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Chemical Bonding and Molecular Structure — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryChemical Bonding and Molecular Structure2 Marks
Question
$\text{N}_2$ molecule has a greater bond dissociation energy than $\text{N}^+_2$ ion whereas $\text{O}_2$ molecule has a lower bond dissociation energy than $\text{O}^+_2$ ion. Explain in terms of molecular orbital theory.
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Answer

$\text{N}_2(14):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{S})^2(\sigma*\text{2s})^2(\pi2\text{p}_\text{x}^2=\pi2\text{p}^2_\text{x})(\sigma2\text{p}_\text{z})^2;$
$\text{B.O.}=\frac12(10-4)=3$
$\text{N}^+_2(13):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\pi2\text{p}_\text{x}^2=\pi2\text{p}_\text{x}^2)(\sigma2\text{p}_\text{z})^1$
$\text{B.O.}=\frac{1}{2}(9-4)=\frac52$
Since $\text{N}_2$ has higher bond order than $\text{N}_2^+,$ therefore $\text{N}_2$ has higher bond dissociation energy than $\text{N}^+_2.$
$\text{O}_2(16):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\sigma*2\text{s})^2(\sigma2\text{p}_\text{z})^2(\pi2\text{p}_\text{x}=\pi2\text{p}_\text{y}^2)(\pi*2\text{p}_\text{x}^1=\pi*2\text{p}_\text{y}^1)$
$\text{B.O.}=\frac12(10-6)=\frac42=2$
$\text{O}^+_2(15):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\sigma*\text{2s})^2(\sigma2\text{p}_\text{z})^2{(\pi2\text{p}_\text{x}^2=\pi\text{p}^2_\text{y}}(\pi*2\text{p}^1_\text{x})$
$\text{B.O.}=\frac{1}{2}(10-5)=\frac52$
$\text{O}^+_2$ has higher bond order than $\text{O}_2$, therefore, it has high bond dissociation energy than $\text{O}_2$.

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