# How do I calculate the bond order for H2- and H2+?

Apr 1, 2017

I'm assuming you mean ${\text{H}}_{2}^{-}$ vs. ${\text{H}}_{2}^{+}$. Well, build the molecular orbital (MO) diagram. Each hydrogen atom contributes one electron, and thus, ${\text{H}}_{2}^{-}$ has three electrons while ${\text{H}}_{2}^{+}$ has one.

Each hydrogen atom contributes one $1 s$ atomic orbital, and thus, the orbitals overlap according to MO theory to form one ${\sigma}_{1 s}$ and one ${\sigma}_{1 s}^{\text{*}}$ MO by conservation of orbitals.

If you calculate their bond order, you get:

"BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding")

$= \frac{1}{2} \left(1 - 0\right) = \frac{1}{2}$

"BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding")

$= \frac{1}{2} \left(2 - 1\right) = \frac{1}{2}$

So, neither is more stable than the other. But of course, they are less stable than ${\text{H}}_{2}$. What does bond order mean in terms of bond strength?