# Draw the MO diagrams of "O"_2, "O"_2^(2-), and "O"_2^(2+)?

Jul 28, 2017

${\text{O}}_{2}^{2 +}$ is more thermodynamically stable, as it has the stronger bond... triple bonds are clearly stronger than $1.5$-bonds.

You can read here if you want to know how to construct an MO diagram.

Given the MO diagram of ${\text{O}}_{2}$, you should be able to deduce the MO diagram of ${\text{O}}_{2}^{-}$ and ${\text{O}}_{2}^{2 +}$.

One trick I like to consider for bond strength is:

• Each added electron in a bonding molecular orbital adds $\textcolor{red}{\underline{0.5}}$ to the bond order. That is, it strengthens the overall bond by half a bond of that type of interaction ($\sigma , \pi , \delta , \phi , . . .$).
• Each added electron in an antibonding molecular orbital adds $\textcolor{red}{\underline{- 0.5}}$ to the bond order. That is, it weakens the overall bond by half a bond of that type of interaction ($\sigma , \pi , \delta , \phi , . . .$).

${\text{O}}_{2}$, of course, has a bond order of $2$... it is $: \stackrel{. .}{\text{O"=stackrel(..)"O}} :$, with a double bond, so it only makes sense...

• ${\text{O}}_{2}^{-}$, or $\stackrel{\left(- 1 / 2\right)}{: \stackrel{\ldots}{\text{O")stackrel(--)("—")stackrel((-1//2))(stackrel(...)"O}} :}$, simply has one more electron in one of the ${\pi}_{2 p x / y}^{\text{*}}$ antibonding molecular orbitals. So, ${\text{O}}_{2}^{-}$ has a bond order of $2 + \left(1 \times - 0.5\right) = \boldsymbol{1.5}$.

• ${\text{O}}_{2}^{2 +}$, or $\stackrel{\left(+\right)}{: \text{O")-=stackrel((+))("O} :}$, simply has two less electrons in the ${\pi}_{2 p x / y}^{\text{*}}$ antibonding molecular orbitals. Thus, it has a bond order of $2 - \left(2 \times - 0.5\right) = \boldsymbol{3}$, isoelectronic with ${\text{N}}_{2}$.

${\text{O}}_{2}^{2 +}$ is thus more thermodynamically stable, as it has the stronger bond... triple bonds are clearly stronger than $1.5$-bonds.