To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.
The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.
When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.
So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.
To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.
EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
$\hspace{75px}$
.
This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.
The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).
By the way, this image is from a very helpful article which should also be helpful for understanding this issue.
I am replying to this because you seem to be a student, and not so clear on the statements.
I have read that during fission and fusion processes, there is some kind of equilibrium between the single nucleus and the disintegration products, so they are constantly being converted into each other.
" I have heard" is not enough, you should give a quote or a link. The statement is wrong. During fission a large nucleus breaks up because its component parts are more stable and the total energy balance is positive.
Look at this binding energy per nucleon in a nucleus curve:
Below the top of the curve, putting together more nucleons gives energy, from the top of the curve to the right, removing a nucleon releases energy.
Furthermore, the energy after the fission of a single parent nucleus into two daughter nuclei is less than the energy required to fuse the two nuclei back together again.
This also is an out of context quote, you should give a link. According to the binding energy curve per nucleon it is a wrong statement
So if there is an equilibrium, how is the fusion energy achieved?
There is no equilibrium in laboratory conditions. Even in the center of the sun, more nuclei fuse than separate, otherwise the sun would non be the source of energy it is.
The fusion energy is released because two deuteron particles tied into one nucleus will have to release energy, as seen in the binding energy curve.
The fission energy happens because heavy nuclei are metastable in the sense that they could be pushed to break up into smaller parts more tightly bound releasing the binding energy of the large system.
Where did the extra energy come from?
The extra energy for fusion comes from the original existence of hydrogen helium atoms.
This happened during the Big Bang, according to the present model of creation of the universe. Atoms up to Fe in the binding energy curve were created in nucleosynthesis time. The heavier atoms were given energy by large explosions of heavy stars, like supernovae explosions, during the early universe days. All matter as we see it now was given its energy content at those early times , from the original impulse that generated the Big Bang.
Best Answer
It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions.
The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 MeV). The total mass of the nucleus is the sum of the masses of its nucleons minus the binding energy, so a tighter binding means a smaller mass. The binding energy of a nucleus can be computed with Weizsäcker's semi-empirical mass formula:
In this plot, you can see the binding energy per nucleon. For isotopes lighter than iron, nuclei can become more stable if they fuse (they move to the right in the plot), and they liberate energy (lower their mass) in the process. In the other hand, fission occurs in heavier nuclei (they go to the left), and also liberates energy. But in both cases the maximum energy liberated is given by the above plot: you can only get in total, no matter what intermediate steps you make, is the 9 MeV/nucleon difference between ${}^{56}$Fe and ${}^1$H, wich corresponds to fuse 26 protons and 30 neutrons together to form an iron nucleus. And obviouslly, iron nucleii do have mass!