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

Aug 29, 2017

The MO diagram for $\text{NO}$ 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 ${\text{N}}_{2}$-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 ${\sigma}_{2 s}^{\text{*}}$ 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 ${\pi}_{2 {p}_{x}}^{\text{*}}$ antibonding MO.
• $2 {b}_{2}$ is the ${\pi}_{2 {p}_{y}}^{\text{*}}$ antibonding MO.
• $4 {a}_{1}$ is the ${\sigma}_{2 {p}_{z}}^{\text{*}}$ 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"^(-))

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

$= \textcolor{b l u e}{2.5}$

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

If paramagnetism occurs due to unpaired electrons, is $\text{NO}$ paramagnetic or diamagnetic?