Question

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

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. 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?