Chapter 10, Problem 10.98QE

(a)

Interpretation Introduction

Interpretation:

Nitrosyl ion is paramagnetic or diamagnetic has to be given.

Explanation of Solution

Nitrosyl ion is given as ${\text{NO}}^{\text{+}}$.  The total number of valence electrons in nitrogen is five and that of oxygen is six.  As there is a positive charge over the nitrosyl ion, there has to be a total of ten electrons.  The molecular orbital electronic configuration of ${\text{NO}}^{\text{+}}$ is given as ${({\sigma }_{2s})}^2{({\sigma }_{2s}^{*})}^2{({\pi }_{2p})}^4{({\sigma }_{2p})}^2$.  As there are no unpaired electrons present in the molecular orbital electronic configuration, nitrosyl ion is diamagnetic.

(b)

Interpretation Introduction

Interpretation:

The highest energy orbital that will be occupied by the electrons has to be given if the molecular orbital diagram is assumed for homonuclear diatomic molecule.

Explanation of Solution

Nitrosyl ion is given as ${\text{NO}}^{\text{+}}$.  The total number of valence electrons in nitrogen is five and that of oxygen is six.  As there is a positive charge over the nitrosyl ion, there has to be a total of ten electrons.  The molecular orbital electronic configuration of ${\text{NO}}^{\text{+}}$ is given as ${({\sigma }_{2s})}^2{({\sigma }_{2s}^{*})}^2{({\pi }_{2p})}^4{({\sigma }_{2p})}^2$.

If the homonuclear diatomic molecule has the same molecular orbital electronic configuration, then the highest energy orbital that will be occupied is ${\sigma }_{2p}$.

(c)

Interpretation Introduction

Interpretation:

Bond order of nitrogen‑oxygen has to be written in nitrosyl ion.

Explanation of Solution

Nitrosyl ion is given as ${\text{NO}}^{\text{+}}$.  The total number of valence electrons in nitrogen is five and that of oxygen is six.  As there is a positive charge over the nitrosyl ion, there has to be a total of ten electrons.  The molecular orbital electronic configuration of ${\text{NO}}^{\text{+}}$ is given as ${({\sigma }_{2s})}^2{({\sigma }_{2s}^{*})}^2{({\pi }_{2p})}^4{({\sigma }_{2p})}^2$.

Bond order is calculated as shown below;

    $\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbital}\\ -\text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals})\\ =\frac{1}{2}(8-2)\\ =3\end{array}$

Thus the bond order of nitrosyl ion is $3$.

(d)

Interpretation Introduction

Interpretation:

The $\text{N}-\text{O}$ bond in ${\text{NO}}^{\text{+}}$ is stronger or not than $\text{NO}$ has to be given.

Explanation of Solution

Nitrosyl ion is given as ${\text{NO}}^{\text{+}}$.  The total number of valence electrons in nitrogen is five and that of oxygen is six.  As there is a positive charge over the nitrosyl ion, there has to be a total of ten electrons.  The molecular orbital electronic configuration of ${\text{NO}}^{\text{+}}$ is given as ${({\sigma }_{2s})}^2{({\sigma }_{2s}^{*})}^2{({\pi }_{2p})}^4{({\sigma }_{2p})}^2$.

Bond order is calculated as shown below;

    $\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbital}\\ -\text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals})\\ =\frac{1}{2}(8-2)\\ =3\end{array}$

Thus the bond order of nitrosyl ion is $3$.

Nitrogen monoxide is given as $\text{NO}$.  The total number of valence electrons present in $\text{NO}$ is $11$.  The molecular orbital electronic configuration of $\text{NO}$ is given as ${({\sigma }_{2s})}^2{({\sigma }_{2s}^{*})}^2{({\pi }_{2p})}^4{({\sigma }_{2p})}^2{({\pi }_{2p}^{*})}^1$.

Bond order is calculated as shown below;

    $\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbital}\\ -\text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals})\\ =\frac{1}{2}(8-3)\\ =2.5\end{array}$

Thus the bond order of $\text{NO}$ is $2.5$.

As the bond order increases, the strength of the bond increases.  Therefore, the nitrogen-oxygen bond present in ${\text{NO}}^{\text{+}}$ is stronger than $\text{NO}$.