Chapter 9, Problem 9.104QP

(a)

Interpretation Introduction

Interpretation: The electronic configuration and the stability of ${\text{He}}_{\text{2}}^{\text{+}}\text{and}{\text{Ne}}_{\text{2}}^{\text{+}}$ as compared to ${\text{He}}_2\text{and}{\text{Ne}}_2$ 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.

To determine: The electronic configuration of ${\text{He}}_2\text{and}{\text{Ne}}_2$.

Answer to Problem 9.104QP

The electronic configuration of ${\text{He}}_2\text{and}{\text{Ne}}_2$ is,

${\text{He}}_{\text{2}}={{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{1s}}\text{)}}^{\text{2}}$

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

Explanation of Solution

Explanation

Helium has two valence electrons.

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

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\text{Number}\text{of}\text{He}\text{atoms}\times \text{Valence}\text{electrons }\pm \text{Charge}\\ =2\times 2-0\\ =4\end{array}$

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

${\text{He}}_{\text{2}}={{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{1s}}\text{)}}^{\text{2}}$

Neon has eight valence electrons.

The total number of valence electrons in ${\text{Ne}}_2$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\text{Number}\text{of}\text{Ne}\text{atoms}\times \text{Valence}\text{electrons }\pm \text{Charge}\\ =2\times 8-1\\ =16\end{array}$

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

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

(b)

Interpretation Introduction

To determine: The stability of the molecular ions formed by removing one electron from ${\text{He}}_2\text{and}{\text{Ne}}_2$.

Answer to Problem 9.104QP

The molecular ions ${\text{Ne}}_{\text{2}}^{\text{+}}$ and ${\text{He}}_{\text{2}}^{\text{+}}$ are more stable than ${\text{He}}_2\text{and}{\text{Ne}}_2$.

Explanation of Solution

Explanation

On removing an electron from ${\text{He}}_{\text{2}}$, it becomes ${\text{He}}_2^{+}$.

The total number of valence electrons in ${\text{He}}_2^{+}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\text{Number}\text{of}\text{He}\text{atoms}\times \text{Valence}\text{electrons }\pm \text{Charge}\\ =2\times 2-1\\ =3\end{array}$

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

${\text{He}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{1s}}\text{)}}^1$

The bond order of ${\text{He}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)\\ =\frac{1}{2}\left(2-1\right)\\ =0.5\end{array}$

The bond order of ${\text{He}}_{\text{2}}^{\text{+}}$ is $0.5$.

On removing an electron from ${\text{Ne}}_2$ it becomes ${\text{Ne}}_{\text{2}}^{\text{+}}$ ion.

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

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\text{Number}\text{of}\text{Ne}\text{atoms}\times \text{Valence}\text{electrons}\pm \text{Charge}\\ =2\times 8-1\\ =15\end{array}$

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

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

The bond order of ${\text{Ne}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)\\ =\frac{1}{2}\left(8-7\right)\\ =0.5\end{array}$

The bond order of ${\text{Ne}}_{\text{2}}^{\text{+}}$ is $0.5$

The stability of an ion or a molecule is directly proportional to its bond order.

Both ${\text{Ne}}_{\text{2}}^{\text{+}}$ and ${\text{He}}_{\text{2}}^{\text{+}}$ ions have non-zero bond order values.

The molecules ${\text{He}}_2\text{and}{\text{Ne}}_2$ have bond order equal to zero.

Therefore, the molecular ions ${\text{Ne}}_{\text{2}}^{\text{+}}$ and ${\text{He}}_{\text{2}}^{\text{+}}$ are more stable than ${\text{He}}_2\text{and}{\text{Ne}}_2$.

The stability of an ion or a molecule is directly proportional to its bond order.