Is #NO^-# Paramagnetic or Diamagnetic?

1 Answer
Apr 7, 2016

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.)

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.

Since this is #"NO"#, if you add an electron to get to #"NO"^(-)#, you add it into the #2b_2# orbital, which is the #\mathbf(pi_(2p_y)^"*")# antibonding MO.

That increases its paramagnetic properties, as there exist two unpaired electrons now instead of just one.

CHALLENGE: What does that do to the #N-O# #pi# bond? Does it weaken or strengthen it? (Hint: Consider the bond order). What about #NO^(+)#? Is that diamagnetic, and how do you know?