Why are attraction forces long range forces and repulsion forces short range forces? When two atoms come close to each other in order to form a covalent bond, then attractive forces initially act on them (due to attraction between nuclei and the electrons). However, when they are very close to each other repulsive forces come into play (due to repulsion between nuclei-nuclei and electron-electron). My question is why do repulsive forces become prominent only when the atoms are very close to each other whereas attractive forces are prominent even when the distance is large (by large, I mean the distance relative to an atom, say 10 atomic diameters)?
 A: AFAIK the assumption is already wrong. Effective attractive forces as well as repulsive forces are usually short ranged due to charge neutrality/mutual shielding of nucleus and electrons. The qualitatively different exchange interaction that is responsible for covalent bonding is only significant in the intermediate range, so it does not make sense to talk about it being long-ranged or short-ranged in my opinion. So what remains as long-ranged candidates are only ions and polar molecules, but that is not a question of repulsion vs. attraction, because they can do both.
A: It's true that attractive and repulsive electrostatic forces are less strength due to the shielding of electrons and nuclei. But i think that the repulsive short interaction the OP is talking about is due to the Paul's exclusion principle. At short range the repulsion is due to the fact that only two electrons can occupy the same orbital with different spin. We can interpret this fact as a repulsion.
We can take the Lennard-Jones potential, that describe intermolecular forces, and we can see that we have a minimum of the potential indicating the length of the bond, but there is a minimum distance where the potential goes up that represent the repulsion due to the Pauli's exclusion principle.
A: In the chapter 7 of the Griffiths book on Quantum Mechanics, the formation of the H2 molecule is analized. The conclusion is a curve energy x distance between the nuclei that matches with your question.
As $F = -\frac{\partial E}{\partial r}$, the attraction force decreases gradually for distance greater than equilibrium, but repulsion increases less gradually when it is smaller.
The assumption is that the wave function of the electron is a function of $r$ in the case of a single atom. It is the well known ground state orbital of the H atom. For 2 nuclei, the wave functions related to the distance of each one are added and normalized.
I can't see an intuitive reason for why the molecule is more stable than 2 atoms (Even if it is only one electron for 2 protons! H2 ion is more stable than a H neutral atom + a proton).
But it is what 4 pages of QM calculations shows, and agrees with experience.
A: Great question. I suspect though that chemists may know more about it than physicists. I am not a big specialist either, but here is an intuitive explanation.
In atomic interactions there are several major factors at play: covalent bonding, shielding, and Van-der-Waals interaction.
Covalent bonding is always attractive (except for nobble gasses) and results from delocalization of electrons between different orbitals. It is very short ranged (the force decays exponentially with the distance), very strong and want to snap atoms together. What prevents them from collapsing, is the Coulomb repulsion between nuclei.
Nuclei are positively charged, and at large distances they are shielded from surroundings by a cloud of electrons. As the atoms get closer, the nuclei of both atoms get into each others electron cloud and start feeling the coulomb repulsion from each other. This prevents atoms from collapsing.
Van-der-Waals forces are the long-range forces you were talking about in your post. The force is weakly attractive at large distances and results from induced polarization by one neutral atom onto another. When electron in one atom is closer to the other atom compared to nuclei, the electron in the second atom prefers to be further away, and such induced dipole-dipole correlations will lower the energy of the system. The Van-der-Waals forces are weakly and long range and come into play when other forces do not matter.
Therefore, at very short distances the Coulomb repulsion dominates, a bit further the  covalent bonding takes over, but drops quickly, leaving only Van-der-Waals attraction as the main interaction.
A: A conceptual explanation without chemistry terminology:

*

*Matter (atoms) consists of fermions. They have to satisfy pauli exclusion principle, so they can't be at the same place. This leads to so-called hard-core interactions. More technically, it's rather that they cannot be in the same quantum state (wavefunction), and the closer the atoms are together, the more difficult it is for them to keep their wavefunctions orthogonal.


*The atoms are like dipoles. With proper orientation, they have lower energy in another electric field. The stronger the field, the lower the energy. The electric field induced by the other dipole gets stronger the closer they are moving together.
In short, atoms want to be close together without their wavefunctions overlapping too much.
A: I think the fundamental answer to the question is in fact "because we're here asking it".
If the repulsive force were not to grow faster that the attractive force when reducing distance, the atoms/particles would invariably collapse into each other (this happens with gravitation, above a threshold, during the collapse of white dwarfs into neutron stars for example).
If the repulsive force were not to decrease faster than the attractive force when increasing distance, then no stable bound states would happen ; no chemistry and no life, hence no question.
About the details of the "how" however, the other answers already cover it.
