Chapter 9, Problem 28PS

The nitrosyl ion. NO⁺, has an interesting chemistry.

Assume the molecular orbital diagram shown in Figure 9.16 applies to NO⁺.

1. (a) Is NO⁺ diamagnetic or paramagnetic? If paramagnetic, how many unpaired electrons does it have?
2. (b) What is the highest-energy molecular orbital (HOMO) occupied by electrons?
3. (c) What is the nitrogen-oxygen bond order?
4. (d) Is the N–O bond in NO* stronger or weaker than the bond in NO?

Interpretation Introduction

Interpretation:

It should be checked that whether the given molecule is diamagnetic or paramagnetic in nature.

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

1. (1) Bonding orbitals
2. (2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O_2$ can be represented as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^2$

The * represent the antibonding orbital

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by  a magnet

Answer to Problem 28PS

The given molecule $N{O}^{+}$ is diamagnetic in nature. It does not have unpaired electrons.

Explanation of Solution

There are $14$ electrons in $N{O}^{+}$ ion.

In accordance with the MO theory, the electron configuration of this molecule can be written as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^2$

The molecular Orbital diagram for the given molecule can be drawn as follows,

  $\begin{array}{l}\text{σ*2p}\\ --\boxed{}--\to (\text{Anti}\text{bonding}\text{electrons})\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }--\\ {\text{π*}}_{\text{2p}}{\text{π*}}_{\text{2p}}\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }-\boxed{\uparrow }----\boxed{\uparrow \downarrow }-\boxed{\uparrow }-\boxed{\uparrow }--\\ {\text{2}}_{\text{px}}{\text{2}}_{\text{py}}{\text{2}}_{\text{pz}}{\text{2}}_{\text{px}}{\text{2}}_{\text{py}}{\text{2}}_{\text{pz}}\\ \\ \\ {\text{σ}}_{\text{2p}}\\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }--\\ {\text{π}}_{\text{2p}}{\text{π}}_{\text{2p}}\end{array}$

                            $\begin{array}{l}\\ --\boxed{\uparrow \downarrow }--\to (\text{Anti}\text{bonding}\text{electrons})\\ \left(\sigma *2s\right)\\ \\ --\boxed{\uparrow \downarrow }----\boxed{\uparrow \downarrow }--\\ 2s2s\\ \\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ {\left(\sigma 2s\right)}^2\end{array}$

    $\begin{array}{l}\\ --\boxed{\uparrow \downarrow }--\to (\text{Anti}\text{bonding}\text{electrons})\\ \left(\sigma *2s\right)\\ \\ --\boxed{\uparrow \downarrow }----\boxed{\uparrow \downarrow }--\\ 2s2s\\ \\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ {\left(\sigma 2s\right)}^2\end{array}$

In the $N{O}^{+}$  molecule, there is no unpaired electron and so the molecule is diamagnetic in nature.

Interpretation Introduction

Interpretation:

The Highest Occupied Molecular Orbital (HOMO) in the given molecule chlorine monoxide $N{O}^{+}$ should be determined.

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

1. (1) Bonding orbitals
2. (2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O_2$ can be represented as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^2$

The * represent the antibonding orbital

HOMO and LUMO: This terms are stands for highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO), respectively. So this energy difference between the HOMO and LUMO is termed the HOMO–LUMO gap.

Answer to Problem 28PS

The $\left( \pi {*}_{\text{2p}}\right)$ electron is the highest occupied molecular orbital shell.

Explanation of Solution

There are $14$ electrons in $N{O}^{+}$ ion.

In accordance with the MO theory, the electron configuration of this molecule can be written as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^2$

The molecular Orbital diagram for the given molecule can be drawn as follows,

  $\begin{array}{l}\text{σ*2p}\\ --\boxed{}--\to (\text{Anti}\text{bonding}\text{electrons})\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }--\\ {\text{π*}}_{\text{2p}}{\text{π*}}_{\text{2p}}\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }-\boxed{\uparrow }----\boxed{\uparrow \downarrow }-\boxed{\uparrow }-\boxed{\uparrow }--\\ {\text{2}}_{\text{px}}{\text{2}}_{\text{py}}{\text{2}}_{\text{pz}}{\text{2}}_{\text{px}}{\text{2}}_{\text{py}}{\text{2}}_{\text{pz}}\\ \\ \\ {\text{σ}}_{\text{2p}}\\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ \\ --\boxed{\uparrow \downarrow }-\boxed{\uparrow \downarrow }--\\ {\text{π}}_{\text{2p}}{\text{π}}_{\text{2p}}\end{array}$

                            $\begin{array}{l}\\ --\boxed{\uparrow \downarrow }--\to (\text{Anti}\text{bonding}\text{electrons})\\ \left(\sigma *2s\right)\\ \\ --\boxed{\uparrow \downarrow }----\boxed{\uparrow \downarrow }--\\ 2s2s\\ \\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ {\left(\sigma 2s\right)}^2\end{array}$

    $\begin{array}{l}\\ --\boxed{\uparrow \downarrow }--\to (\text{Anti}\text{bonding}\text{electrons})\\ \left(\sigma *2s\right)\\ \\ --\boxed{\uparrow \downarrow }----\boxed{\uparrow \downarrow }--\\ 2s2s\\ \\ --\boxed{\uparrow \downarrow }--\to (\text{Bonding}\text{electrons})\\ {\left(\sigma 2s\right)}^2\end{array}$

In the ($N{O}^{+}$) molecule, four electrons were occupied in two p-orbitals, mainly in $\left( \pi {*}_{\text{2p}}\right)$ molecular orbital. Therefore, it is the highest occupied molecular orbital.

Interpretation Introduction

Interpretation:

Nitrogen – oxygen bond order in the molecule $N{O}^{+}$ has to be calculated.

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

1. (1) Bonding orbitals
2. (2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O_2$ can be represented as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^2$

The * represent the antibonding orbital

Bond order: It is the measure of number of electron pairs shared between two atoms.

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

Explanation of Solution

There are $14$ electrons in $N{O}^{+}$ ion.

In accordance with the MO theory, the electron configuration of this molecule can be written as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^2$

$\begin{array}{l}\text{The bond order of can}\text{be }\text{calculated}\text{as}\text{follows,}\\ \\ \text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs-}\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \text{Bond}\text{order}\text{=}\frac{10-4}{2}=3\end{array}$

Interpretation Introduction

Interpretation:

It should be checked that whether the $N-O$ bond in $N{O}^{+}$ stronger or weaker than the bond in $NO$  

Concept Introduction:

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

1. (1) Bonding orbitals
2. (2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O_2$ can be represented as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^2$

The * represent the antibonding orbital

Bond order: It is the measure of number of electron pairs shared between two atoms.

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

Bond strength is directly proportional to bond order value.

Explanation of Solution

There are $14$ electrons in $N{O}^{+}$ ion.

In accordance with the MO theory, the electron configuration of this molecule can be written as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^2$

$\begin{array}{l}\text{The bond order of can}\text{be }\text{calculated}\text{as}\text{follows,}\\ \\ \text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs-}\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \text{Bond}\text{order}\text{=}\frac{10-4}{2}=3\end{array}$

There are $15$ electrons in $NO$ molecule.

In accordance with the MO theory, the electron configuration of this molecule can be written as follows,

${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^1$

$\begin{array}{l}\text{The bond order of can}\text{be }\text{calculated}\text{as}\text{follows,}\\ \\ \text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs-}\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \text{Bond}\text{order}\text{=}\frac{10-5}{2}=2.5\end{array}$

Bond strength is directly proportional to bond order value.

Therefore,

$N-O$ bond in $N{O}^{+}\left(Bondorder=3\right)$ stronger than the bond in $NO\left(Bondorder=2.5\right)$.