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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?

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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.

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  • $\begingroup$ I want to know how would you justify it theoretically. Because, we can't remember all the experimental data. $\endgroup$ – Prashant Kumar Apr 15 '17 at 20:00
  • $\begingroup$ You can run the calculation in some electronic structure programme (for example Gaussian) $\endgroup$ – santimirandarp Apr 15 '17 at 20:02
  • $\begingroup$ Sorry, but I am still a student. And how would I determine if this question comes in an exam. $\endgroup$ – Prashant Kumar Apr 15 '17 at 20:07
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    $\begingroup$ Sorry for this question, but are you sure it was not about Be$_2$ ? $\endgroup$ – santimirandarp Apr 15 '17 at 20:49
  • $\begingroup$ Yes, I am sure. $\endgroup$ – Prashant Kumar Apr 16 '17 at 15:26

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