Question

Question #2a65a

Answer 1

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