\text{7}\text{.28}\times {10}^3\text{kJ/mol}\text{.}$