Chapter 5, Problem 5.103QA

Interpretation Introduction

To find:

The Bond order of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- from the MO theory and predict the bond orders.

Answer to Problem 5.103QA

Solution:

1) Bond order of N₂⁺ = 2.5

2) Bond order of O₂⁺ = 2.5

3) Bond order of C₂⁺ = 1.5

4) Bond order of Br₂^2- = 0

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Explanation of Solution

The Molecular orbital diagram of N₂⁺, O₂⁺, C₂⁺, and Br₂^2- are below

1) Bond order of N₂⁺

:

Electronic configuration of N₂⁺ is

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

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(9-4)}{2}=\mathrm{ }2.5$

2) Bond order of O₂⁺:

Electronic configuration of O₂⁺ is

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

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(8-3)}{2}=\mathrm{ }2.5$

3) Bond order of C₂⁺:

Electronic configuration of C₂⁺ is

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

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(7-4)}{2}=\mathrm{ }1.5$

4) Bond order of Br₂^2- :

In case of Br₂^2- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br₂^2- is

${(\sigma }_{4p_z}{)}^2{(\pi }_{4p_x}{)}^2{(\pi }_{4p_y}{)}^2{({\pi }^{*}}_{4p_x}{)}^2{({\pi }^{*}}_{4p_y}{)}^2{({\sigma }^{*}}_{4p_z}{)}^2$

$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{ }=\left(\frac{Bonding electrons – Nonbonding electrons}{2}\right)= \frac{(6-6)}{2}=\mathrm{ }0$

The bond order of Br₂^2- is 0 because bonding and antibonding electrons are equal.

All species with nonzero bond order may exist, therefore, N₂⁺, O₂⁺ and C₂⁺ may exist.

Conclusion:

The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.