Chapter 5, Problem 79QRT

Use electron configurations to explain why

1. (a) sulfur has a lower electron affinity than chlorine.
2. (b) boron has a lower first ionization energy than beryllium.
3. (c) chlorine has a lower first ionization energy than fluorine.
4. (d) oxygen has a lower first ionization energy than nitrogen.
5. (e) iodine has a lower electron affinity than bromine.

Interpretation Introduction

Interpretation:

The reason for sulfur having lower electron affinity than chlorine has to be given using electron configurations.

Explanation of Solution

Electron affinity is the energy required to add an electron to the neutral atom.

  $\text{A}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\to }{\text{A}}^{\text{-}}$

The electronic configuration of neutral chlorine and chlorine anion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Cl:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Cl}}^{\text{-}}\text{:}\\ \left[{}_{\text{18}}\text{Ar}\right]\end{array}$

The electronic configuration of neutral sulfur and sulfuranion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{S:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{S}}^{\text{-}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}\end{array}$

Addition of one electron to chlorine gives a noble gas configuration.  Hence, it has greater electron affinity than sulfur.

Interpretation Introduction

Interpretation:

The reason for boron having lower first ionization energy than beryllium has to be given using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral boron and boroncation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{B:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{B}}^{\text{+}}\text{:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}\end{array}$

The electronic configuration of neutral beryllium and beryllium cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Be:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Be}}^{\text{+}}\text{:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{1}}\end{array}$

The beryllium atom has all subshells filled.  So, removing an electron requires more energy.  Hence, boron has lower first ionization energy.

Interpretation Introduction

Interpretation:

The reason for chlorine having lower first ionization energy than fluorine has to be given using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral chlorine and chlorine cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Cl:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Cl}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}\end{array}$

The electronic configuration of neutral fluorine and fluorine cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{F:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{F}}^{\text{+}}\text{:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}\end{array}$

The electron from chlorine is removed from $\text{3p}$ orbital whereas in fluorine the electron is removed from $\text{2p}$ orbital.  Hence, chlorine has lower first ionization energy that fluorine.

Interpretation Introduction

Interpretation:

The reason for oxygen having lower first ionization energy than nitrogen has to be given using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral oxygen and oxygen cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{O:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{O}}^{\text{+}}\text{:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}\end{array}$

The electronic configuration of neutral nitrogen and nitrogen cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{N:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{N}}^{\text{+}}\text{:}\\ \left[{}_{\text{2}}\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}\end{array}$

The nitrogen atom attains half-filled $\text{p}$ subshell, so removing an electron requires more energy.  Hence, oxygen has lower first ionization energy than nitrogen.

Interpretation Introduction

Interpretation:

The reason for iodine having lower electron affinity than bromine has to be given using electron configurations.

Explanation of Solution

Electron affinity is the energy required to add an electron to the neutral atom.

  $\text{A}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\to }{\text{A}}^{\text{-}}$

The electronic configuration of neutral iodine and iodine anion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{I}\text{(Z}\text{=}\text{53):}\\ \left[{}_{\text{36}}\text{Kr}\right]{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{I}}^{\text{-}}\text{:}\\ \left[{}_{\text{54}}\text{Xe}\right]\end{array}$

The electronic configuration of neutral bromine and bromine anion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Br (Z}\text{=}\text{35):}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{10}{\text{4p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Br}}^{\text{-}}\text{:}\\ \left[{}_{\text{36}}\text{Kr}\right]\end{array}$

The electron added to bromine enters into $\text{4p}$ orbital and in iodine it enters into $\text{5p}$ orbital.  Hence, iodine has lower electron affinity than bromine.