# Are N_2 and N_2^+ paramagnetic or diamagnetic? Which one has the stronger bond?

Jun 18, 2016

Recall that paramagnetic means it contains at least one unpaired electron and diamagnetic is the lack thereof.

${\text{O}}_{2}$ is paramagnetic, with one electron each in its ${\pi}_{2 {p}_{x}}^{\text{*}}$ and ${\pi}_{2 {p}_{y}}^{\text{*}}$ antibonding molecular orbitals. When we go back over to ${\text{N}}_{2}$, since $\text{N}$ has one less electron than $\text{O}$ in its atomic orbitals, ${\text{N}}_{2}$ has two less electrons than ${\text{O}}_{2}$ in its molecular orbitals.

Furthermore, going from ${\text{O}}_{2}$ to ${\text{N}}_{2}$ doesn't change the energy ordering for the ${\pi}_{2 {p}_{x}}^{\text{*}}$ and ${\pi}_{2 {p}_{y}}^{\text{*}}$ relative to the ${\sigma}_{2 {p}_{z}}$ or ${\pi}_{2 {p}_{x}}$ or ${\pi}_{2 {p}_{y}}$, so this assumption is valid.

That means ${\text{N}}_{2}$ is diamagnetic, with no unpaired electrons.

In fact, its highest energy occupied molecular orbital (HOMO) is its ${\sigma}_{2 {p}_{z}}$ bonding orbital, which currently contains two electrons. (Due to orbital mixing effects between the ${\sigma}_{2 s}$ and ${\sigma}_{2 {p}_{z}}$ from ${\text{Li}}_{2}$ to ${\text{N}}_{2}$, the ${\sigma}_{2 {p}_{z}}$ is higher in energy relative to if the effects were not present, and the ${\sigma}_{2 s}$ is lower in energy instead. ${\text{N}}_{2}$ and ${\text{O}}_{2}$ mark the border-crossing for when these effects are not significant; that is, when the ${\sigma}_{2 s}$ and ${\sigma}_{2 {p}_{z}}$ are too far in energy to interact.)

${\text{N}}_{2}^{+}$ therefore involves the removal of one ${\sigma}_{2 {p}_{z}}$ electron.

Thus, ${\text{N}}_{2}^{+}$ has a paramagnetic configuration due to the unpaired ${\sigma}_{2 {p}_{z}}$ electron.

Think about it; if you lose an electron in a bonding molecular orbital, does the bond get weaker or stronger? ${\text{N}}_{2}^{+}$ has less bonding character than ${\text{N}}_{2}$, which means it is less thermodynamically stable.

Does that mean it's a weaker bond? You should be able to figure this out from here.