In the previous post, I said I was going to give a "practical justification" for making analytical distinctions on the "input" and "output" sides of the transform, but that may not happen in this post. Hopefully the next few posts will implicitly make that more justified, however.
Anyway, for some reason it occured out of the blue the other night that the sum totals of electrons as atomic subshells are filled (according to the Aufbau principle: 1s,2s,2p,3s,..), contain the numbers 10, 20, and 30 - viz they are 2,4,10,12,18,20,30,.. . (The subshells themselves contain 2,2,6,2,6,2,10,.. electrons.) This could be handy, since these initial 30 elements are the most essential to supporting complex life on Earth (in fact, element 30 is zinc, and then immediately after it follow elements that are much less biologically desirable for humans in anything more than very trace amounts). Even more strangely, the arrangement of the 20 consonants in the related conalphabet seems to reflect the electron filling sequence somewhat (but I hope to get to that later).
(Also possibly worth mentioning, the subshell capacities are 2,6,10,14,18,22,26. If we added 1 to each of these, we would get 3+4*m, and remember that n^3=n^(3+4*m) (mod 10).)
First, let's look back at the Möbius object comprised of two linked loops of paper (one large and one small) with numbers written on them. Recall that the larger loop had 2 sides (with 10 numbers on each side), and the smaller was a true Möbius strip with only one side (with 10 numbers). If we consider a "pair" of numbers to be made up of those numbers that are on the same position of the large loop strip, but written on opposite sides of the paper, then we can list all the numbers as the following:
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0 5*
1 4*
2 3*<-
3 2*<-
4 1*
5 0*
6 9*
7 8*<-
8 7*<-
9 6*
SMALL
0**
1**
2**
3**
4**
5**
6**
7**
8**
9**
(<-The arrows indicate the {2,3,7,8}-positions, which are used by the transform mentioned in the previous posts.)
(* **Asterisks are used to keep track of the sides of the paper.)
Let the 20 numbers on the large loop represent the first 20 chemical elements, viz H to Ca. Let the 10 elements on the small loop represent the next elements, viz the transition metals Sc to Zn.
In the above list, the 10 pairs each sum to 5 (mod 10). In chemical reactions, however, we often would like to sum the number of valence electrons to 2 or 8 (eg Al (3) + P (5) = AlP, aluminum phosphide). This is what I mean when I say valence, although the terminology may be technically imprecise. Let's set up these pairs as though each were a chemical bond producing a neutral compound:
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Valence Corresponding elements
1 1* H K* (potassium hydride)
2 8* He Ar* (noble helium,noble argon)
1 7* Li Cl* (lithium chloride)
2 6* Be S* (beryllium sulfide)
3 5* B P* (boron phosphide)
4 4* C Si* (carborundum)
5 3* N Al* (aluminum nitride)
6 2* O Mg* (magnesium oxide)
7 1* F Na* (sodium fluoride)
8 2* Ne Ca* (noble neon,calcium ion)
You might notice that in each column of numbers above, we have a repetition of (1,2), viz the common sequence to both columns is 1,2,1,2,3,4,5,6,7,8. Let's redefine this as 9,0,1,2,3,4,5,6,7,8, calling these numbers val instead of valence:
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Val Corresponding elements
0 8* He Ar* (noble helium,noble argon)
1 7* Li Cl* (lithium chloride)
2 6* Be S* (beryllium sulfide)
3 5* B P* (boron phosphide)
4 4* C Si* (carborundum)
5 3* N Al* (aluminum nitride)
6 2* O Mg* (magnesia)
7 1* F Na* (sodium fluoride)
8 0* Ne Ca* (noble neon,calcium ion)
9 9* H K* (potassium hydride)
So valence=val (mod 8) except for the cases of
Val Element
0...He
0...Ca
Helium is easily explained, because it lies in the first period of the element table, so is rightly mod 2, not mod 8.
Ca is not quite as simple. Like He, Ca also has valence 2. Since Ne is a noble gas, Ne and Ca cannot react anyhow, so it seems possibly acceptable to assign Ca a different value. But that's not a good plan, because Ca must be able to react with other elements; for this reason and to create symmetry of the neutrals (He, Ar, Ne), Ca was replaced with CaH2, calcium hydride (or hydrolith). Ca has an oxidation number of +2, and H has an oxidation number of -1 in this molecule, so CaH2 is neutral overall. Also, CaH2 reacts strongly with water to produce hydrogen. In this way, it sort of brings us full circle back to element H.
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Val Corresponding elements
0 8* He Ar* (noble helium,noble argon)
1 7* Li Cl* (lithium chloride)
2 6* Be S* (beryllium sulfide)
3 5* B P* (boron phosphide)
4 4* C Si* (carborundum)
5 3* N Al* (aluminum nitride)
6 2* O Mg* (magnesia)
7 1* F Na* (sodium fluoride)
8 0* Ne CaH2* (noble neon,hydrolith)
9 9* H K* (potassium hydride)
The subshells are associated with val as the following (ellipses (..) indicate that the left edge is connected to the right edge):
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Filling order: 1s,2s,2p,3s,3p,4s
| | | |
V V V V
0 1 2 3 4 5 6 7 8 9
..__ _____ _________________ __..
1s 2s 2p
| | | |
V V V V
8* 7* 6* 5* 4* 3* 2* 1* 0* 9*
_________________ _____ _____
3p 3s 4s
Notice that the {2,3,7,8}-positions (those used in the transform) appear symmetrical, one set in shell 2, and the other in shell 3.
Let's try to add some more symmetry. Consider what happens if we "cut in half" the p subshell, defining subshell pa to contain the first 3 electrons in the p-subshell, and the other "half" of the p-subshell will be called subshell pb. So now pa and pb contain 3 electrons each, and the s-subshells still contain 2 each. If we add together an s-subshell with either pa or pb, we contain a total 5 electrons, viz we get the following:
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Filling order: 1s,2s,2pa,2pb,3s,3pa,3pb,4s
| | | |
V V V V
0 1 2 3 4 5 6 7 8 9
..__ ______________ ___________..
2s2pa 1s2pb
| | | |
| || | | || |
V |V | V |V |
8* 7* 6* 5* 4* 3* 2* 1* 0* 9*
..________ ______________ _____..
4s3pb 3s3pa
In the above figure (sorry if it looks a bit busy), I find it interesting that each break in the line segments representing our newly defined "subshells" split the {2,3} and {7,8}-positions in the other sequence. I've tried to mark where these splits are by drawing lines in the figure.
What rationale might we have for defining these new "subshells"? I don't exactly know.