H2Te dipole and electronegativities of H and Te (high school level chemistry)
Information in this question is all over the place and often out of context. Let me get a few things straight:


Van der Waals forces, also known as London interactions, occur between every pair of molecules regardless of polarity. They are the only intermolecular force observed for non-polar molecules such as $\ce{I2}$. They are explained by induced and spontaneous dipoles at a molecular level and femtosecond timescale.
Dipole interactions are present between macroscopic dipoles and account for the boiling point difference between e.g. tetrahydrofurane and cyclopentane ($\Delta \vartheta_\mathrm{b} \approx 15~\mathrm{^\circ C}$). For those, you need a notable difference in electronegativity resulting in an overall dipole (so $\ce{CO2}$ is not a dipole because the dipole moments of both $\ce{C=O}$ bonds cancel each other out). For strong dipoles, these interactions are orders of magnitude larger than the underlying van der Waals forces.
Electronegativities are, of course, atomic properties (and also depend on things like oxidation state if one is truly accurate). For no two different elements is the electronegativity difference ever zero. Every $\ce{X-Z}$ bond is always polarised ever so slightly in one of the two directions. Sometimes, as in the case of $\ce{NCl3}$, this is best determined experimentally. ($\ce{N-Cl}$ bonds in $\ce{NCl3}$ are polarised towards nitrogen.)
However, only bonds which are polarised to a certain extent (usually a cutoff value of $\approx 0.4$) are called polar on a high school level as per convention.
$\ce{H2Te}$ has non-polar bonds as per convention of the fourth bullet point. Therefore, you should consider it a non-dipole and hence only consider van der Waals forces/London dispersion forces. (DavePhD’s source is likely correct, but the dipole moment is neglegible at high school level.)
I would love to see a link to the pages that say $\ce{H2Te}$ be notably polar/display dipole dipole interactions. (Compare the $0.2~\mathrm{D}$ to the $1.78~\mathrm{D}$ dipole moment of ethyl acetate.) Calling van der Waals forces dipole interactions is usually considered wrong (although they build up upon the same principles: electrostatic interaction).




I tried to stick to a high-school level as well as possible. Hence a number of simplifications in this answer. At research levels, every molecule without improper rotation (of which inversion and symmetry planes are a special case) should be considered to have a non-zero dipole moment as DavePhD’s answer states. Thus, every molecule of that type will display dipole-dipole interactions however weak they may be.