Question

Is `NO^-` Paramagnetic or Diamagnetic?

Answer 1

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?