What is the bond order of "H"_2^+?

Aug 7, 2017

$1 / 2$. What does this mean? Is the bond stronger or weaker than in ${\text{H}}_{2}$? Is ${\text{H}}_{2}^{+}$ more stable or less stable than ${\text{H}}_{2}$?

Well, we begin from the ${\text{H}}_{2}$ molecule, with the simplest molecular orbital (MO) diagram you would ever find (I am not exaggerating):

The $1 s$ atomic orbitals had overlapped (and being spherical, they can only overlap head-on) to form:

• one ${\sigma}_{1 s}$ bonding MO.
• one ${\sigma}_{1 s}^{\text{*}}$ antibonding MO.

Changes to bond order (corresponding to bond strength) can be summarized as follows:

• Putting electrons into a bonding MO increases the bond order by $\frac{1}{2}$ per electron (strengthening the bond).
• Putting electrons into an antibonding MO decreases the bond order by $\frac{1}{2}$ per electron (weakening the bond).

And vice versa for taking out electrons.

Therefore, the ${\text{H}}_{2}^{+}$ cation, with one less electron (examine the charge!), would lose one from the $\boldsymbol{{\sigma}_{1 s}}$ bonding MO.

Thus, the bond order is $\boldsymbol{\underline{1 / 2}}$. What does this mean? Is the bond stronger or weaker than in ${\text{H}}_{2}$?