Chapter 10, Problem 10.102SP

Interpretation Introduction

Interpretation: The given species have to be arranged in the increasing order of their bond order.

Concept Introduction:

Bond order:

Bond order determines the number of bonds in the pair of two atoms. So, it is the quantitative measure of a bond.

The bond order can be calculated as follows:

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

Trends in the bond order:

The species with the negative charge means that electrons are getting added in the anti-bonding molecular orbital. So the number of electrons in the anti-bonding molecular orbital increases accordingly. Ultimately, the bond order decreases. Whereas, the species with the positive charge means that electrons are being removed from the anti-bonding molecular orbital. So the number of electrons in the anti-bonding molecular orbital decreases accordingly. Ultimately, the bond order increases.

Answer to Problem 10.102SP

The given species can be arranged in the increasing order of their bond order as follows:

${\text{NO}}^{\text{2-}}<{\text{NO}}^{\text{-}}<\text{NO}={\text{NO}}^{2+}<{\text{NO}}^{+}$

Explanation of Solution

Given: The order of molecular orbitals of $\text{NO}$ is similar to that of ${\text{O}}_{\text{2}}$.

The molecular orbitals of ${\text{O}}_{\text{2}}$:

The molecular orbital configuration of ${\text{O}}_{\text{2}}$ molecule is:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^{\text{1}}\text{=}{\text{π*2p}}_{\text{y}}^{\text{1}}\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

Bond order of $NO$:

The molecular orbitals of $\text{NO}$:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^{\text{1}}\right)\left({\text{π*2p}}_{\text{y}}^0\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of $\text{NO}$ can be calculated as follows:

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

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-\text{5}}{\text{2}}\\ =\frac{5}{2}\\ =2.5\end{array}$

Similarly, the bond orders can be calculated for each of the given species as follows:

Bond order of $N{O}^{2-}$:

The molecular orbitals of ${\text{NO}}^{\text{2-}}$:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^2\right)\left({\text{π*2p}}_{\text{y}}^1\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of ${\text{NO}}^{\text{2-}}$ can be calculated as follows:

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

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-\text{7}}{\text{2}}\\ =\frac{3}{2}\\ =1.5\end{array}$

Bond order of $N{O}^{-}$:

The molecular orbitals of ${\text{NO}}^{\text{-}}$:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^{\text{1}}\right)\left({\text{π*2p}}_{\text{y}}^1\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of ${\text{NO}}^{\text{-}}$ can be calculated as follows:

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

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-6}{\text{2}}\\ =\frac{4}{2}\\ =2\end{array}$

Bond order of $N{O}^{+}$:

The molecular orbitals of $\text{N}{\text{O}}^{\text{+}}$:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^0\right)\left({\text{π*2p}}_{\text{y}}^0\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of $\text{N}{\text{O}}^{\text{+}}$ can be calculated as follows:

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

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-4}{\text{2}}\\ =\frac{6}{2}\\ =3\end{array}$

Bond order of $N{O}^{2+}$:

The molecular orbitals of $\text{N}{\text{O}}^{\text{2+}}$:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\right)\left({\text{π2p}}_{\text{y}}^1\right)\left({\text{π*2p}}_{\text{x}}^0\right)\left({\text{π*2p}}_{\text{y}}^0\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of $\text{N}{\text{O}}^{\text{2+}}$ can be calculated as follows:

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

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(9\right)-4}{\text{2}}\\ =\frac{5}{2}\\ =2.5\end{array}$

The calculated bond orders can be tabulated as follows:

Given molecule	Bond order.
${\text{NO}}^{\text{2-}}$	1.5
${\text{NO}}^{\text{-}}$	2
$\text{NO}$	2.5
${\text{NO}}^{+}$	3
${\text{NO}}^{\text{2+}}$	2.5

Based on this tabulation, the given species can be arranged in the increasing order of their bond order as follows:

${\text{NO}}^{\text{2-}}<{\text{NO}}^{\text{-}}<\text{NO}={\text{NO}}^{2+}<{\text{NO}}^{+}$

Conclusion

The given species have been arranged in the increasing order of their bond order.