Do intermolecular forces due to Pauli exclusion happen in reality?
Yes. Despite the fact that repulsive electromagnetic forces alone are more than enough to stop us from making a neutron star, Pauli exclusion is needed to explain why we don't fall through the floor. Short-distance effects like these are sensitive to the fact that atoms (of either integer or half-integer total spin) are made of fermions. Modelling an integer spin atom as a boson with no further constituents would only be appropriate at long distances.
Should the Pauli exclusion principle be considered a type of Van der Waals force?
No. The Van der Waals interaction should be defined as the asymptotic form of the potential at a large distance compared to the typical length scale (like the Bohr radius). The reason why this is qualitatively different from the potential at intermediate distances is not related to the Pauli exclusion principle.
Consider two Hydrogen atoms where protons are at $\textbf{r}_{A,B}$ and electrons are at $\textbf{r}_{1,2}$. The free and interaction Hamiltonians are
\begin{align}
H_0 &= \frac{p_1^2}{2m} + \frac{p_2^2}{2m} - \frac{e^2}{r_{1A}} - \frac{e^2}{r_{2B}} \\
H_\text{int} &= e^2 \left ( \frac{1}{r_{AB}} + \frac{1}{r_{12}} - \frac{1}{r_{1B}} - \frac{1}{r_{2A}} \right )
\end{align}
respectively. We can then Taylor expand the latter at large inter-atomic distance $r_{AB}$ to write
\begin{equation}
H_\text{int}/e^2 = \frac{\textbf{r}_{1A} \cdot \textbf{r}_{2B} - 3(\textbf{r}_{1A} \cdot \hat{\textbf{r}}_{AB})(\textbf{r}_{2B} \cdot \hat{\textbf{r}}_{AB})}{r_{AB}^3} + O(r_{AB}^{-5})
\end{equation}
which is the multipole expansion. Only the dipole term has been shown but the quadrupole would be easy to compute.
At typical distances, the first order energy shift $\left < \psi | H_\text{int} | \psi \right > = O(e^2)$ will dominate over higher orders. What about larger distances? Then there are two possible cases.
- If the dipole and/or quadrupole terms are non-zero, the potential continues to be well approximated by first-order perturbation theory.
- If these moments vanish, long distances will give a second order i.e. $O(e^4)$ result which is more important than the first order.
The two-neutral atom problem is symmetric enough that it falls under case 2. This derives the Van der Waals' contribution to the inter-atomic potential without explicitly referring to Pauli exclusion. Moreover, if you actually compute the ground state wavefunction $\left | \psi \right >$ to be used in the calculation above, you will find that it is symmetric. In other words, this part of the potential would be the same even if you pretended that the electrons were bosons. The only change you would have to make is symmetrizing the spin degrees of freedom instead of anti-symmetrizing them.
What happened on Wikipedia?
The person who wrote "typically contributions (1) and (4) are considered Van der Waals forces" didn't cite any sources. The next sentence even makes it sound like this was written in error. When saying "the totality of forces, including repulsion", it only makes sense to emphasize that repulsion is not being skipped if the previous point of view did skip it. But it didn't since (1) is a repulsive force. I will change this if no one argues with me or changes it first.
