# Identical bond angles [closed]

Is it possible to incorrectly predict that two bond angles are different from van der Waals repulsions alone?

I'm thinking that because bond lengths differ when different atoms are involved that we could, at least theoretically, have two identical bond angles involving different molecules.

For example, say we are considering the bond angles of molecule $\ce{AXY}$ and $\ce{BZA}$.

X and Y are highly electron dense and also big. The first suggests a big bond angle; the second suggests a long bond length, which in turn suggests the possibility of a smaller bond angle.

Now consider $\ce{BZA}$. Z and A are not electron dense but also large. The first suggests a small bond angle. The second suggests a long bond length and thus a smaller bond angle.

Is it still possible for the two to have identical bond angles?

## closed as unclear what you're asking by Pritt Balagopal, Tyberius, aventurin, airhuff, Todd MinehardtMar 26 '18 at 16:26

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Is "A" meant to be the same in both molecules? Are "A" and "B" the central atoms in AXY and BZA respectively? Can they be linear? – ron Jul 18 '14 at 22:27
• I think the answer to this question is, yes, it is POSSIBLE, at least in non-ideal situations (i.e. non-gas phase, non-vacuum, non-0K, etc.). In ideal situations? I would err on the side of saying that it was POSSIBLE but I don't know of a concrete example that I could provide. Of course, in reality, molecules are not stationary states and are constantly in motion (vibrations, rotations, and/or translations) so it is sort of a moot point from this perspective. – LordStryker Sep 12 '14 at 12:42