Question

What is the bond order of any molecule containing equal numbers of bonding and antibonding electrons?

Answer 1

Bond order really describes the "degree" of the bond. For instance, an ideal triple bond has a bond order of `3`, while an ideal double bond has a bond order of `2`.

Each electron can be thought of as contributing to the bonding or antibonding character.

• Each electron in a bonding molecular orbital contributes `1/2` to the bonding character.
• Each electron in an antibonding molecular orbital contributes `1/2` to the antibonding character.

When bonding and antibonding character are equal, the bond order is `0`.

So, when the number of bonding and antibonding electrons is equal, the bond order is `bb(0)`. The physical meaning is that the compound doesn't exist.

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It can also be shown mathematically. You may have been taught:

`"Bond Order" = 1/2("bonding electrons - antibonding electrons")`

but if there are equal numbers of each kind, then the bond order is `1/2(x - x) = 0`, as we predicted above.

Take `"Be"_2` as an example.

The bond order for this is `0`, and therefore, MO theory predicts that `"Be"_2` doesn't exist. It pretty much doesn't, as its theoretical bond length is about `"245 pm"^([1][2])` (2.45 angstroms), while the average bond distance for a typical diatomic molecule is about `"100 pm"`:

`""^([1])` http://www.sciencedirect.com/science/article/pii/0009261489800521

`""^([2])` https://arxiv.org/pdf/1408.5090.pdf