Question #35197

1 Answer
Jan 8, 2018

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


For example, in acetylene, which can be written as
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 #1/2#(no. of bonding electrons - no. of antibonding electrons).

For example, for #H_2^+# the bond order is #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: