I want to ask a question about the fragment orbitals of the $\ce{B2H6}$ system, specifically when I combine the $\ce{B2H4}$ with the $\ce{H2}$ fragments.

I am halfway through my inorganic chemistry molecular orbitals course and today we were discussing the application of MO theory in figuring out the structure of $\ce{B2H6}$.

So far, I have been able to combine the two $\ce{B2H4}$ FOs but I am struggling to justify the placement of the $\ce{H-H}$ FO to complete the task.

So far, I've been given this diagram.

enter image description here

And the professor said the following:

the $\ce{b2}$ orbitals are non-bonding FOs so it does has nothing to bond with in the $\ce{B2H6}$ molecule. We assume the H $\ce{1s}$ orbital is the reference and line up the hydrogen $\ce{H-H}$ FO with the $\ce{b2}$ FO, giving the symmetry groups $\ce{ag}$ and $\ce{b_{3u}}$ respectively.

I understand the use of Hydrogen as a reference and the assignment of point groups in this situation, but I can't seem to work out why Hydrogen lines up with the $\ce{b2}$ FOs and not, for instance, with the $\ce{a1}$ FOs.

I've searched across the internet for the past day and failed to find any conclusive answer.

Why is $\ce{H-H}$ given to have the same energy as the $\ce{b2}$ FOs of $\ce{B2H4}$?


This question has an open bounty worth +100 reputation from vik1245 ending in 7 days.

Looking for an answer drawing from credible and/or official sources.

Just a reasonable answer to the question about diborane, any help would be appreciated! It's an unanswered question for a while and still haven't found an answer that solves this matter!

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