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

1 Answer
Aug 29, 2017

The MO diagram for "NO" is as follows (Miessler et al., Answer Key):

(The original was this; I added the orbital depictions and symmetry labels. For further discussion on the orbital energy ordering being "N"_2-like, see here and comments.)

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 "NO"^(+) is isoelectronic with "CO", which has a bond order of 3. With one additional electron in an antibonding orbital (2b_2), the bond order decreases by 1/2 relative to "NO"^(+).

If paramagnetism occurs due to unpaired electrons, is "NO" paramagnetic or diamagnetic?