# Is NO^- Paramagnetic or Diamagnetic?

Apr 7, 2016

The MO diagram for $\text{NO}$ is as follows (Miessler et al., Answer Key):

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.

Since this is $\text{NO}$, if you add an electron to get to ${\text{NO}}^{-}$, you add it into the $2 {b}_{2}$ orbital, which is the $\setminus m a t h b f \left({\pi}_{2 {p}_{y}}^{\text{*}}\right)$ 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 $N {O}^{+}$? Is that diamagnetic, and how do you know?