## Using the MO diagram of \(NO,\) calculate the bond order. Compare it to \(\mathrm{NO}^{+} ?\)

Answer:

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 \(\mathrm{N}_{2}\)-like, see here and comments.)

Quick overview of what the labels correspond to what MOs:

- \(1 a_{1}\) is the \(\sigma_{2 s}\) bonding MO.
- \(2 a_{1}\) is the \(\sigma_{2 s}^{*}\) antibonding MO.
- \(1 b_{1}\) is the \(\pi_{2 p_{x}}\) bonding MO.
- \(1 b_{2}\) is the \(\pi_{2 p_{y}}\) bonding MO.
- \(3 a_{1}\) is the \(\sigma_{2 p_{z}}\) bonding MO, but it’s relatively nonbonding with respect to oxygen.
- \(2 b_{1}\) is the \(\pi_{2 p_{x}}^{*}\) antibonding MO.
- \(2 b_{2}\) is the \(\pi_{2 p_{x}}^{*}\) antibonding MO.
- \(4 a_{1}\) is the \(\pi_{2 p_{x}}^{*}\) antibonding MO.

To obtain the bond order, look at the molecular orbitals formed and decide whether they are bonding or antibonding.

\(\mathrm{BO}=\frac{1}{2}\left(\text { bonding } \mathrm{e}^{-}-\text {antibonding } \mathrm{e}^{-}\right)\)

\(=\frac{1}{2}[(2+2+2+2)-(2+1)]\)

\(=2.5\)

And this should make sense because \(\mathrm{NO}^{+}\) is isoelectronic with \(CO\) which has a bond order of \(3\). With one additional electron in an antibonding \(\left(2 b_{2}\right)\), orbital the bond order decreases by \(\frac{1}{2}\) relative to \(\mathrm{NO}^{+}.\)

If paramagnetism occurs due to unpaired electrons, is \(NO\) paramagnetic or diamagnetic?