CHEM 121 glossary

Bond order

In MO theory, half the difference between the electrons in bonding and in antibonding MOs: $\text{bond order} = \tfrac{1}{2}(\text{bonding} - \text{antibonding})$ . It matches the Lewis count (1 single, 2 double, 3 triple) but can be a half-integer; a higher bond order means a shorter, stronger bond, and 0 means no stable molecule.

\ce{F}

atom

\ce{F2}

MOs

\ce{F}

atom

$\ce{F2}$ : 8 bonding and 6 antibonding valence electrons, bond order 1.

Example

$\ce{F2}$ has 14 valence electrons, 8 bonding and 6 antibonding: bond order ½(8 − 6) = 1, the F–F single bond. $\ce{F2+}$ loses one antibonding $\pi^*_{2p}$ electron, so its bond order rises to 1.5 and its bond is shorter and stronger.