First of all, the Hamiltonian in question is defined on $L^2(\mathbb R^3)$, not on $L^2(\mathbb R)$. This is important because in the one-dimensional case the potential would have a non-integrable singularity which complicates things seriously. On $L^2(\mathbb R^3)$, the operator, defined as a closure from $C_0^\infty$, is selfadjoint. This is proved, for example, in the book by

 - T. Kato, *Perturbation theory for linear operators*, Springer, 1966. 

Thus the residual spectrum is impossible. A rigorous calculation of eigenvalues and eigenfunctions can be found in the books

- L. D. Faddeev and O. A. Yakubovskii, *Lectures on quantum mechanics for mathematics students*. American Mathematical Society, 2009;

 - L. A. Takhtajan, *Quantum mechanics for Mathematicians*, American Mathematical Society, 2008.

The point 0 is an accumulation point of negative eigenvalues and the limit point of continuous spectrum, thus it belongs to the essential spectrum.