In reply to Edmund Storms's message of Sat, 19 Nov 2005 15:19:06 -0700: Hi, [snip] >> Why? In a "perfect" ionic compound, solidity results from the >> binding energy of positive and negative ions. IOW the attractive >> force between ions of opposite charge pulls the ensemble together. >> There is no real need for electrons to be interchanged at a local >> level as would be the case in a covalent bond. Granted, with >> normal substances there is more often a "polar" bond than a pure >> ionic bond. In short, the hyh "bond" with a positive ion would be >> the most extreme ionic bond imaginable. You may calculate the >> degree of electron sharing if you wish, but given an ionization >> potential of around 70 eV for hyh[n=1/16], I think you will find >> that it is so negligible as to be immeasurable. > >A "normal" ionic compound results from electrons being moved from one >atom to the other. For example, in making NaCl, the electron moves from >the Na atom to reside for most of the time at the Cl atom. This is >different from the situation with Hy, which I'm trying to understand, so >be patient. When Hy is involved, the situation involves a preionized >atom, so to speak, which as a negative charge that can not be removed by >chemical interaction. Consequently for it to form a bond, the other >atom must also be preionized to form a positive ion.
Essentially correct, but be careful not to confuse Hy (neutral) with what I have been designating "hyh" (Hydrinohydride) which carries a negative charge (or Hy- if you prefer that notation). >Of course, this is >easily done. You would propose that if Hy were bubbled through a >solution of Na+ Cl- in H2O, a compound should form having the formula >NaHy. >In a similar fashion, Hy bubbled through an acid should result in >HHy. This would then be the neutral dihydrino molecule. Some time back I asked Mills directly whether he thought that Hy + proton -> Hy2+ and he said that he thought it would. >In addition to ionic bonding, both compounds have the potential for >some covalent bonding as electrons from the "normal" atom briefly occupy >"normal" energy levels in the Hy structure. I suggest this addition of >pure ionic and covalent components makes the bond exceptionally strong, >not just the ionic part. Only experiment will tell. >I suggest the ionic part can not be any >stronger than the stability of the positive ion respect to regaining its >electron from other sources in the environment. Does this fit with your >understanding? I find myself forced to agree, if the Hyh is not buried within the shell structure of the positive ion. Furthermore, the statement you make must also be true of all ordinary ionic compounds. So, while reducing the positive ion with a free electron determines the upper limit of the bond strength[1], the Hyh compounds should nevertheless be stronger than other ionic compounds, because of the small size of the Hyh. [snip] >> Bonds are formed by two forces. Electrostatic, and magnetic. Pure >> covalent bonds are pure magnetic bonds. Pure Ionic bonds are pure >> electrostatic bonds. Polar bonds are a mixture of the two. In the >> case of hyh, because the second electron is so tightly bound, the >> bond is the purest electrostatic bond of all compounds. IMO. > >No, I don't agree. The positive ion can get an electron from many >sources other than the Hy. Consequently, the bond is no more stable >than any other pure ionic bond. But *no* ionic bond is stronger than the upper limit implied by neutralizing the positive ion. In short, sorted in order of "increasing bond strength", we have: 1) Normal ionic bond. 2) Hyh ionic bond. 3) Neutralization energy of positive ion by free electron(s). [snip] >This is an interesting possibility. The question is, can a highly >reduced Hy actually act like a neutron that is stuck to an atom outside >of the nucleus? That's the general concept, though it would essentially be a negatively charged "neutron", effectively reducing the atomic number by 1. This is because it would orbit the nucleus inside the K shell, so from the point of view of the electrons, the nuclear charge would be reduced by 1. (Actually I'm guessing here. The size of the hydrino is still "up in the air" somewhat as far as I'm concerned). Besides it will depend on which Hy- combines with which positive ion. >On the other hand, I would expect such a structure to >be so close to being neutral that interaction with the electron quantum >states would not be possible. This seems to be an idea worth exploring. > Would this explain the Fisher-Oriani super heavy carbon? Never heard of it. Reference? [snip] >> I doubt it, because, while the hyh may not be able to leak away, >> the electron it replaces can leak away. This would still leave a >> neutral charge over all. > >Yes, over all. But immediately at the surface of the metal, the >positive changes left behind would generate a voltage gradient. Would >this gradient be large enough to do something unusual? I suspect that here you may be confusing Hy with Hy-. If not, then I'm confused. [snip] >Yes, but I'm trying to show that my high is equivalent 22 and my low is >equivalent to 1. [snip] Ok, I can work with that. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
