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