The #sigma# bond is made by a head-on overlap between two compatible atomic orbitals that are symmetric about the internuclear axis. Two common examples are #s"/"s# and #s"/"p_z# combinations, where let's say the #z# axis is the horizontal, internuclear axis.
The #pi# bonds are made by the sidelong overlap between two compatible atomic orbitals, which are specifically the #p_x"/"p_x# and #p_y"/"p_y# combinations. Let's say that #x# is vertical and #y# is towards or away from you.
Let us consider a diatomic molecule for simplicity. How about #"CO"#?
The lewis structure for #"CO"# is #""^((-)):"C"-="O":^((+))#
The MO diagram is:
From bottom to top, where again, #z# is horizontal, #x# is vertical, and #y# is in or out of the screen/paper, we have:
- #sigma_(2s)# bonding MO, from the #2s(C)"/"2s(O)# head-on combination
- #sigma_(2s)^"*"# antibonding MO, from the #2s(C)"/"2s(O)# head-on combination
- #pi_(2p_x)# and #pi_(2p_y)# bonding MOs, from the #2p_x(C)"/"2p_x(O)# and #2p_y(C)"/"2p_y(O)# sidelong combination
- #sigma_(2p_z)# bonding MO, from the #2p_z(C)"/"2p_z(O)# head-on combination. This is the HOMO.
- #pi_(2p_x)^"*"# and #pi_(2p_y)^"*"# antibonding MOs (empty), from the #2p_x(C)"/"2p_x(O)# and #2p_y(C)"/"2p_y(O)# sidelong combination. This is the LUMO.
- #sigma_(2p_z)^"*"# antibonding MO (empty), from the #2p_z(C)"/"2p_z(O)# head-on combination
The #sigma_(2s)# and #sigma_(2s)^"*"# cancel out and contribute zero to the bond order because they contribute two bonding, but also two antibonding electrons. That is 2 bonding and 2 antibonding.
The #sigma_(2p_z)# contributes two bonding electrons, and the #pi_(2px)# and #pi_(2p_y)# contribute two bonding electrons each. That is 6 bonding.
That gives a bond order of #([2+6] - 2)/2 = 3#, as expected.
For a diatomic molecule whose bond order is #3#, we can conclude that there is one #sigma# bond and two #pi# bonds.
CHALLENGE: If #NO^(+)# ion is isoelectronic with #CO#, what about neutral #NO#; what is the bond order on that, and from that, how many #pi# bonds on average does it have? Hint: It still has one #sigma# bond.
