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29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed

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  29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-05-23 22:03 -0700
    Re: 29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed Uncle Hal <derp@hotmail.com> - 2018-05-23 23:08 -0700
      Re: 29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed benj <benj@nobody.net> - 2018-05-24 12:04 -0400
    29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-05-24 00:24 -0700
      Re: 29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-05-25 20:20 +0000

#686868 — 29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed

29-Maximum Electricity Coil theory// Internal parts of atoms and how the chemical bond is formed


CHEM ONE, 2nd ed., Waser, Trueblood, Knobler, 1980, page 327, Figure 16-3 Schematic Energy-Level Diagram for a Many-Electron Atom

So let me try to duplicate what that Figure 16-3 looks like

                                      5d                   4f
                                   - - - - -             - - - - - - -
   __6s
              _ _ _5p
                                     4d
                                  - - - - -
   __5s
              _ _ _4p
                                     3d
                                  - - - - -
   __4s
              _ _ _3p


   __3s

 
              _ _ _2p


 __2s





 __1s

y-axis low to high energy

x-axis  L = 0    1,    2,      3

Now the reason I went ahead and drew the above of the Aufbau, is that it is a 2dimensional diagram of what I believe is a 3 dimensional torus. I believe the shape of each atom is a torus, where the highest energy level is the largest overall torus and all others are nested inside one another. So for example neon is 1s^2 2s^2 2p^6 meaning that the 2p^6 is the biggest and outer torus of neon, having the 1s nested inside the 2s toruses, and these two toruses nested inside the 2p where the 2p is the coil of neon and the bar-magnet of neon is the two nested toruses of 1s and 2s, going around the bigger torus as a bar magnet and creating new magnetic monopoles.

Now this provides us with some geometry figure when atoms bond into molecules, take for example O2. We have two oxygen atoms of 1s^2 2s^2 2p^4 and when they bond together we have 1s^2 2s^2 2p^6 + two valence electrons. Where do those two valence electrons go? They could go to forming a 3s torus and having the O to O bond be a 1s^2 2s^2 2p^6 torus inside that of a 3s torus. So in this situation you have for O2 bonding

two 1s toruses nested inside
two 2s torusus nested inside 
one 2p^6 torus nested inside
one 3s torus formed from the two left over muons

Or, alternatively, the two left over muons with there attendent protons can be a third 2s torus, all of which these toruses are nested inside one another.

Now, does this work with the experiment result of Rutherford, Geiger, Marsden? A alpha particle is the same as a helium atom in New Physics, for the .5MeV particle is not the electron but a surface particle-- a magnetic monopole, so for all purposes, alpha particles are still helium atoms. And those that bounced off gold atoms and can hurdling back in the experiment, were hitting not some nucleus of a atom, but rather, they were hitting up against the toruses that keep traveling around in circles inside the largest torus. They, the alpha particles were hitting the bar magnets inside the largest torus as a coil with windings. Now the shape of the O2 molecule with its extra two valence muon would make either a larger 3s torus or make a 2p^6 torus with one more monopole inside the big torus.

Experiments should be able to decide the question. But modern physics is already moving into this direction with what are called "halo neutrons" outside the nucleus. And I think by re-doing the old gold foil Rutherford experiment with other target atoms and molecules, we should be able to confirm the above.

And slowly but surely I am getting back to the question of what does CO2 look like and whether it has at least to very different isomers. So, CO2 would have 4 extra valence muons, once the 2p^6 is filled. And where do these 4 extra muons with attendant protons go? Do they make a 3s torus then start making a 3p torus? Or do the four extra muons with attendant protons go back to making two more 1s and 2s toruses?

Can molecules have duplicate 1s and 2s orbitals? Maybe atoms by themselves cannot duplicate a 1s and 2s , but these are molecules of CO2.

If they can duplicate, then I see at least 4 different isomers for CO2,

one being duplicates of 1s and 2s
one being a duplicate of 2s and a new 3s
one being a duplicate of 1s and a new 3s
one being a new 3s and starting 3p toruses

And then we can say that animal CO2 is probably the lowest energy CO2 molecule.

Very hopeful, very reassuring because if animal CO2 is what plants require and cannot use fire CO2 of burning fossil fuels, means that it is imperative we save all large wild animals, save them from extinction as we thus-- save ourselves.

Now, recently, and inadvertently I have done some rather minor experiments. Whenever I see a insect in the house, butterfly, moth, ant, spider, millipede, lots of lady bugs, I trap them in a jar and release outdoors, by next day if not sooner. Sometimes they get pretty stressed out in the glass jar--peanut butter, if left too long. So I am guessing they have taken up most of the oxygen and replaced with CO2. So I leave a crack open in the jar. Which gives me the idea of having a jar with a plant in it
and somehow having some insect inside the jar and see if the plant flourishes compared to a jar with plant with no CO2 replenishment and another jar with plant that is fed fire CO2. Perhaps we can prove Animal CO2 versus fire CO2 with just simple easy jars.


AP

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#686878

1) A brain

2) Get one

3) LOL!

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#686915

On 5/24/2018 2:08 AM, Uncle Hal wrote:
> 1) A brain
> 
> 2) Get one
> 
> 3) LOL!
> 
He has one, it's just a little bit diseased.

Ok. It's a LOT diseased!

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#686880

Alright the CO molecule is a perfect test case of torus as atomic and molecular geometry, for the CO fills the 2P orbital exactly with 6, and a wonderment why CO is not more stable of a bond than CO2.

Now i cannot draw large circles so pretend these rectangles are toruses
           |———
———|          |
|          |          |
|          |          |
———|          |
           |———|

Where the smaller is C and larger is O

And by giving up C two muons to O makes the 2P of O a complete 2P torus, leaving behind in C two toruses of 2s and 1s and whether inside O torus of 2P are now two 1s tori and two 2s tori all four of which keep circling the O 2P torus producing magnet monopoles via Faraday law.

AP

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#687087

Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:

>29-Maximum Electricity Coil theory// Internal parts of atoms and how the 
>chemical bond is formed


>CHEM ONE, 2nd ed., Waser, Trueblood, Knobler, 1980, page 327, Figure 16-3
>Schematic Energy-Level Diagram for a Many-Electron Atom

Does it involve conic sections such as ellipses?

Below you will find a simple *proof* that shows that certain conic 
sections are ellipses.

Some preliminaries:

Top view of the conic section and depiction of the coordinate system used 
in the proof:

              ^ x
              |
             -+- <= x=h
         .'   |   `.
        .     |     .
        |     |     |
        '     |     '      
         `.   |   .'
 y <----------+ <= x=0
             
Cone (side view):
                 .
                /|\
               / | \
              /b |  \
             /---+---' <= x = h
            /    |'   \
           /   ' |     \
          / '    |      \
x = 0 => '-------+-------\
        /    a   |        \

Proof:

r(x) = a - ((a-b)/h)x  and  d(x) = a - ((a+b)/h)x,  hence

y(x)^2 = r(x)^2 - d(x)^2 = ab - ab(2x/h - 1)^2 = ab(1 - 4(x - h/2)^2/h^2.

Hence (1/ab)y(x)^2 + (4/h^2)(x - h/2)^2 = 1  ...equation of an ellipse

qed

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