So I come from the math community (so please disregard my ignorance on the topic).

If I have a water molecule and write the potential energy surface associated with it, my understanding is that I will find that the atoms configuration minimizes the potential energy function, so it will be a critical point of the potential energy surface.

My question is: suppose I have the building blocks of water, i.e 2 hydrogens and an oxygen, if it were the case by some magic, I added energy to the system that resulted the correct bond length and angles that correspond to water. Does that mean I have the water? That is to say ,geometrically, that I have taken a trajectory from the original potential minimum of just the original building blocks to a new potential minimum by adding energy. Since this new configuration has the correct bond length and angles, and potential energy, does that mean this atom configuration must indeed be water?

Thank you.

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    $\begingroup$ A water molecule (in its equilibrium geometry) has far less energy than three isolated atoms. So you should be taking energy away, not adding it. Now, if you’re saying that you hurl three atoms together and they coincidentally happen to form water, then that’s ok, but your water molecule would need to have a lot of kinetic energy to ensure that energy is conserved. In practice it would likely fall apart right after that. I know I’m not addressing your actual question, but I am slightly confused by why you’re adding energy. $\endgroup$ – orthocresol Dec 30 '20 at 1:41
  • $\begingroup$ hey @orthocresol thanks for answering me - I was thinking of the classical experiment where you apply a flame to hydrogen and oxygen to form water, for example. In that instance I would think you would be adding energy to the system, and for energy to be conserved, the hydrogen explodes causing water. $\endgroup$ – Vogtster Dec 30 '20 at 1:55
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    $\begingroup$ Now, I think, if I am understanding your question correctly, that the answer is "yes". The water molecule is defined precisely by it having three atoms in that particular geometry, so "having a molecule of water" $\Leftrightarrow$ "having two hydrogen atoms and an oxygen atom in that particular geometry". I don't know if any chemistry book would explicitly state that, but that is definitely the understanding amongst chemists. $\endgroup$ – orthocresol Dec 30 '20 at 2:04
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    $\begingroup$ It is not quite as simple. Geometry plus electron density defines the molecule; geometry alone doesn't. If your orbitals aren't set right, then the molecule with correct bond lengths and angles might be in a forced and unstable state, which will fall apart the moment you turn off the magic. $\endgroup$ – Ivan Neretin Dec 30 '20 at 5:58
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    $\begingroup$ The point is that by some magic makes everything possible. But in this case you won't be able to minimise anything. It is just that you, by chance, putted the right energy to the undefined dots. If you use real atoms, no. It is a bit hard to follow your reasoning. See Ivan Neretin above, too. If they are real H and O than it can be a water so excited that it does not exist but seen in a geometrical minimum. $\endgroup$ – Alchimista Dec 30 '20 at 8:21

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