# Question 2a65a

Feb 20, 2014

To determine the bond order, you subtract the antibonding electrons from the bonding electrons and divide by 2.

The formula for bond order is

Bond Order = ½(b - a)

where b is the number of bonding electrons and a is the number of antibonding electrons.

You first draw the molecular orbital diagram for the molecule. Let’s use C₂ as an example.

You put the eight valence electrons in the lowest orbitals: 2sσ, 2sσ^✳, 2pσ, 2pπ.

You have six bonding and two antibonding electrons. The bond order is

Bond Order = ½(b - a), = ½(6 – 2) = ½ × 4 = 2

EXAMPLE:

Calculate the bond order in C₂²⁺.

Solution:

Here we have only six valence electrons. They are in the 2sσ, 2sσ^✳, and 2pσ# orbitals. This makes four bonding and two antibonding electrons.

Bond Order=½(b - a), = ½(4 – 2) = ½ × 2 = 1