Chapter 8, Problem 8.22E

(a)

Interpretation Introduction

Interpretation:

The molecular orbital diagram of ${\text{O}}_{\text{2}}$ has to be drawn, the bond order and magnetic properties of ${\text{O}}_{\text{2}}$ has to be given.

Explanation of Solution

The electronic configuration is,

  ${\text{O}}_{\text{2}}\text{=}{\left({\text{σ}}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2pz}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2py}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{1}}{\left({\text{π}}^{\text{*}}{}_{\text{2py}}\right)}^{\text{1}}$

The molecular orbital diagram of ${\text{O}}_{\text{2}}$ is,

Figure 1

The bond order of ${\text{O}}_{\text{2}}$ is,

  $\begin{array}{l}\text{B}\text{.O=}\frac{{\text{N}}_{\text{b}}{\text{-N}}_{\text{a}}}{\text{2}}\\ \text{B}\text{.O=}\frac{\text{10-6}}{\text{2}}\\ \text{B}\text{.O=2}\end{array}$

The bond order of ${\text{O}}_{\text{2}}$ is two.

The magnetic nature of ${\text{O}}_{\text{2}}$ is paramagnetic since it shows the presence of two unpaired electrons.

(b)

Interpretation Introduction

Interpretation:

The molecular property of oxygen that is explained by its molecular orbital diagram but not by Lewis structure has to be given.

Explanation of Solution

The molecular property of oxygen that is explained by its molecular orbital diagram is magnetic property.

The Lewis structure of oxygen is,

From the Lewis structure, it can be seen that all electrons are paired, and it is diamagnetic in nature.  However, the molecular orbital diagram of oxygen (figure 1) shows the presence of two unpaired electrons and therefore, oxygen is paramagnetic.

(c)

Interpretation Introduction

Interpretation:

The nature of highest occupied molecular orbital in oxygen has to be given.

Explanation of Solution

The electronic configuration of oxygen in the ground state is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}\text{.}$  The electronic configuration is ${\text{O}}_{\text{2}}\text{=}{\left({\text{σ}}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2pz}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2py}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{1}}{\left({\text{π}}^{\text{*}}{}_{\text{2py}}\right)}^{\text{1}}\text{.}$  The highest occupied molecular orbital is ${\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{1}}{\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{1}}\text{.}$  Both these orbitals are anti-bonding in nature.

(d)

Interpretation Introduction

Interpretation:

The bond order and magnetic property in peroxide ion and superoxide ion has to be given.

Explanation of Solution

The formula of peroxide ion is ${\text{O}}_{\text{2}}{}^{\text{2-}}.$

The formula of superoxide ion is ${\text{O}}_{\text{2}}{}^{\text{-}}.$

Superoxide ion $\left({\text{O}}_{\text{2}}{}^{\text{-}}\right)$ has three electrons in ${\text{π}}^{\text{*}}$ electrons.  The electronic configuration is ${\text{O}}_{\text{2}}{}^{\text{-}}\text{=}{\left({\text{σ}}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2pz}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2py}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2py}}\right)}^{\text{1}}.$

The molecular orbital diagram of ${\text{O}}_{\text{2}}{}^{\text{-}}$ is,

Figure 2

The bond order of ${\text{O}}_{\text{2}}{}^{\text{-}}$ is,

  $\begin{array}{l}\text{B}\text{.O=}\frac{\text{1}}{\text{2}}\left({\text{N}}_{\text{b}}{\text{-N}}_{\text{a}}\right)\\ \text{B}\text{.O=}\frac{\text{1}}{\text{2}}\left(\text{10-7}\right)\\ \text{B}\text{.O=1}\text{.5}\end{array}$

Superoxide ion $\left({\text{O}}_{\text{2}}{}^{\text{-}}\right)$ is paramagnetic in nature since it shows the presence of one unpaired electron.

Peroxide ion ${\text{O}}_{\text{2}}{}^{\text{2-}}$ has four electrons in ${\text{π}}^{\text{*}}$ electrons.  The electronic configuration is ${\text{O}}_{\text{2}}{}^{\text{2-}}={\left({\text{σ}}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{1s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}{}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2pz}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}_{\text{2py}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2px}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{}_{\text{2py}}\right)}^2.$

The molecular orbital diagram of ${\text{O}}_{\text{2}}{}^{\text{2-}}$ is,

Figure 3

The bond order of ${\text{O}}_{\text{2}}{}^{\text{2-}}$ is,

  $\begin{array}{l}\text{B}\text{.O=}\frac{\text{1}}{\text{2}}\left({\text{N}}_{\text{b}}{\text{-N}}_{\text{a}}\right)\\ \text{B}\text{.O=}\frac{\text{1}}{\text{2}}\left(\text{10-8}\right)\\ \text{B}\text{.O=1}\end{array}$

Peroxide ion ${\text{O}}_{\text{2}}{}^{\text{2-}}$ is diamagnetic in nature because all the electrons are paired.