# MO diagram of "B"_2"H"_6^(2-)?

Dec 18, 2016

If you wish to see a true MO diagram, see this derivation:

MO diagram of ${\text{B"_2"H}}_{6}$

Then, add two valence electrons, if you want to see the $2 -$ ion.

I would anticipate that the relatively nonbonding ${b}_{1 u}$ MO in the bridging interactions MO diagram is the one that is filled, since it is probably slightly higher in energy than the already-filled ones in the terminal interactions MO diagram.

(In that case, it would only slightly stabilize the molecule, and these electrons would probably get delocalized into the bridging bonds so that the bond order increases from $\frac{1}{2}$ to $\frac{3}{4}$.)

However, it does depend on the extent of orbital mixing, since the terminal interactions also have a nonbonding ${b}_{1 u}$ MO (with slightly more bonding character), which should lower the energy of the ${b}_{1 u}$ MO participating in the bridging interactions.

(This is similar to the orbital mixing effects in ${\text{N}}_{2}$ that lower the ${\sigma}_{2 s}$ energy and raise the ${\sigma}_{2 {p}_{z}}$ energy relative to no orbital mixing, which is what gives rise to the flipped $\sigma \text{/} \pi$ MO energy ordering when going to ${\text{O}}_{2}$.)