Chapter 9, Problem 9.94QP

(a)

Interpretation Introduction

Interpretation: The molecular ions which have electrons in $\text{π}$ antibonding molecular orbitals from the given molecules is to be predicted.

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 ${\text{O}}_{\text{2}}^{-}$ has electrons in $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{O}}_{\text{2}}^{-}$ has three electrons in $\text{π}$ antibonding molecular orbital.

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,

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

Three electrons are present in the $\text{π}$ antibonding orbital in ${\text{O}}_{\text{2}}^{-}$.

(b)

Interpretation Introduction

To determine: If the molecular ion ${\text{O}}_{\text{2}}^{2-}$ has electrons in $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{O}}_{\text{2}}^{2-}$ has four electrons in $\text{π}$ antibonding molecular orbitals

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,

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

Three electrons are present in the $\text{π}$ antibonding orbital in ${\text{O}}_{\text{2}}^{2-}$.

(c)

Interpretation Introduction

To determine: If the molecular ion ${\text{N}}_{\text{2}}^{2-}$ has electrons in $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{N}}_{\text{2}}^{2-}$ has two electrons in $\text{π}$ antibonding molecular orbitals

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,

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

Two electrons are present in the $\text{π}$ antibonding orbital in ${\text{N}}_{\text{2}}^{2-}$.

(d)

Interpretation Introduction

To determine: The molecular ion ${\text{F}}_{\text{2}}^{+}$ has $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{F}}_{\text{2}}^{+}$ has three electrons in $\text{π}$ antibonding molecular orbitals

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,

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

Three electrons are present in the $\text{π}$ antibonding orbital in ${\text{F}}_{\text{2}}^{+}$.

(e)

Interpretation Introduction

To determine: The molecular ion ${\text{N}}_{\text{2}}^{+}$ has $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{N}}_{\text{2}}^{+}$ has no electron in $\text{π}$ antibonding molecular orbitals.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

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

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

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

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

No electron is present in the $\text{π}$ antibonding molecular orbital in ${\text{N}}_{\text{2}}^{+}$.

(f)

Interpretation Introduction

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

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{O}}_{\text{2}}^{\text{+}}$ has one electron in $\text{π}$ antibonding molecular orbitals

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 $\text{π}$ antibonding molecular orbital in ${\text{O}}_{\text{2}}^{\text{+}}$.

(g)

Interpretation Introduction

To determine: If the molecular ion ${\text{C}}_{\text{2}}^{\text{+}}$ has $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{C}}_{\text{2}}^{\text{+}}$ has no electron in $\text{π}$ antibonding molecular orbitals

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,

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

No electron is present in $\text{π}$ antibonding molecular orbital in ${\text{C}}_{\text{2}}^{\text{+}}$.

(g)

Interpretation Introduction

To determine: If the molecular ion ${\text{Br}}_{\text{2}}^{2-}$ has electrons in $\text{π}$ antibonding molecular orbital.

Answer to Problem 9.94QP

Solution

The molecular ion ${\text{Br}}_{\text{2}}^{2-}$ has four electrons in $\text{π}$ antibonding molecular orbitals

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,

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

Four electrons are present in $\text{π}$ antibonding molecular orbitals in ${\text{Br}}_{\text{2}}^{\text{2-}}$.

Conclusion

Electrons in antibonding molecular orbitals of the molecule decrease the strength of the bond. All the given molecular ion species except ${\text{N}}_{\text{2}}^{\text{+}}$ and ${\text{C}}_{\text{2}}^{\text{+}}$ have electrons in $\text{π}$ antibonding molecular orbitals.