Chapter 5, Problem 5.98QA

Interpretation Introduction

To find:

if removing an electron will create molecular ions (He₂⁺ and Ne₂⁺) more stable than He₂ and Ne₂.

Answer to Problem 5.98QA

Solution:

a) Bond order of He₂⁺ = 0.5

b) Bond order of Ne₂⁺ = 0.5

All species with nonzero bond order may exist. Therefore removing of and electron create He₂⁺ and Ne₂⁺ and are more stable than He₂ and Ne₂.

Explanation of Solution

Molecular orbital diagrams of He₂, He₂⁺, Ne₂ and Ne₂⁺ are shown below:

A) Comparison of He₂ and He₂⁺:

Bond order of He₂:

Electronic configuration of He₂⁺ is,

${(\sigma }_{1s}{)}^2 {({\sigma }^{*}}_{1s}{)}^2$

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(2-2)}{2}=\mathrm{ }0$

Bond order of He₂⁺:

Electronic configuration of He₂⁺ is,

${(\sigma }_{1s}{)}^2 {({\sigma }^{*}}_{1s}{)}^1$

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(2-1)}{2}=\mathrm{ }0.5$

B) Comparison of Ne₂ and Ne₂⁺:

Electronic configuration of Ne₂ is,

${(\sigma }_{1s}{)}^2 {({\sigma }^{*}}_{1s}{)}^2{(\sigma }_{2s}{)}^2 {({\sigma }^{*}}_{2s}{)}^2 {(\sigma }_{2p_z}{)}^2{(\pi }_{2p_x}{)}^2{(\pi }_{2p_y}{)}^2 {({\pi }^{*}}_{2p_x}{)}^2 {({\pi }^{*}}_{2p_y}{)}^2{({\sigma }^{*}}_{2p_z}{)}^2$

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(10-10)}{2}=\mathrm{ }0$

Electronic configuration of Ne₂⁺ is,

${(\sigma }_{1s}{)}^2 {({\sigma }^{*}}_{1s}{)}^2{(\sigma }_{2s}{)}^2 {({\sigma }^{*}}_{2s}{)}^2 {(\sigma }_{2p_z}{)}^2{(\pi }_{2p_x}{)}^2{(\pi }_{2p_y}{)}^2 {({\pi }^{*}}_{2p_x}{)}^2 {({\pi }^{*}}_{2p_y}{)}^2{({\sigma }^{*}}_{2p_z}{)}^1$

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(10-9)}{2}=\mathrm{ }0.5$

All species with nonzero bond order may exist, therefore, He₂⁺, Ne₂⁺ may exist and are more stable than He₂ and Ne₂. The strength of bond and stability of the molecule increase as the bond order increases.

Conclusion:

We used molecular orbital diagram to find the bond order and the stability of the compound.