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

##### 1 Answer
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 $\text{100 pm}$:

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

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