Chapter 11, Problem 11.45P

(a)

Interpretation Introduction

Interpretation:

The bond orders of the given molecules and the ions are to be compared with the help of molecular orbital theory.

Concept introduction:

A molecular orbital diagram is a tool that is used to describe the chemical bonding formed between different molecules. It is used to predict the bond strength and the electronic transitions that a molecule can undergo.

The formula to calculate the bond order of any species is as follows:

  $\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbitals}-\\ \text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals}\end{array}\right)$        (1)

Answer to Problem 11.45P

The bond orders of $\text{NO}$, ${\text{O}}_2$ and ${\text{N}}_2$ are 2.5, 2 and 3 whereas the bond orders of ${\text{NO}}^{+}$, ${\text{O}}_2^{+}$ and ${\text{N}}_2^{+}$ are 3, 2.5 and 2.5.

Explanation of Solution

The molecular orbital configuration of $\text{NO}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1$.

Substitute 8 for bonding and 3 for the antibonding orbitals in equation (1) to calculate the bond order of $\text{NO}$.

  $\begin{array}{c}\text{Bond order of NO}=\frac{1}{2}\left(8-3\right)\\ =2.5\end{array}$

The molecular orbital configuration of ${\text{O}}_2$ is

${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1{\left({\pi }_{2p}^{*}\right)}^1$.

Substitute 8 for bonding and 4 for the antibonding orbitals in equation (1) to calculate the bond order of ${\text{O}}_2$.

  $\begin{array}{c}{\text{Bond order of O}}_2=\frac{1}{2}\left(8-4\right)\\ =2\end{array}$

The molecular orbital configuration of ${\text{N}}_2$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2$.

Substitute 8 for bonding and 2 for the antibonding orbitals in equation (1) to calculate the bond order of ${\text{N}}_2$.

  $\begin{array}{c}{\text{Bond order of N}}_2=\frac{1}{2}\left(8-2\right)\\ =3\end{array}$

The molecular orbital configuration of ${\text{NO}}^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2$.

Substitute 8 for bonding and 2 for the antibonding orbitals in equation (1) to calculate the bond order of ${\text{NO}}^{+}$.

  $\begin{array}{c}{\text{Bond order of NO}}^{+}=\frac{1}{2}\left(8-2\right)\\ =3\end{array}$

The molecular orbital configuration of ${\text{O}}_2^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1$.

Substitute 8 for bonding and 3 for the antibonding orbitals in equation (1) to calculate the bond order of ${\text{O}}_2^{+}$.

  $\begin{array}{c}{\text{Bond order of O}}_2^{+}=\frac{1}{2}\left(8-3\right)\\ =2.5\end{array}$

The molecular orbital configuration of ${\text{N}}_2^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^1$.

Substitute 7 for bonding and 2 for the antibonding orbitals in equation (1) to calculate the bond order of ${\text{N}}_2^{+}$.

  $\begin{array}{c}{\text{Bond order of N}}_2^{+}=\frac{1}{2}\left(7-2\right)\\ =\mathrm{2.5.}\end{array}$

Conclusion

The bond order of the species depends on the number of electrons that are present in bonding and antibonding molecular orbitals.

(b)

Interpretation Introduction

Interpretation:

Whether the magnetic behavior of all the species change or not when ions are formed is to be determined.

Concept introduction:

When a neutral atom gains electrons, it acquires a negative charge on it due to the presence of excess electrons as compared to the protons. This results in the formation of the anion. When a neutral atom loses electrons, it acquires a positive charge on it due to the presence of excess protons as compared to the electrons. This results in the formation of the cation.

Paramagnetism is the property of materials due to which they are weakly attracted by an externally applied magnetic field. It arises due to the presence of unpaired electrons in the atoms so the atoms with incompletely filled atomic orbitals are paramagnetic. The unpaired electrons have magnetic dipole moment and therefore act like tiny magnets.

Diamagnetism is the property of materials due to which they are slightly repelled by an externally applied magnetic field. It occurs due to the presence of paired electrons so the atoms with all the filled orbitals are diamagnetic.

Answer to Problem 11.45P

The magnetic behavior of all the molecules does not change when ions are formed.

Explanation of Solution

The molecular orbital configuration of $\text{NO}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1$ and it is paramagnetic in nature due to the presence of an unpaired electron.

The molecular orbital configuration of ${\text{O}}_2$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1{\left({\pi }_{2p}^{*}\right)}^1$ and it is paramagnetic due to the presence of two unpaired electrons.

The molecular orbital configuration of ${\text{N}}_2$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2$ and it is diamagnetic due to the absence of unpaired electrons.

The molecular orbital configuration of ${\text{NO}}^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^2$ and it is diamagnetic in nature due to the absence of an unpaired electron.

The molecular orbital configuration of ${\text{O}}_2^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\sigma }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}^{*}\right)}^1$ and it is paramagnetic due to the presence of an unpaired electron.

The molecular orbital configuration of ${\text{N}}_2^{+}$ is ${\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\pi }_{2p}\right)}^2{\left({\sigma }_{2p}\right)}^1$ and it is paramagnetic due to the presence of an unpaired electron.

The magnetic behavior of $\text{NO}$ and ${\text{N}}_2$ changes when their respective ions are formed but that of ${\text{O}}_2$ remains the same.

Conclusion

The magnetic behavior depends on the availability of the unpaired electrons.