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