Chapter 9, Problem 56E

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy.

a. CN⁺

b. CN

c. CN⁻

Interpretation Introduction

Interpretation: The electronic configuration for the given diatomic species is to be determined and their bond orders have to be calculated. The paramagnetic species have to be identified. The given molecules have to be placed in the correct order of increasing bond length and bond energy.

Concept introduction: The electronic configuration for multi-electron diatomic is written using the molecular orbitals, derived from the ${\text{H}}_{\text{2}}^{\text{+}}$ molecular ion. A molecule having an unpaired electron in the valence shell is termed as a paramagnetic in nature, while the one that has no unpaired electrons is termed as diamagnetic. The bond order is calculated by difference between the anti-bonding electrons and the bonding electrons by two. This can be stated as,

$\text{Bond}\text{order}=\frac{\left[\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{bonding}\text{orbitals}\end{array}\right)-\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{anti-bonding}\text{orbitals}\end{array}\right)\right]}{2}$

The bond order is directly proportional to the bond energy and inversely proportional to the bond length.

To determine: The electronic configuration of ${\text{CN}}^{+}$, its bond order and its paramagnetic or diamagnetic nature.

Answer to Problem 56E

Answer

The configuration of ${\text{CN}}^{+}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4$ and its bond order is $\underset{\_}{2}$. It is diamagnetic in nature.

Explanation of Solution

The electronic configuration of the involved atoms is,

$\begin{array}{l}\text{C}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}\\ \text{N}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^3\end{array}$

The number of valence electrons present in ${\text{CN}}^{+}$ $\begin{array}{l}=4+5-1\\ =8\end{array}$

The molecular orbital configuration of ${\text{CN}}^{+}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4$

Number of bonding electrons $=6$

Number of non-bonding electrons $=2$

Hence,

Bond order $\begin{array}{c}=\frac{1}{2}\left(6-2\right)\\ =\underset{\_}{2}\end{array}$

No unpaired electron is present; hence it is a diamagnetic molecule.

Interpretation Introduction

Interpretation: The electronic configuration for the given diatomic species is to be determined and their bond orders have to be calculated. The paramagnetic species have to be identified. The given molecules have to be placed in the correct order of increasing bond length and bond energy.

Concept introduction: The electronic configuration for multi-electron diatomic is written using the molecular orbitals, derived from the ${\text{H}}_{\text{2}}^{\text{+}}$ molecular ion. A molecule having an unpaired electron in the valence shell is termed as a paramagnetic in nature, while the one that has no unpaired electrons is termed as diamagnetic. The bond order is calculated by difference between the anti-bonding electrons and the bonding electrons by two. This can be stated as,

$\text{Bond}\text{order}=\frac{\left[\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{bonding}\text{orbitals}\end{array}\right)-\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{anti-bonding}\text{orbitals}\end{array}\right)\right]}{2}$

The bond order is directly proportional to the bond energy and inversely proportional to the bond length.

To determine: The electronic configuration of $\text{CN}$, its bond order and its paramagnetic or diamagnetic nature.

Answer to Problem 56E

Answer

The configuration of $\text{CN}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4{\left({\sigma }_{2\text{p}}\right)}^1$ and its bond order is $\underset{\_}{2.5}$. It is paramagnetic in nature.

Explanation of Solution

The electronic configuration of the involved atoms is,

$\begin{array}{l}\text{C}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}\\ \text{N}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^3\end{array}$

The number of valence electrons present in $\text{CN}$ $\begin{array}{l}=4+5\\ =9\end{array}$

The molecular orbital configuration of $\text{CN}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4{\left({\sigma }_{2\text{p}}\right)}^1$

Number of bonding electrons $=7$

Number of non-bonding electrons $=2$

Hence,

Bond order $\begin{array}{l}=\frac{7-2}{2}\\ =\underset{\_}{2.5}\end{array}$.

The $\text{CN}$ molecule contains one unpaired electron. Hence, it is a paramagnetic molecule.

Interpretation Introduction

Interpretation: The electronic configuration for the given diatomic species is to be determined and their bond orders have to be calculated. The paramagnetic species have to be identified. The given molecules have to be placed in the correct order of increasing bond length and bond energy.

Concept introduction: The electronic configuration for multi-electron diatomic is written using the molecular orbitals, derived from the ${\text{H}}_{\text{2}}^{\text{+}}$ molecular ion. A molecule having an unpaired electron in the valence shell is termed as a paramagnetic in nature, while the one that has no unpaired electrons is termed as diamagnetic. The bond order is calculated by difference between the anti-bonding electrons and the bonding electrons by two. This can be stated as,

$\text{Bond}\text{order}=\frac{\left[\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{bonding}\text{orbitals}\end{array}\right)-\left(\begin{array}{l}\text{Electrons}\text{in}\\ \text{anti-bonding}\text{orbitals}\end{array}\right)\right]}{2}$

The bond order is directly proportional to the bond energy and inversely proportional to the bond length.

To determine: The electronic configuration of ${\text{CN}}^{-}$, its bond order and its paramagnetic or diamagnetic nature.

Answer to Problem 56E

Answer

The configuration of ${\text{CN}}^{-}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4{\left({\sigma }_{2\text{p}}\right)}^2$ and its bond order is $\underset{\_}{3}$. It is diamagnetic in nature. The required order is, ${\text{CN}}^{-}<\text{CN}<{\text{CN}}^{\text{+}}$

Explanation of Solution

The electronic configuration of the involved atoms is,

$\begin{array}{l}\text{C}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}\\ \text{N}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^3\end{array}$

The number of valence electrons present in ${\text{CN}}^{-}$ $\begin{array}{l}=4+5+1\\ =10\end{array}$

The molecular orbital configuration of ${\text{CN}}^{-}$ is ${\left({\sigma }_{2\text{s}}\right)}^2{\left({\sigma }_{2\text{s}}{}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^4{\left({\sigma }_{2\text{p}}\right)}^2$

Number of bonding electrons $=8$

Number of non-bonding electrons $=2$

Hence,

Bond order $\begin{array}{l}=\frac{8-2}{2}\\ =\underset{\_}{3}\end{array}$.

No unpaired electron is present; hence it is a diamagnetic molecule.

Conclusion

The diatomic configuration of a diatomic species can be determined using the molecular orbital diagram. The difference between the bonding electrons and the non-bonding electrons divided by two gives the bond order of the molecule.

The bond order is inversely proportional to bond length. The molecule having the least bond order value has the greatest bond length.

The bond order is directly proportional to bond energy. The molecule having the least bond order value has the least bond energy.

The bond order is directly proportional to the bond energy and inversely proportional to the bond length.