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

Nov 23, 2016

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 $\frac{1}{2}$ to the bonding character.
• Each electron in an antibonding molecular orbital contributes $\frac{1}{2}$ 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 $\boldsymbol{0}$. The physical meaning is that the compound doesn't exist.

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 $\frac{1}{2} \left(x - x\right) = 0$, as we predicted above.

Take ${\text{Be}}_{2}$ as an example.

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

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

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