# Is B_2 diamagnetic or paramagnetic?

Apr 5, 2016

It depends on the MO diagram.

ORBITAL MIXING EFFECTS

${\text{B}}_{2}$ will experience orbital mixing, just like ${\text{N}}_{2}$ and every homonuclear diatomic molecule on period 2 to the left of ${\text{N}}_{2}$, where the ${\sigma}_{2 {p}_{z}}$ MO increases in energy and the ${\sigma}_{2 s}$ decreases in energy relative to if there is NOT orbital mixing.

So, for ${\text{B}}_{2}$, the ${\sigma}_{2 {p}_{z}}$ MO will be higher in energy than the ${\pi}_{2 {p}_{x}}$ and ${\pi}_{2 {p}_{y}}$ MOs, NOT lower!

As predicted, the ${\sigma}_{2 {p}_{z}}$ MO (labeled ${\sigma}_{g} \left(2 p\right)$) is higher in energy than the ${\pi}_{2 {p}_{x}}$ and ${\pi}_{2 {p}_{y}}$ MOs (labeled as ${\pi}_{u} \left(2 p\right)$).

Notice the trend in how the ${\sigma}_{g} \left(2 p\right)$ MO has been moving upwards as we move from ${\text{Ne}}_{2}$ to ${\text{Li}}_{2}$; that is indicating that the orbital mixing is greater towards ${\text{Li}}_{2}$, and becomes less than significant for ${\text{O}}_{2}$, ${\text{F}}_{2}$, and ${\text{Ne}}_{2}$.

RATIONALIZING THE ELECTRON CONFIGURATION OF $\setminus m a t h b f \left({B}_{2}\right)$

Since boron contributes three valence electrons, two borons give a total of six valence electrons. So, the $\sigma$ and $\pi$ MOs generated by the $2 s$ and $2 p$ atomic orbitals in forming ${\text{B}}_{2}$ will be filled with six valence electrons total.

Two fill the ${\sigma}_{2 s}$ (labeled ${\sigma}_{g} \left(2 s\right)$) and two fill the ${\sigma}_{2 s}^{\text{*}}$ (labeled ${\sigma}_{u}^{\text{*}} \left(2 s\right)$). Finally, one electron goes in the ${\pi}_{2 {p}_{x}}$ and the other goes into the ${\pi}_{2 {p}_{y}}$.

That accounts for all six valence electrons.

CONCLUSION

Therefore, ${\text{B}}_{2}$ contains two unpaired electrons and is paramagnetic.

CHALLENGE: Had the ${\pi}_{u} \left(2 p\right)$ MOs been higher in energy than the ${\sigma}_{g} \left(2 p\right)$ MO, ${B}_{2}$ would have been diamagnetic. Why?