Chapter 2, Problem 2.53E

(a)

Interpretation Introduction

Interpretation:

Changes in bond order, bond distance, and magnetic properties expected when ${\text{C}}_{\text{2}}$ changes to ${\text{C}}_{\text{2}}{}^{+}$ has to be determined.

Concept Introduction:

Bond order is calculated by the expression given as follows:

  $\text{Bond Order}=\frac{1}{2}\left[\begin{array}{c}\left(\text{number of electrons in antibonding orbital}\right)-\\ \left(\text{number of electrons in antibonding orbital}\right)\end{array}\right]$

The magnetic properties are related to terms such as diamagnetism and paramagnetism. Paramagnetism defines the ability of elements to be weakly attracted in an external magnetic field. It arises due to the presence of unpaired electrons. Diamagnetism defines the ability to be repelled in the external magnetic field environment. This is because diamagnetic species have paired electrons.

The bond length is estimated to be an average of covalent radii of two atoms within a bond. When bond order increases bond becomes stronger and bond length reduces. This accounts for shorter bond length in the case of unsaturated compounds while longer bonds in saturated compounds.

Explanation of Solution

For ${\text{C}}_{\text{2}}$ the total electron to be filled in various molecular orbital is calculated as follows:

  $\begin{array}{c}\mathrm{Total}valenceelectrons=6\left(2\right)\\ =12\end{array}$

The corresponding molecular orbitals in ${\text{C}}_{\text{2}}$ can be filed as follows:

Bond order is calculated by the expression given as follows:

  $\text{Bond Order}=\frac{1}{2}\left[\begin{array}{c}\left(\text{number of electrons in bonding orbital}\right)-\\ \left(\text{number of electrons in antibonding orbital}\right)\end{array}\right]$        (1)

Substitute 4 for anti-bonding electrons and 8 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[8-4\right]\\ =2\end{array}$

Thus bond order is 2 in ${\text{C}}_{\text{2}}$.

For ${\text{C}}_{\text{2}}{}^{+}$, one of the electrons is removed from the highest occupied molecular orbital ${\pi }_{2p}$. Hence the bonding electron is reduced to 7.

Substitute 4 for anti-bonding electrons and 7 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[7-4\right]\\ =1.5\end{array}$

Thus bond order is 1.5 in ${\text{C}}_{\text{2}}{}^{+}$. Therefore bond order reduces by 0.5 when ${\text{C}}_{\text{2}}$ loses an electron to form ${\text{C}}_{\text{2}}{}^{+}$.

Since the reduction in bond implies bond is longer thus the bond length is more in case of ${\text{C}}_{\text{2}}{}^{+}$ than in ${\text{C}}_{\text{2}}$. Further in ${\text{C}}_{\text{2}}$, it is evident from the molecular orbital diagram that all the electron are paired so it is diamagnetic while in case of ${\text{C}}_{\text{2}}{}^{+}$ one of paired lection is lost from ${\pi }_{2p}$ and consequently ${\text{C}}_{\text{2}}{}^{+}$ is paramagnetic.

(b)

Interpretation Introduction

Interpretation:

Changes in bond order, bond distance, and magnetic properties expected when ${\text{N}}_{\text{2}}$ changes to ${\text{N}}_{\text{2}}{}^{+}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

For ${\text{N}}_{\text{2}}$ the total electron to be filled in various molecular orbital is calculated as follows:

  $\begin{array}{c}\mathrm{Total}valenceelectrons=7\left(2\right)\\ =14\end{array}$

The corresponding molecular orbitals in ${\text{N}}_{\text{2}}$ can be filed as follows:

Bond order is calculated by the expression given as follows:

  $\text{Bond Order}=\frac{1}{2}\left[\begin{array}{c}\left(\text{number of electrons in bonding orbital}\right)-\\ \left(\text{number of electrons in antibonding orbital}\right)\end{array}\right]$        (1)

Substitute 4 for anti-bonding electron and 10 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[10-4\right]\\ =3\end{array}$

Thus bond order is 3 in ${\text{N}}_{\text{2}}$.

For ${\text{N}}_{\text{2}}{}^{+}$, one of the electrons is removed from the highest occupied molecular orbital ${\sigma }_{2p}$. Hence the bonding electron is reduced to 9.

Substitute 4 for anti-bonding electrons and 9 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[9-4\right]\\ =2.5\end{array}$

Thus bond order is 2.5 in ${\text{N}}_{\text{2}}{}^{+}$. Therefore bond order reduces by 0.5 when ${\text{N}}_{\text{2}}$ loses an electron to form ${\text{N}}_{\text{2}}{}^{+}$.

Since the reduction in bond order implies bond is longer thus the bond length is more in case of ${\text{N}}_{\text{2}}{}^{+}$ than in ${\text{N}}_{\text{2}}$.

Further in ${\text{N}}_{\text{2}}$, it is evident from the molecular orbital diagram that all the electron are paired so it is diamagnetic while in case of ${\text{N}}_{\text{2}}{}^{+}$ one of these paired electrons is lost from ${\sigma }_{2p}$ and consequently ${\text{N}}_{\text{2}}{}^{+}$ is paramagnetic.

(b)

Interpretation Introduction

Interpretation:

Changes in bond order, bond distance, and magnetic properties expected when ${\text{O}}_{\text{2}}$ changes to ${\text{O}}_{\text{2}}{}^{+}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

For ${\text{O}}_{\text{2}}$ the total electron to be filled in various molecular orbital is calculated as follows:

  $\begin{array}{c}\mathrm{Total}valenceelectrons=8\left(2\right)\\ =16\end{array}$

The corresponding molecular orbitals in ${\text{O}}_{\text{2}}$ can be filed as follows:

Bond order is calculated by the expression given as follows:

  $\text{Bond Order}=\frac{1}{2}\left[\begin{array}{c}\left(\text{number of electrons in bonding orbital}\right)-\\ \left(\text{number of electrons in antibonding orbital}\right)\end{array}\right]$        (1)

Substitute 6 for anti-bonding electrons and 10 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[10-6\right]\\ =2\end{array}$

Thus bond order is 2 in ${\text{O}}_{\text{2}}$.

For ${\text{O}}_{\text{2}}{}^{+}$, one of the electrons is removed from the highest occupied molecular orbital $\pi {*}_{2p}$. Hence the antibonding electron is reduced to 5.

Substitute 5 for anti-bonding electrons and 10 for bonding electrons in equation (1).

  $\begin{array}{c}\text{Bond Order}=\frac{1}{2}\left[10-5\right]\\ =2.5\end{array}$

Thus bond order is 2.5 in ${\text{O}}_{\text{2}}{}^{+}$. Therefore bond order increases by 0.5 when ${\text{O}}_{\text{2}}$ loses an electron to form ${\text{O}}_{\text{2}}{}^{+}$.

Since the reduction in bond order implies bond is shorter thus the bond length is more in case of ${\text{O}}_{\text{2}}{}^{+}$ than in ${\text{O}}_{\text{2}}$.

Further in ${\text{O}}_{\text{2}}$, it is evident from the bond order of 2 and the molecular orbital diagram that all the electron are not paired so it is paramagnetic. In the case of ${\text{O}}_{\text{2}}{}^{+}$, one of paired electron is lost from $\pi {*}_{2p}$ and consequently ${\text{O}}_{\text{2}}{}^{+}$ is paramagnetic with an unpaired electron.