Chapter 9, Problem 9.105QP

(a)

Interpretation Introduction

Interpretation: The molecular ions with one or more unpaired electron from the given molecules are to be identified.

Concept introduction: When two atomic orbitals come close to each other they lose their identity and form new pair of orbitals knows as molecular orbitals. Among the two molecular orbitals formed, one has energy lower than the atomic orbitals is known as bonding molecular orbital and the other has energy higher than the atomic orbitals and is known as antibonding molecular orbital. The filling electrons in molecular orbitals follow Aufbau’s principle and Hund’s rule.

To determine: If the molecular ion ${\text{N}}_{\text{2}}^{\text{+}}$ has unpaired electrons.

Answer to Problem 9.105QP

Solution

The ${\text{N}}_{\text{2}}^{\text{+}}$ ion has one unpaired electron.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

The total number of valence electrons in ${\text{N}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{N}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{nitrogen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 5-1\\ =9\end{array}$

According to the molecular orbital theory, the electronic configuration of ${\text{N}}_{\text{2}}^{\text{+}}$ is

${\text{N}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{1}}$

One unpaired electron is present in ${\text{σ}}_{\text{2p}}$ orbital in ${\text{N}}_{\text{2}}^{\text{+}}$.

(b)

Interpretation Introduction

To determine: If the molecular ion ${\text{O}}_{\text{2}}^{\text{+}}$ has unpaired electrons.

Answer to Problem 9.105QP

Solution

The ${\text{O}}_{\text{2}}^{\text{+}}$ ion has one unpaired electron.

Explanation of Solution

Explanation

Oxygen has six valence electrons.

The total number of valence electrons in ${\text{O}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{O}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{oxygen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 6-1\\ =11\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{O}}_{\text{2}}^{\text{+}}$ is

${\text{O}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{2p}}\text{)}}^{\text{1}}$

One unpaired electron is present in the ${\text{π}}_{\text{2p}}^{\text{*}}$ orbital in ${\text{O}}_{\text{2}}^{\text{+}}$.

(c)

Interpretation Introduction

To determine: If the molecular ion ${\text{C}}_{\text{2}}^{\text{+}}$ has unpaired electrons.

Answer to Problem 9.105QP

Solution

The ${\text{C}}_{\text{2}}^{\text{+}}$ ion has one unpaired electron.

Explanation of Solution

Explanation

Carbon has four valence electrons.

The total number of valence electrons in ${\text{C}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{C}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{carbon}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 4-1\\ =7\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{C}}_{\text{2}}^{\text{+}}$ is

${\text{C}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{\text{(π}}_{{\text{2p}}_{\text{x}}}{}^2{\text{=π}}_{{\text{2p}}_{\text{y}}}{}^1\text{)}$

One unpaired electron is present in the ${\pi }_{2{\text{p}}_{\text{y}}}$ orbital in ${\text{C}}_{\text{2}}^{\text{+}}$.

(d)

Interpretation Introduction

To determine: If the molecular ion ${\text{Br}}_{\text{2}}^{2-}$ has unpaired electron.

Answer to Problem 9.105QP

Solution

The ${\text{Br}}_{\text{2}}^{2-}$ ion has no unpaired electron.

Explanation of Solution

Explanation

Bromine has seven valence electrons.

The total number of valence electrons in ${\text{Br}}_{\text{2}}^{2-}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{Br}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{bromine}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 7-(-2)\\ =16\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{Br}}_{\text{2}}^{\text{2-}}$ is

${\text{Br}}_{\text{2}}^{\text{2-}}={{\text{(σ}}_{\text{4s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{4s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{4p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{4p}}\text{)}}^4\text{}{{\text{(σ}}^{\text{*}}{}_{\text{4p}}\text{)}}^{\text{2}}$

No unpaired electron is present in any orbital of ${\text{Br}}_{\text{2}}^{\text{2-}}$.

(e)

Interpretation Introduction

To determine: If the molecular ion ${\text{O}}_{\text{2}}^{-}$ has unpaired electron.

Answer to Problem 9.105QP

Solution

The ${\text{O}}_{\text{2}}^{-}$ ion has one unpaired electron.

Explanation of Solution

Explanation

Oxygen has six valence electrons.

The total number of valence electrons in ${\text{O}}_{\text{2}}^{-}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{O}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{oxygen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 6+1\\ =13\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{O}}_{\text{2}}^{-}$ is

${\text{O}}_{\text{2}}^{-}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{2p}}\text{)}}^3$

One unpaired electron is present in the ${\text{π}}_{\text{2p}}^{\text{*}}$ orbital in ${\text{O}}_{\text{2}}^{-}$.

(f)

Interpretation Introduction

To determine: If the molecular ion ${\text{O}}_{\text{2}}^{2-}$ has unpaired electron.

Answer to Problem 9.105QP

Solution

The ${\text{O}}_{\text{2}}^{2-}$ ion has no unpaired electron.

Explanation of Solution

Explanation

Oxygen has six valence electrons.

The total number of valence electrons in ${\text{O}}_{\text{2}}^{2-}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{O}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{oxygen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 6+2\\ =14\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{O}}_{\text{2}}^{2-}$ is

${\text{O}}_{\text{2}}^{-}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{2p}}\text{)}}^4$

No unpaired electron is present in any orbital of ${\text{O}}_{\text{2}}^{2-}$.

(g)

Interpretation Introduction

To determine: If the molecular ion ${\text{N}}_{\text{2}}^{2-}$ has unpaired electron.

Answer to Problem 9.105QP

Solution

The ${\text{N}}_{\text{2}}^{2-}$ ion has two unpaired electrons.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

The total number of valence electrons in ${\text{N}}_{\text{2}}^{2-}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{N}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{nitrogen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 5-(-2)\\ =12\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{N}}_{\text{2}}^{2-}$ is

${\text{N}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}\left({\text{π}}_{{\text{2p}}_{\text{y}}}^{\text{*}}{}^{\text{1}}\text{=}{\text{π}}_{{\text{2p}}_{\text{z}}}^{\text{*}}{}^{\text{1}}\right)$

Two unpaired electrons are present in $\left({\text{π}}_{{\text{2p}}_{\text{y}}}^{\text{*}}{}^{\text{1}}\text{=}{\text{π}}_{{\text{2p}}_{\text{z}}}^{\text{*}}{}^{\text{1}}\right)$ orbitals of ${\text{N}}_{\text{2}}^{2-}$.

(h)

Interpretation Introduction

To determine: If the molecular ion ${\text{F}}_{\text{2}}^{+}$ has unpaired electron.

Answer to Problem 9.105QP

Solution

The ${\text{F}}_{\text{2}}^{+}$ ion has one unpaired electron.

Explanation of Solution

Explanation

Fluorine has seven valence electrons.

The total number of valence electrons in ${\text{F}}_{\text{2}}^{+}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{F}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{fluorine}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 7-1\\ =13\end{array}$

According to the molecular orbital theory the electronic configuration of ${\text{F}}_{\text{2}}^{+}$ is

${\text{F}}_{\text{2}}^{+}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^3$

One unpaired electron is present in ${\text{π}}_{\text{2p}}^{\text{*}}$ orbital in ${\text{F}}_{\text{2}}^{+}$.

Conclusion

The molecules which contain unpaired electrons in their molecular orbitals are paramagnetic in nature. All the molecular ion species except ${\text{Br}}_{\text{2}}^{\text{2-}}\text{and}{\text{O}}_{\text{2}}^{\text{2-}}$, have unpaired electrons.