# diborane fragment orbitals energy of dihydrogen

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.

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!