Chapter 9, Problem 57QP

Interpretation Introduction

Interpretation:

The bond order and magnetic properties of the given molecules are to be determined.

Concept introduction:

Two atomic orbitals combine to form a bonding molecular orbital and an antibonding molecular orbital. Orbitals that lie on internuclear axis combine to form $\sigma$

(sigma) molecular orbitals, and orbitals parallel to each other combine to form $\pi$

molecular orbitals.

The molecular orbitals formed by the combination of $\text{1s}$

orbitals are a bonding molecular orbital, designated by ${\sigma }_{1s}$, and an antibonding molecular orbital, designated by ${\sigma }^{\ast }{}_{1s}$. The $\text{2s}$

orbital forms corresponding molecular orbitals.

The molecular orbitals formed by the combination of ${\text{2p}}_{\text{x}}$

orbitals area bonding molecular orbital, designated by ${\sigma }_{2{\text{p}}_{\text{x}}}$, and an antibonding molecular orbital, designated by ${\sigma }^{\ast }{}_{2{\text{p}}_{\text{x}}}$.

The molecular orbitals formed by combining ${\text{2p}}_{\text{y}}$ and ${\text{2p}}_{\text{z}}$

orbitals are bonding molecular orbitals, designated by ${\text{π}}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}_{{\text{2p}}_{\text{z}}}$, and antibonding molecular orbitals, designated by ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{z}}}$.

Electrons are filled in the molecular orbitals in increasing order of energy.

Bond order is determined by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals, and dividing by two.

Higher the bond order, more stable is the molecule.

A diamagnetic substance contains paired electrons whereas a paramagnetic substance contains unpaired electrons.

Answer to Problem 57QP

Solution:

Bond order of ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{+}{\text{, O}}_{\text{2}}{}^{-},{\text{ and O}}_{\text{2}}{}^{2-}$ is 2, 2.5, 1.5, and 1, respectively. ${\text{O}}_{\text{2}}{}^{2-}$

is diamagnetic and ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{+}{\text{, and O}}_{\text{2}}{}^{-}$

are paramagnetic.

Explanation of Solution

The electronic configuration of an oxygen atom is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$. The molecular orbital diagram for ${\text{O}}_{\text{2}}$

is as follows:

Now, remove one electron from ${\text{O}}_{\text{2}}$

to get the molecular orbital diagram for ${\text{O}}_{\text{2}}{}^{+}$.

Add one electron to ${\text{O}}_{\text{2}}$ to get the molecular orbital diagram for ${\text{O}}_{\text{2}}{}^{-}$.

Add two electrons to ${\text{O}}_{\text{2}}$ to get the molecular orbital diagram for ${\text{O}}_{\text{2}}{}^{2-}$.

In ${\text{O}}_{\text{2}}$, there are $\text{8}$ electrons in bonding molecular orbitals and $\text{4}$ electrons in antibonding orbital. Thus, the bond order of ${\text{O}}_{\text{2}}$ is calculated as follows:

$\begin{array}{c}\text{b}\text{.o}=\frac{8-4}{2}\\ =2\end{array}$

In ${\text{O}}_{\text{2}}{}^{+}$, there are $\text{8}$ electrons in bonding molecular orbitals and $\text{3}$ electrons in antibonding orbital. Thus, the bond order of ${\text{O}}_{\text{2}}{}^{+}$

is calculated as follows:

$\begin{array}{c}\text{b}\text{.o}=\frac{8-3}{2}\\ =2.5\end{array}$

In ${\text{O}}_{\text{2}}{}^{-}$, there are $\text{8}$ electrons in bonding molecular orbitals and $\text{5}$ electrons in antibonding orbital. Thus, the bond order of ${\text{O}}_{\text{2}}{}^{+}$ is calculated as follows:

$\begin{array}{c}\text{b}\text{.o}=\frac{8-5}{2}\\ =1.5\end{array}$

In ${\text{O}}_{\text{2}}{}^{2-}$, there are $\text{8}$ electrons in bonding molecular orbitals and $\text{5}$ electrons in antibonding orbital. Thus, the bond order of ${\text{O}}_{\text{2}}{}^{2-}$

is calculated as follows:

$\begin{array}{c}\text{b}\text{.o}=\frac{8-6}{2}\\ =1\end{array}$

Conclusion

A molecule of ${\text{O}}_{\text{2}}$

contains two unpaired electrons, ${\text{O}}_{\text{2}}{}^{+}$

contains one unpaired electron, ${\text{O}}_{\text{2}}{}^{-}$

contains one unpaired electron, and ${\text{O}}_{\text{2}}{}^{2-}$

contains no unpaired electron. Thus, ${\text{O}}_{\text{2}}{}^{2-}$

is diamagnetic and ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{+}{\text{, and O}}_{\text{2}}{}^{-}$

are paramagnetic.