Why do two hydrogen atoms bond? I was wondering (with my limited classical physics knowledge) why two hydrogen atoms tend naturally to a bonding configuration, I mean, given two hydrogen atoms with zero relative velocity between eachoter, and given that a monoatomic hydrogen is representable as a positive $+Q$ charge sphere surrounded by a spherically symmetric negatively charged layer with spherically symmetric charge density and overall $-Q$ charge, if so, for Gauss theorem, there should be a net zero electric field in the entire space, given that biatomic molecule configuaration is more stable than two monoatomic infinitely distant H atoms configuration, there should be a net force that drives towards the more stable configuration, in contrast with the conclusions that I get starting with my hypothesized model. How should I change the way I think about the monoatomic hydrogen spatial charge configuration, in order to give sense to what happens experimentally?
 A: With two hydrogen atoms you have two protons and two electrons. The question amounts to asking: what arrangement of two protons and two electrons has the lowest energy? We need quantum mechanics to answer, but a knowledge of classical mechanics and electromagnetism can give some useful pointers.
First it is clear that if the electrons are near the protons then they can have lower electrostatic potential energy. On a classical model, they would spiral right in and hit the protons. Quantum mechanics says this does not happen, because as any electron gets more tightly confined it also gets more kinetic energy, so there is a trade-off: as the electron is bound more tightly to a proton its potential energy goes down but its kinetic energy goes up. Somewhere there is a compromise (which for electrons and protons happens at distances of order $5 \times 10^{-11}$ m.)
To estimate this you can use the Heisenberg Uncertainty Principle. If the position is confined within a distance $\Delta x$ then the momentum is uncertain by at least
$$
\Delta p \simeq \hbar / \Delta x
$$
so the kinetic energy is at least
$$
\frac{\Delta p^2}{2m} \simeq \frac{\hbar^2}{2 m \Delta x^2}.
$$
For a single electron bound to a proton the total energy is roughly
$$
\frac{\hbar^2}{2 m \Delta x^2} - \frac{e^2}{4\pi \epsilon_0 \Delta x}.
$$
and the electron settles at the value of $\Delta x$ which minimises this.
Now let's think about the molecule. The protons won't get very close because they repel each other, but if they are quite close to each other then each electron can be attracted to both, which makes its potential energy lower. Meanwhile each electron repels the other, which roughly balances the effect of the further proton to which each electron is attracted. But by forming a single molecule rather than two atoms, each electron can move around both the protons, which gives it more space to move in while still having a low potential energy. This added space leads to a lower $\Delta p$ and so a lower kinetic energy, hence overall this configuration has lower energy than two separate atoms.
A: If the two atoms are many diameters apart, they mainly attract via van der Waals forces.
When they reach proximity and their electron orbits overlap appreciably, more energy can be gained through spin interactions which eventually makes both electrons occupy the same orbital.
A: Consider the reaction:
$$\text{H.}+\text{.H}\to \text{H-H}$$
i.e. the formation of dihydrogen from monoatomic hydrogen, to be a chemical reaction, then like all reactions that proceed 'spontaneously' it is exo-energetic, meaning that:
if the energy of the state on the left is $E_L$ and the energy of the state on the right is $E_R$, then:
$$\Delta E=E_R-E_L<0$$

Very very simply put, a dihydrogen molecule ($\text{H}_2$) forms when the wavefunctions $\Psi$ of the individual hydrogen atoms positively interfere, that is form a bonding molecular orbital (BMO):

When negative interference occurs it results in anti-bonding (no bonding)
A: The electron of an (ideal, nonrelativistic) hydrogen atom has nonzero probability density to be found throughout all space. So there is no such thing as "outside" a hydrogen atom, where you would expect to find zero electric field (which requires the further assumption of spherical symmetry, which is a good enough approximation for now). In your model, the negatively charged layer should not contain the full -1 e charge of the electron.
Of course, the probability of the electron being more than a couple Bohr radii (what is usually quoted as the radius of hydrogen) from the atom is minuscule, and so the electric field far from the atom goes to zero. But the bond length in H2 is .74 angstrom, while the Bohr radius is $a_0\approx.54\,\textrm{angstrom}.$ So two hydrogen atoms placed a bond length apart are each in the region of strong electric field around the other atom, and so they should definitely interact somehow. (This does not explain the actual formation of the bond or its properties, which requires more serious quantum mechanics, just argues that we should expect something to happen).
Explicitly, for an isolated hydrogen atom, the probability $P(R<xa_0)$ for the electron to be found within $x$ Bohr radii of the center is $$P(R<xa_0)=1-e^{-2x}(1+2x+2x^2).$$
So the expectation value for the amount of charge $\bar Q(r)$ within the sphere of radius $r$ is (with $q$ the elementary charge) $$\bar Q(r)=q-qP(R<r),\\\bar Q(xa_0)=qe^{-2x}(1+2x+2x^2).$$
And this produces an average electric field (directed radially inwards) of magnitude $$\bar E(r)=\frac{\bar Q(r)}{4\pi\varepsilon_0r^2}\\\bar E(xa_0)=\frac{qe^{-2x}(1+2x+2x^2)}{4\pi\varepsilon_0{a_0}^2x^2}\approx\left(5.14\cdot10^{11}\,\frac{\mathrm{N}}{\mathrm{C}}\right)e^{-2x}\left(2+\frac2x+\frac1{x^2}\right).$$
That constant ($5.14\cdot10^{11}\,\frac{\mathrm{N}}{\mathrm{C}}$) is the atomic unit for electric field strength, commonly used in computational chemistry calculations. The term multiplying it is (by design) on the order of 1 for radii near to the Bohr radius, so it sets the scale for the electric field strengths in the chemically relevant region of a hydrogen atom. E.g. at the previously quoted bond length $1.4a_0,$ $\bar E$ is $1.23\cdot10^{11}\frac{\mathrm{N}}{\mathrm{C}}.$ Certainly not zero!
A: When thinking classicaly, even taking the electrons as a cloud, each atom has a finite size, and outside it any interaction would need an electric field. But, outside a spherical symmetric neutral body the electric field is zero. So, there is no interaction between them in this model.
One difference in the QM approach is that the interaction here is not through electric field between separated bodies, but between fermionic fields. When we talk of the addition of wave functions, it is a kind of interaction different of the electric field between charges.
In the Griffiths' book on QM, there is a detailed calculation of this issue in chapter 7.3. He uses 2 nuclei and only one electron to simplify the problem, but the graph illustrating the dependence of the potential energy from the distance between protons is interesting. The gradient of this potential (that is the force in modulus), is still meaningful far from the Bohr radius, that is considered the atom size.
