Question

The increasing order of bond order of $O_2, O_2^+ , O_2^-$ and $O_2^{-}$ is

• A. $O _{2}^{+}, O _{2}, O _{2}^{-}, O _{2}^{--}$
• B. $O _{2}^{--}, O _{2}^-, O _{2}^{+}, O _{2}$
• C. $O _{2}, O _{2}^{+}, O _{2}^{-}, O _{2}^{--}$
• D. $O _{2}^{2-}, O _{2}^{-}, O _{2}, O _{2}^{+}$

Answer 1

The Correct Option is D

For $O _{2}$ molecule, electronic configuration is

$\sigma\, 1 s^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \overset{*}{\pi }2 p_{x}^{1}=\overset{*}{\pi} 2 p_{y}^{1}$
$\therefore$ Bond order $=\frac{10-6}{2}=2$

For $O _{2}^{+}$ molecule, an electron is removed from $\overset{*}{\pi }2 p_{y}$ orbital. $\therefore$ Bond order $=\frac{10-5}{2}=2.5$

For $O _{2}^{-}$ molecule, an electron is added to $\overset{*}{\pi} 2 p_{x}^{1}$

orbital.

$\therefore$ Bond order $=\frac{10-7}{2}=1.5$

For $O _{2}^{2-}$ molecule, two electrons are added to $\overset{*}{\pi} 2 p_{x}^{1}$

and $\overset{*}{\pi} 2 p_{y}^{1}$ orbitals.

$\therefore$ Bond order $=\frac{10-8}{2}=1$

So, the order is :

$O _{2}^{2-}< \,O _{2}^{-}