# Using the MO diagram of #"NO"#, calculate the bond order. Compare it to #"NO"^(+)#?

##### 1 Answer

The MO diagram for

(The original was this; I added the orbital depictions and symmetry labels. For further discussion on the *orbital energy ordering* being

Quick overview of what the labels correspond to what MOs:

#1a_1# is the#sigma_(2s)# bonding MO.#2a_1# is the#sigma_(2s)^"*"# antibonding MO.#1b_1# is the#pi_(2p_x)# bonding MO.#1b_2# is the#pi_(2p_y)# bonding MO.#3a_1# is the#sigma_(2p_z)# bonding MO, but it's relatively nonbonding with respect to oxygen.#2b_1# is the#pi_(2p_x)^"*"# antibonding MO.#2b_2# is the#pi_(2p_y)^"*"# antibonding MO.#4a_1# is the#sigma_(2p_z)^"*"# antibonding MO.

To obtain the **bond order**, look at the molecular orbitals formed and decide whether they are bonding or antibonding.

#"BO" = 1/2 ("bonding e"^(-) - "antibonding e"^(-))#

#= 1/2[(2+2+2+2) - (2+1)]#

#= color(blue)(2.5)#

And this should make sense because

If paramagnetism occurs due to unpaired electrons, is