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Molecular potential energy in dependence with atomic distance for bonding orbital is given by following graph/picture. graph

We can see that at large distances force between atoms is attractive and potential energy drops to minimum which corresponds to bond energy and length. This part of the curve looks very similiar to interaction between charges of opposite sign at large enough distances (Coulomb interactions)

After that interatomic potential energy starts increasing and at some distance force becomes repulsive.

My question is why does molecular potential energy curve have that shape? Why is it that potential energy first drops to minimum and than starts increasing?

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The curve in your picture depicts the $H_2^+$ potential curves. Note that the ion consists of 3 particles so it can not be solved analytically. In the Born-Oppenheimer approximation you assume a constant separation $R$ of the two protons. This allows you to solve the resulting Schrödinger equation for a given value of $R$. The energy depicted in the diagram is then the expectation value of energy of the resulting state $E = \langle \psi | H |\psi\rangle$.

A different approach which is slightly more illustrative is to just take the hydrogen orbitals and just use a linear combination as an approximation for the orbital of the electron in the $H_2^+$ ion. As a first approximation assume then that the electron is in a 1s orbital at proton 1 or 2. This wavefunction would look like $$ |\psi^{\pm}\rangle \propto |1s\rangle_1 \pm |1s\rangle_2 $$ If you imagine the plus combination $\psi^{+}$ since the 1s orbitals from the two protons will show slight overlap there will be an increased likelihood of finding the electron inbetween the protons. This means that the coulomb repulsion between them is reduced and leads to a bound state. Note here that if you bring the protons close together this won't be the case anymore and lead to the expected repulsion you see in the diagram. On the otherhand if you go from the bound state and increase the distance the coulomb force will fall of and you will simply be left with the a hydrogen atom and a proton.

The minus $\psi^{-}$ wavefunction will lead to a decreased likelihood of finding the electron between the protons and as such won't reduce the energy of the system. This state is therefore antibonding, because it will naturally dissociate.

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  • $\begingroup$ Yes I think I got it, interaction between atoms is a sum of attractive and repulsive force. Attractive force is dominant at larger distances and its origin is randomly induced dipoles in atoms because of electron movement. This force causes chemical bonding. At shorter distances, repulsive force becomes significant as orbitals of atoms get closer to each other when Pauli exclusion principle doesn't allow orbitals to occupy same space. $\endgroup$
    – Dario Mirić
    Commented Aug 6, 2021 at 9:48
  • $\begingroup$ That's not entirely correct. Chemical bonds are typically ascribed to the overlap of electron wavefunctions (kovalent), metallic- or ionic-bonds (monopole-monople/Coulomb/$r^{-2}$). The van der Waals force acts between molecules, so it's more important when considering for example if something is a liquid or a solid at a given temperature. So you'd apply van der Waals potentials to gases etc. $\endgroup$
    – Wihtedeka
    Commented Aug 6, 2021 at 10:01
  • $\begingroup$ As you can see on the graph and question I posted I am asking about chemical bonding. Graph represents how molecular potential energy changes with distance in bonding and antibonding orbital in hydrogen molecule. I was told that attractive force between two hydrogen atoms in bonding orbital is of Van der Waals origin which was strange to me at first as I know Van der Waals is as you said intermolecular force. If attractive force in bonding orbital which causes chemical bonding is not Van der Waals, what is it? $\endgroup$
    – Dario Mirić
    Commented Aug 6, 2021 at 10:20
  • $\begingroup$ $H_2$ forms because the wavefunction of the electrons overlap i.e. a covalent bond is formed. For $H_2^+$ it's the same really, but I find it more illustrative to look at the one electron which is now at both nuclei. The probability density in the bound state will be higher inbetween the protons, which means the Coloumb energy is reduced and as such the energy is lowered. $\endgroup$
    – Wihtedeka
    Commented Aug 6, 2021 at 10:55
  • $\begingroup$ I edited my post to focus more on your original question, I apologize for not looking at the diagram more closely. $\endgroup$
    – Wihtedeka
    Commented Aug 6, 2021 at 12:11
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Its because it is the sum of two individual components. One component is attractive (coulombic) and minimizes energy by minimizing the interatomic separation. the other is repulsive and kicks in when the other orbitals get crowded against each other.

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  • $\begingroup$ Attractive force between atoms may not be coulombic (falls with r^2 at large enough distances) as atoms don't need to be charged. In this example we have two hydrogen atoms. I am not sure what is the origin of this attractive force between two hydrogen atoms? $\endgroup$
    – Dario Mirić
    Commented Aug 5, 2021 at 22:06
  • $\begingroup$ One electron attracts the other's proton. $\endgroup$
    – niels nielsen
    Commented Aug 6, 2021 at 3:33

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