In a previous paper I promised to diagram the Oxygen molecule, to show why the Oxygen nucleus bonds to itself to make O2 and O3, and then fails to bond to itself anymore. Although Oxygen doesn't bond to itself above the triatomic level, it does bond easily with many other nuclei. This mystery has never really been solved until now.
The blue disks are alphas, so they have two protons in each. The black disks are protons. You need to have read my previous papers on nucleus building to understand my method. Each disk is understood to have a hole in its center, like a CD. We are looking at the CD's from on edge, since they are easier to diagram that way. Since the blue disks are double, their holes can accept two protons each. And so the bond in the center of the Oxygen molecule is made with two protons. I could have drawn those two black disks as one blue disk.
It looks like that Oxygen molecule should be able to bond with other Oxygen molecules in the same way, creating a solid. But it can't. It can't because of the way charge is channeled through the molecule. To see this, we have to study each alpha separately.
Previously I simplified the field by telling you that the protons took in charge through the holes and emitted it radially, keeping charge from passing between the two protons. But we have seen in my recent papers that isn't strictly true. It is now time for us to hit the next stage of rigor in our diagramming.
In the larger elements, we have seen that the nucleus channels charge in several channels, not just one. Even here in the Helium nucleus, we have crossing channels. The black disks channel charge in the x,y plane, flinging it off like a lawn sprinkler. But charge is also moving in the z plane, going from one hole to the other. Since I labeled the spin as clockwise, the z motion is down through the holes (right hand rule).
This pole acts like a pole in other ways, since, as we saw in the diagrams of the larger nuclei, the nucleus tends to spin about this vertical axis (it doesn't have to be vertical, of course, I have just chosen to diagram it that way). The carousel level or c-level of the nucleus spins about this axis, for a start. The nucleus may also spin end-over-end in some situations, but the carousel level spin is the defining spin of the nucleus, due to the configuration of protons.
This means that the alpha acts like a dipole, with potential differences at each pole. Even though the alpha has one proton at each “end”, and therefore no potential difference by my method in a previous paper, it does have a small potential difference just due to the spin of the protons. They are spinning one way and not the other, you see.
Without this dipole motion, and the movement of charge through the alpha, we would instead get charge trapped in the alpha interior. We don't want that, since it doesn't match data. We need charge to move on through, since this creates the channel and protects the nucleus from charge dissolution.
But if we study this charge motion even more closely, we see that the two protons have to be spinning the same direction. This is only to be expected, since opposite spinning protons would interfere with eachother, pull charge into the alpha, and blast the two neutrons out.
Well, if we start stacking alphas to create large elements, we start stacking a lot of same spins in a small area. These spins aren't additive, but they do tend to create resonances. We have seen before that the three-stack of alphas is the limit. Yes, we give Oxygen a proton on each end here not only to create the known reactivity (although it does that, too): we have to put the 7th and 8th proton out there, because we cannot stack four alphas to make 8. Why not? I am not completely sure, but I have found it to be true by diagramming a lot of nuclei. The nuclei can't be diagrammed using a four-alpha base. It may have to do with the neutrons in the alphas. I have shown that neutrons can come in 16 forms (including anti-neutrons). Each form will work equally well to act as a post keeping the protons from turning in the alpha, but once we start spinning the alpha, these different neutrons will act a bit differently. Due to internal spins, these neutrons will have different inertias.
This means that unless the alphas have the same neutrons, they will not spin the same. The likelihood of having neutrons with the same inertia decreases the more alphas we stack. Apparently this likelihood goes to zero with four alphas.
Whatever is the cause of that limit, I assume it is the same limit that prevents Oxygen from linking above O3. As you see from the first diagram above, we have the same general problem, since Oxygen links in a long line, where all the alphas are spinning in the same direction. The double links between the nuclei seem to help stabilize the molecule up to O3, but after that the instabilities become too pronounced, and the links break. This is why Oxygen is a gas at most temperatures.
[You can now go to my newest paper on the Hydrogen molecule, to see how charge bonding replaces covalent bonding between atoms.]