Chapter 7, Problem 7.129QP

The BO⁺ ion is paramagnetic. Determine (a) whether the order of molecular-orbital energies is like that in B₂ or in O₂; (b) the bond order; and (c) the number of unpaired electrons in the ion.

Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of ${\text{BO}}^{+}$ is like that in ${\text{O}}_2$ or ${\text{B}}_2$. Also the bond order and the number of unpaired electrons in ${\text{BO}}^{+}$ should be found.

Concept Introduction:

• In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
• The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
• There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
• If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
• Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

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

Explanation of Solution

In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals and only contain a maximum of two electrons. The number of newly formed molecular orbital is equal to the number of atomic orbitals involved in the bonding.

There are two types of molecular orbitals,

1. a) Bonding molecular orbitals: sharing of electron density is between the nuclei and has comparatively lower energy and fills first.
2. b) Antibonding molecular orbitals: Two nuclei are pulled by the electrons density in opposite direction and have higher energy comparing to bonding molecular orbital.

The orbital with lower energy gets the electron first.

Each orbital only contain a maximum of two electrons.

It obeys Hund's Rule: The orbital having maximum number of electrons with same spin in separate orbitals is the most stable arrangement in an orbital.

Electronic configuration of $\text{O}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$ and the electronic configuration of $\text{B}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The total charge of ${\text{BO}}^{+}$ is +1 so an electron is reduced from the total valence electrons. Hence the total number of valence electrons in ${\text{BO}}^{+}$ is 8. By molecular orbital theory, it has unpaired electrons and thus is a paramagnetic species.

To compare the order of molecular orbital energies, the valence electrons in ${\text{O}}_2$ and ${\text{B}}_2$ is taken as eight. ${\text{BO}}^{+}$ is paramagnetic so the orbitals are arranged as that of ${\text{O}}_2$.

Molecular orbital diagram of ${\text{B}}_2$

${\text{σ}}_{{\text{2p}}_{\text{z}}}{}^{\text{*}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{}^{\text{*}}{\text{π}}_{{\text{2p}}_{\text{y}}}{}^{\text{*}}$

${\text{σ}}_{{\text{2p}}_{\text{z}}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{\text{π}}_{{\text{2p}}_{\text{y}}}$

${\text{σ}}_{\text{2s}}{}^{\text{*}}$

${\text{σ}}_{\text{2s}}$

Molecular orbital diagram of ${\text{O}}_2$

${\text{σ}}_{{\text{2p}}_{\text{z}}}{}^{\text{*}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{}^{\text{*}}{\text{π}}_{{\text{2p}}_{\text{y}}}{}^{\text{*}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{\text{π}}_{{\text{2p}}_{\text{y}}}$

${\text{σ}}_{{\text{2p}}_{\text{z}}}$

${\text{σ}}_{\text{2s}}{}^{\text{*}}$

${\text{σ}}_{\text{2s}}$

The molecular orbital diagram of ${\text{O}}_2$ is similar to ${\text{BO}}^{+}$.

Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of ${\text{BO}}^{+}$ is like that in ${\text{O}}_2$ or ${\text{B}}_2$. Also the bond order and the number of unpaired electrons in ${\text{BO}}^{+}$ should be found.

Concept Introduction:

• In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
• The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
• There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
• If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
• Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

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

Explanation of Solution

Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

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

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

Interpretation Introduction

Interpretation: To find whether the order of molecular-orbital energies of ${\text{BO}}^{+}$ is like that in ${\text{O}}_2$ or ${\text{B}}_2$. Also the bond order and the number of unpaired electrons in ${\text{BO}}^{+}$ should be found.

Concept Introduction:

• In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule.
• The orbital with lower energy gets the electron first and each orbital only contain a maximum of two electrons. It obeys Hund's Rule.
• There are two types of molecular orbitals- bonding molecular orbitals and antibonding molecular orbitals.
• If molecular species contains at least one unpaired electron, then they are paramagnetic. They tend to get attracted towards the magnetic field. Whereas in diamagnetic species there is no unpaired electrons and are slightly repels the magnetic field.
• Stability of a molecule can be obtained from the bond order value. As the bond order increases, stability increases. Bond order can be obtained by,

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

Explanation of Solution

The total charge of ${\text{BO}}^{+}$ is +1 so an electron is reduced from the total valence electrons. Hence the total number of valence electrons in ${\text{BO}}^{+}$ is 8. The molecular orbital diagram of ${\text{O}}_2$ is similar to ${\text{BO}}^{+}$ since it is paramagnetic.

Molecular orbital diagram of ${\text{BO}}^{+}$

${\text{σ}}_{{\text{2p}}_{\text{z}}}{}^{\text{*}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{}^{\text{*}}{\text{π}}_{{\text{2p}}_{\text{y}}}{}^{\text{*}}$

${\text{π}}_{{\text{2p}}_{\text{x}}}{\text{π}}_{{\text{2p}}_{\text{y}}}$

${\text{σ}}_{{\text{2p}}_{\text{z}}}$

${\text{σ}}_{\text{2s}}{}^{\text{*}}$

${\text{σ}}_{\text{2s}}$

${\text{BO}}^{+}$ has two unpaired electrons.