Robin van Spaandonk wrote:
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).
So, we have three possible combinations:
1: A proton with one electron in the normal Bohr quantum state (i.e. the
usual hydrogen atom)
2. A proton with one or more electrons in Mills quantum states, which
would be a very stable negative Hy ion, which I would designate Hy-,
Hy--, etc, depending on the number of electrons.
3. A proton with one or more electrons in Mills quantum states and zero
or one electron in the usual Bohr quantum state, which would be a stable
molecule. In this case the other ion would have a positive charge equal
to the total number of electrons associated with the proton. If the
other ion is a proton, I would designate this as HyH, which would be
neutral. A corresponding compound with a Hy-- would be HyH2, or HyNa2
or HyCa, if other elements are used. Does this describe your understanding?
The presence of electrons in the Hy states would, I expect, alter the
energy level of the Bohr quantum states. As a result, a range of
properties would be expected depending on the energy level of electrons
in the other atom, how many Hy electrons were present, and what quantum
levels they occupied.
snip
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?
Oriani, R.A. Anomalous Heavy Atomic Masses Produced by Electrolysis.
in The Seventh International Conference on Cold Fusion. 1998. Vancouver,
Canada: ENECO, Inc., Salt Lake City, UT.
Fisher, J.C., Polyneutrons as agents for cold nuclear reactions.
Fusion Technol., 1992. 22: p. 511.
Regards,
Ed
snip
