# Question #35197

Jan 8, 2018

In the simplest sense, the bond order is equal to one-half the number of shared electrons between two atoms.

#### Explanation:

For example, in acetylene, which can be written as
$H - C \equiv C - H$
the bond order between each H and C is 1 because two electrons are shared (1 from H and 1 from C).

However, there are 6 electrons shared between the two C atoms (3 from each C atom), so this bond order is 3. This simple picture covers most stable molecules.

However, with radicals and ions, a more complicated approach is required.

When considering the molecular orbital diagram, the bond order is equal to $\frac{1}{2}$(no. of bonding electrons - no. of antibonding electrons).

For example, for ${H}_{2}^{+}$ the bond order is $\frac{1}{2}$ because there is only 1 electron in the system.

For ${O}_{2}^{+}$ the bond order is 2.5 because there are 11 valence electrons (not counting the core 1s electrons on each atom). 8 of these are in bonding orbitals and 3 in anti-bonding orbitals:

$\frac{1}{2} \left(8 - 3\right) = 2.5$