Question

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

Answer 1

`1//2`. What does this mean? Is the bond stronger or weaker than in `"H"_2`? Is `"H"_2^(+)` more stable or less stable than `"H"_2`?

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Well, we begin from the `"H"_2` molecule, with the simplest molecular orbital (MO) diagram you would ever find (I am not exaggerating):

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

• one `sigma_(1s)` bonding MO.
• one `sigma_(1s)^"*"` 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 `1/2` per electron (strengthening the bond).
• Putting electrons into an antibonding MO decreases the bond order by `1/2` per electron (weakening the bond).

And vice versa for taking out electrons.

Therefore, the `"H"_2^(+)` cation, with one less electron (examine the charge!), would lose one from the `bb(sigma_(1s))` bonding MO.

Thus, the bond order is `bbul(1//2)`. What does this mean? Is the bond stronger or weaker than in `"H"_2`?