Why is $\ce{B2}$ more stable than $\ce{Li2}$?

Both $\ce{B2}$ and $\ce{Li2}$ have bond order equal to 1. Moreover, $\ce{Li2}$ has only 2 electrons in the antibonding molecular orbital while $\ce{B2}$ has 4 electrons in the antibonding molecular orbital. Thus, $\ce{Li2}$ should have been more stable than $\ce{B2}$ if we consider the number of electrons in the antibonding molecular orbital as the parameter of stability. Does the fact that $\ce{B2}$ is smaller than $\ce{Li2}$ makes it more stable?


1 Answer 1


Firstly, the use of molecular orbital diagrams allows you to predict the electronic configuration, if a molecule would exist or not (as you say, evaluating the bond order), and some magnetic properties. Moreover, if you have "the same molecule" as O$_2$ and its ion O$_2^-$, etc, you can say the higher bond order, the bigger stability.

But if the bond order is the same (your case), you can not talk about stability using those diagrams.

Coming back to the question, looking the dissociation energy (the energy relative to separated atoms) for B$_2$, you will see it is bigger in module. So it will be more stable. I do not have a qualitative way to estimate which of those molecules is more stable.

The way to know it theoretically is using electronic structure programmes.

  • $\begingroup$ I want to know how would you justify it theoretically. Because, we can't remember all the experimental data. $\endgroup$ Commented Apr 15, 2017 at 20:00
  • $\begingroup$ You can run the calculation in some electronic structure programme (for example Gaussian) $\endgroup$
    – user43021
    Commented Apr 15, 2017 at 20:02
  • $\begingroup$ Sorry, but I am still a student. And how would I determine if this question comes in an exam. $\endgroup$ Commented Apr 15, 2017 at 20:07
  • 1
    $\begingroup$ Sorry for this question, but are you sure it was not about Be$_2$ ? $\endgroup$
    – user43021
    Commented Apr 15, 2017 at 20:49
  • $\begingroup$ Yes, I am sure. $\endgroup$ Commented Apr 16, 2017 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.