[bessieboy521]

Although I understand (I think) most of what you're saying orgopete, I do not quite get the comparison of the density of the electrons around the nucleus of Iodide vs Fluorine. I get the vast difference in number, and the greater positive charge pulling ect.... My question is what role/effects do the valence electrons play differently than the descending closed shells. Does the radius of the over all shell have any role to play? Overall this is some dense information to absorb, especially since I've always been taught that only the valence electrons really matter, with exception to the "lightly touched" idea of sheer electron density being able to stabilize a true charge just by being in proximity. I'm gonna pop out some books on the subject before I post again, I just don't quite get what it matters if in comparison ratio-wise iodide's electrons are held closer to the nucleus when they're in such a further out shell.... The true distance of bonding electrons is further away from the nucleus. Also what proportion does the positive field increase with proton density increase? Maybe the distance and positive increase at different rates??? allowing for a still predictable, however, less intuitive bond strength as electron mass increases proportionally to proton mass increase?

[orgopete]

I need to take a step back on electron density. This is what I had thought. If you take the bulk density of the halogens (solid or liquid forms) and adjust for the number of electrons, then it appears as though the electron density increases as you move down the periodic table. However, if you take an atomic radius and calculate the volume per electron, then the density is reasonably consistent. It doesn't matter whether it is fluorine or iodine.

In the atomic model that I have been thinking about, the atomic properties emanate rather more simply from Coulomb's Law. As the nuclear charge is increased, the proton field grows proportionally with the charge. As the distance increases, the field decreases with the inverse square of the distance. In this model, distance is more important than (formal) charge. That is, as one moves across the periodic table from C to F, the bonds become shorter, consistent with the nuclear charge effect. NH3 is more basic than H2O (or HF) because the electrons of oxygen are closer to the nucleus of oxygen. Ammonia is also more basic than chloride even though chloride has an excess of electrons. It isn't the net charge that matters because the charge of electrons is constant, minus one. The difference is the distance or rather the field of the electrons. The further they are from the nuclear proton field, the greater their attraction for protons. In this model, electrons have microscopic fields consistent with the orbitals they exist in. Their fields are dependent on the distance from the nucleus.

As a consequence, in this model, compounds like HF or H2O can form intermolecular bonds consistent with the tetrahedral orientation of their orbitals. Compounds like NaF are in cubic close packing crystals in which no orbital orientation is maintained and is consistent with a larger separation between the ions. This greater distance is also consistent with weak ionic bonds. In this case, weak and ionic are redundant. Bonds are the result of a Coulombic attraction and weak is a function of distance. When we refer to ionic bonds, we mean weak bonds. A greater separation seems rational as sodium and fluoride have the same number of electrons surrounding each nucleus. A Coulombic repulsion at short distances is present. However, a differential in charges creates an ionic attraction which in this case is qualitatively different from the short range attraction making covalent bonds.