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.
In a nutshell, atoms decay because they're unstable and radioactive.
Ununoctium (or Oganesson) has an atomic number of 118. That means that there are 118 protons in the nucleus of one atom of Oganesson, and that isn't including the number of neutrons in the nucleus. We'll look at the most stable isotope of Oganesson, $\mathrm{{}^{294}Og}$. The 294 means that there are 294 nucleons, or a total of 294 protons and neutrons in the nucleus. Now, the largest stable isotope of an element known is $\mathrm{{}^{208}Pb}$, or lead-208.
Beyond that many nucleons, the strong nuclear force begins to have trouble holding all those nucleons together. See, normally, we'd think of the nucleus as impossible because the protons (all having a positive charge) would repel each other, because like charges repel. That's the electromagnetic force. But scientists discovered another force, called the strong nuclear force. The strong nuclear force is many times stronger than the electromagnetic force (there's a reason it's called the strong force) but it only operates over very, very small distances. Beyond those distances, the nucleus starts to fall apart. Oganesson and Uranium atoms are both large enough that the strong force can't hold them together anymore.
So now we know why the atoms are unstable and decay (note that there are more complications to this, but this is the general overview of why). But why the difference in decay time? First, let me address one misconception. Quantum mechanics says that we don't know exactly when an atom will decay, or if it will at all, but for a collection of atoms, we can measure the speed of decay in what's called an element's half-life. It's the time required for the body of atoms to be cut in half.
So, to go back to decay time, it's related (as you might expect) again to the size of the nucleus. Generally, isotopes with an atomic number above 101 have a half-life of under a day, and $\mathrm{{}^{294}Og}$ definitely fits that description. (The one exception here is dubnium-268.) No elements with atomic numbers above 82 have stable isotopes. Uranium's atomic number is 92, so it is radioactive, but decays much more slowly than Oganessson for the simple reason that it is smaller.
Interestingly enough, because of reasons not yet completely understood, there may be a sort of "island" of increased stability around atomic numbers 110 to 114. Oganesson is somewhat close to this island, and it's half-life is longer than some predicted values, lending some credibility to the concept. The idea is that elements with a number of nucleons such that they can be arranged into complete shells within the atomic nucleus have a higher average binding energy per nucleon and can therefore be more stable. You can read more about this here and here.
Hope this helps!
Best Answer
Protons are positively charged, and neutrons are neutral, so large nuclei are highly positively charged. A postively charged sphere will energetically prefer to break up into two separate charged droplets which move far apart, this reduces the electrostatic energy, since the electrostatic field does work during this process.
This thing, spontaneous fission, is usually phase-space unlikely, since you need to have a large chunk of the nucleus tunnel away from another large chunk, and it's unlikely for all those particles to tunnel out together. But at large atomic numbers, you are unstable even just to shooting out an alpha-particle, and this doesn't require a conspiracy, so large Z nuclei are alpha unstable, usually with long half-lives.
The positive charge on nuclei puts a limit to the stable ones. The reason is simply that the electrostatic force is long range, while the cohesion force is short range. The same phenomenon causes the instability of water droplets, so that if you charge one up, it will break into a fine mist. The cohesion of the droplets is local, while the electrostatic repulsion is long range.
The scale at which you get a fission instability directly can be estimated from surface-tension considerations. If you break a sphere into two adjacent spheres of same total volume, the radius is reduced by the cube-root of two, so that the surface area is decreases by the square of this, and you multiply by 2 (since there are two spheres) so the net factor is the cube-root of 2, which is around 1.3. So the extra surface tension energy is increased by a factor of 1.3, or 30%.
But in separating the two spheres, you have taken one ball of charge, with an energy of $Q^2\over R$ and separated it into two adjacent balls of reduced radius and half the charge. Adding up the electrostatic energy, it is about 80% of the original electrostatic energy in the single sphere.
So spontaneous droplet fission will happen when you have a charged ball for which 30% of the surface tension energy is less than 20% of the charge energy. Since charge goes up almost as the volume (not quite, but close) while the surface tension goes up as the area, there is a crossover, and charged droplets will spontaneously separate when they are too big.
The surface tension can be found from the binding energy curve of nuclei, and these simple considerations limit stable nuclear size to about that of Uranium. The U nucleus can spontaneously fission at an extremely low rate, but the transuranics become progressively more unstable because their electrostatic energy is increasing as the volume to a power greater than 2/3, while their surface tension energy is increasing as the surface area, which grows as the 2/3 power of the volume.
These considerations, in much more sophisticated form, are due to Niels Bohr in the seminal liquid drop model of the 1940s. This model explained the nuclear binding energy curve quantitatively, and accounted well for fission phenomena. The only major thing left out of this was the shell model and magic numbers, which was supplied by Mayer.