**bond order**is the difference between the number of bonding and antibonding electrons, divided by two:

Note that the bond order equation can also yield a fractional bond order, which has not been possible with other theories. Any bond order greater than zero, fractional or not, reflects a stable molecule.

Consider three proposed species of hydrogen: an H_{2}^{–} ion, an H_{2} molecule, and an H_{2}^{+} ion. Calculating the bond order can predict whether these species are stable. First, look at the H_{2}^{–} ion. This structure has three electrons. The first two electrons enter the lower-energy $\sigma_{1s}$ bonding orbital, and the last electron enters the ${\sigma_{1s}}^*$ antibonding orbital. The bond order is therefore $(\rm 2-1)/2=0.5$. Because 0.5 is greater than zero, this molecule is stable and has a "half" bond.

Next, consider the H_{2} molecule. H_{2} has two electrons, both in the $\sigma_{1s}$ bonding orbital with no antibonding electrons. The bond order is $(2-0)/2=1$. The bond order indicates a stable single bond.

_{2}

^{+}. This molecule has one electron in the $\sigma_{1s}$ bonding orbital and no antibonding electrons. The bond order is $(1-0)/2=0.5$. So, like the first ion, H

_{2}

^{+}is stable with a "half" bond. The bond order of the H

_{2}molecule is 1, and the bond orders of the H

_{2}

^{–}ion and the H

_{2}

^{+}ion are 0.5. The H

_{2}molecule has the highest bonding order and is the most stable of these species.