4
$\begingroup$

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 $\ce{B2H4}$?

$\endgroup$

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!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.