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 $\ce{B2H4}$?

  • $\begingroup$ I learned this as the precise ordering is somewhat arbitrary. You know the energy of the fragment is roughly near the energy of the fragment it's mixing with—otherwise they wouldn't mix to form a bond—and then place it within that range for whatever is convenient to make the drawing readable. These diagrams are qualitative, their purpose to serve as accessible, conceptual overviews, so the priority isn't precision. Only calculations can reveal more rigorously precise orderings. $\endgroup$
    – Blaise
    Nov 22, 2019 at 11:08


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