Bond Stability

Because every atomic orbital pairing results in two molecular orbitals—one which pulls atoms together (bonding) and one which pushes atoms apart (antibonding)—there must be a way to determine whether a covalent bond is stable. This is accomplished by calculating the bond order. The bond order is the difference between the number of bonding and antibonding electrons, divided by two: If the bond order is equal to 1, there are by definition two more bonding electrons than antibonding electrons, and this corresponds to a single bond. A bond order equal to 2 is a double bond, and a bond order of 3 is a triple bond.

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 H₂^– ion, an H₂ molecule, and an H₂⁺ ion. Calculating the bond order can predict whether these species are stable. First, look at the H₂^– 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 H₂ molecule. H₂ 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.

Finally, consider H₂⁺. 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, H₂⁺ is stable with a "half" bond. The bond order of the H₂ molecule is 1, and the bond orders of the H₂^– ion and the H₂⁺ ion are 0.5. The H₂ molecule has the highest bonding order and is the most stable of these species.