This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.
If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.
Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.
Response to comment:
The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).
However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:
Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a qualitative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements.
So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.
The extreme stability of He-4.
Look at the decay modes of the the eights and they produce two alphas and if it is necessary convert a neutron/proton to a proton/neutron with an appropriate beta decay.
The production of two alphas is energetically favourable.
Li-6 and Li-7 lack nucleons to form two alphas.
Li-5 kicks out a proton and He-5 kicks out a neutron to form He-4.
More about 5 in LubošMotl's answer.
Best Answer
The main difference is gonna be the stability of the various isotopes. Most elements technically have a very large number of isotopes (carbon isotopes range from carbon 8 to carbon 22), but most of these have a very short half-life due to the poor stability of a number of neutrons too large (or too small). The list of isotopes will usually be either somewhat stable isotopes or, for more complete lists, experimentally detected isotopes.
The stability of those isotopes can be determined by the various nuclear models, such as the liquid drop model, nuclear shell model, etc etc. The liquid drop model is particularly useful to get a good intuition of the various factors contributing to nuclear stability.