New Bohr's single and double bonds (C-C, C=C, O=O)

Top page (correct Bohr model including the two-electron atoms).
Electron spin is an illusion !

Our new Bohr model has succeeded in calculating the Helium ionization energy more correctly than the quantum mechanical variational methods as shown in the Top page.
In this new successful Bohr model, the two electrons of the helium atom (He) are moving on the orbits of just one de Broglie's wavelength, which are perpendicular to each other.
Furthermore, we have succeeded in visualizing the four valence electrons of carbon , and six valence electrons of oxygen.
(See Carbon page and Oxygen page).

Bohr's methane (CH4) - 2nd version.

In this page (Virial theorem of methane), we have succeeded in expressing the methane molecule by new Bohr model.
When we consider the C-C bond, we need to use the second version of methane (CH4) molecule, too.

Fig. 1. Periodic motions of methane's electrons
motion

As shown in this Carbon page, there are two versions when the four valence electrons of carbon atom are arranged "tetrahedrally".

1st version --- the four electrons of carbon are the closest to each hydrogen's nucleus.
2nd version --- the four electrons move to the opposite positions of the carbon nucleus from the original positions.

Of course, the ground state energies of the 1st and 2nd versions of methane are the same (= -219.675 eV).
In the 1st version of methane, the hydrogens' electrons are repelled outward by the carbon's electrons.
In the 2nd version, the hydrogens' electrons become closer to the carbon nucleus, because the carbon's electrons go away from the hydrogen nuclei.
(On the other hand, the oxygen atom has only one version of "octahedral" structure.)
Here we try to express this 2nd version of methane using the next sample JAVA program.

Sample JAVA program of CH4 (second version)
If you copy and paste the program source code inside into a text editor, you can easily compile and run this.
(Unit of force: 1000 = force between +e nucleus and -e electron which are Bohr radius apart.)

According to the Virial theorem, the total potential energy (tV) of methane is -219.675 × 2 = -439.3 eV
When the total potential energy is about -439.3 eV, how are the eight electrons of methane arranged ?
About the detailed methods, see this page.

Fig. 2. The 2nd version of methane's (CH4) electrons
second

Table 1. Parameters of the 2nd version CH4 ( tV = -439.44 eV ).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -5260 0 -3719 6442 2107 0 -12 0 2.007
ele 1 5260 0 -3719 6442 2107 0 12 0 2.007
ele 2 0 -5260 3719 6442 2107 12 0 0 2.007
ele 3 0 5260 3719 6442 2107 -12 0 0 2.007
ele 4 -838 -3530 -2000 4142 1574 328 6 208 1.000
ele 5 837 3530 -2000 4142 1574 -328 -6 208 1.000
ele 6 3530 -838 2000 4142 1574 -6 328 -208 1.000
ele 7 -3530 837 2000 4142 1574 6 -328 -208 1.000

As shown in Table 1, the hydrogen electrons 4-7 are attracted to each hydrogen nucleus.
So, the hydrogen nuclei are attracted to these electrons, too. (Cn is big.)
As shown in this page, the distance between electron (4-7) and each hydrogen nucleus of the 1st version methane is 4896 MM.
So the average distance (nuc) is ( 4896 + 4142 ) / 2 = 4519 MM (about 4500 MM).
We can know the electron distribution of the 2nd version of methane.

New Bohr model Ethane (H3C-CH3)

Breaking all bonds ( 6 × C-H + C-C ) of ethane requires 2826.1 kJ/mol (= 29.29 eV)
So the ground state energy of ethane (8 carbon valence electrons + 6 hydrogen atoms) becomes -29.29 -148.024 × 2 -13.606 × 6 = -406.974 eV
(148.024 eV is the sum of 1-4th ionization energies of the carbon atom. )
According to the Virial theorem, the total potential energy (tV) becomes 2 × -406.974 = -813.948 eV.
The C-C and C-H bond lengths are 15351 MM, and 10940 MM, respectively. (1 MM = 10-14 meter).
And C-C-H angle is 111.7 degrees.

Fig. 3. Periodic motions of ethane's electrons (H3C-CH3)
ethane

As shown in Fig.3, the ethane molecule consists of a pair of the 1st and 2nd versions of methane.
Between two carbon nuclei, the four electrons are attracted by another nucleus avoiding each other.
Basically, each orbit of the electron is extended by the positive nuclei such as hydrogen and another carbon.
But here we are using the Virial theorem, so the both sides of each electron's orbit becomes the average length.
Here we try to express the ethane molecule using the next sample JAVA program.
About the detailed methods, see this page.

Sample JAVA program of ethane (Virial theorem)
(If you copy and paste the program source code inside into a text editor, you can easily compile and run this.)
In this program, the central charge (+4e = +6e -2e ) is supposed to be +4.217.
So de Broglie's wave is just 2.000, if the "tetrahedral" structure of carbon is completely kept.

Fig. 4. distribution of ethane's electrons (H3C-CH3)
ethane

Table 2. Parameters of ethane ( C-C: 15351 MM, = control.).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -6285 0 0 6285 1880 0 0 0 2.006
ele 1 2305 6019 0 6445 1267 -6 2 0 1.995
ele 4 2235 6045 0 6444 2064 33 -6 -43 1.999
ele 5 2235 -3022 -5235 6444 2064 33 -34 27 1.999
ele 6 2235 -3022 5235 6444 2064 33 40 16 1.999
ele 7 -6360 0 0 6360 2116 0 0 0 2.006
ele 8 1800 1825 -3705 4505 1017 -247 -614 -2 1.000
ele 10 1800 2296 3432 4505 1017 -247 309 -530 1.000
ele 11 800 1975 3900 4444 1416 -172 -298 0 1.000

In Table 2, the total potential energy becomes -813.37 eV, which is almost same as the experimental value -813.948eV.
The average distance between electron (8-13) and each hydrogen nucleus is (4505 + 4444) / 2 = 4474.5 MM, which is almost 4500 MM.
Due to the repulsive force of another carbon, the "tetrahedral" structures of carbons are a little broken.
de Broglie's waves are a little different from 2.000. (The error is ± 0.005 (or 0.006))

Next we choose the value 16500 in the scrollbar and click the C-C (MM) button.

Table 3. Parameters of ethane (C-C 16500 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -6300 0 0 6300 2049 0 0 0 2.010
ele 1 2315 6020 0 6449 1261 -8 3 0 1.995
ele 4 2290 6050 0 6468 2082 25 -4 -42 2.000
ele 5 2290 -3025 -5239 6468 2082 25 -34 25 2.000
ele 6 2290 -3024 5239 6468 2082 25 39 17 2.000
ele 7 -6380 0 0 6380 2119 0 0 0 2.016
ele 8 1900 1600 -3750 4498 1042 -233 -567 -3 1.000
ele 10 1900 2447 3260 4498 1042 -233 287 -489 1.000
ele 11 800 1876 4110 4588 1358 -158 -234 1 1.000

Basically when the repulsive forces of the four nuclei (three hydrogen nuclei + another carbon nucleus) are balanced, the carbon nucleus becomes stable.
But when the C-C length is longer than the balanced length, the influence of another carbon nucleus becomes weaker. (Fig. 5, upper panel)
As a result, the balance of the four repulsive forces breaks down, the "tetrahedral structure" is more broken to make the carbon nuclei stable (= FX, FY, FZ of C0,1 are small).
(In Table 3, the maximum error is 0.016, which is bigger than the control of Table 2.)
And the total potential energy (tV) is -806.74 eV, which is higher than the experimental value (-813.948 eV)
So the C-C bond in Table 3 is weaker than the control.

Fig. 5. balance of forces of ethane (H3C-CH3)
balance

Next we choose the value 14000 in the scrollbar and click the C-C (MM) button.

Table 4. Parameters of ethane (C-C 14000 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -6190 0 0 6190 1592 0 0 0 1.996
ele 1 2350 6070 0 6509 1192 -9 3 0 2.003
ele 4 2235 6100 0 6496 1993 42 -8 -43 2.007
ele 5 2235 -3050 -5282 6496 1993 42 -33 29 2.007
ele 6 2235 -3049 5282 6496 1993 42 42 14 2.007
ele 7 -6280 0 0 6280 2188 0 0 0 1.991
ele 8 1800 1825 -3705 4505 1011 -246 -609 -2 1.002
ele 10 1800 2296 3432 4505 1011 -246 307 -526 1.002
ele 11 800 1945 3800 4343 1470 -174 -327 0 0.999

when the C-C length is shorter (=Table 4) than the balanced length, C0 nucleus is repelled by another nucleus, but on the other hand, C1 nucleus is closer to e0 electron.
So C1 nucleus is attracted to the right, and becomes unstable.(Fig. 5, lower panel)
As a result, the "tetrahedral structure" is more broken to make the carbon nuclei stable (= FX, FY, FZ of C0,1 are small).
(In Table 4, the maximum error is 0.009, which is bigger than the control of Table 2.)

Next we choose the value 13500 in the scrollbar and click the C-C (MM) button, which makes C-C length much shorter.

Table 5. Parameters of ethane (C-C 13500 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -6180 0 0 6180 1354 0 0 0 1.987
ele 1 2305 6095 0 6516 1190 2 0 0 2.007
ele 4 2200 6200 0 6578 1934 55 -10 -46 2.024
ele 5 2200 -3100 -5369 6578 1934 55 -35 32 2.024
ele 6 2200 -3099 5369 6578 1934 55 45 14 2.024
ele 7 -6300 0 0 6300 2173 0 0 0 1.984
ele 8 1800 1850 -3680 4494 1011 -247 -623 -1 1.000
ele 10 1800 2261 3442 4494 1011 -247 312 -538 1.000
ele 11 850 1980 3840 4403 1446 -173 -315 0 0.999

In Table 5, C0 nucleus is more repelled by another nucleus, and C1 is more attracted by e0 electrons. (Fig.5, lower)
So, to make the carbon nuclei stable (= FX, FY, FZ of C0,1 are small), "tetrahedral" structures need to be more broken than the upper case of Table 4.
In Table 5, the maximum error is 0.024, which is much bigger than the control of Table 2.

Here, the hydrogen electrons 11,12,13 are supposed to be closer to C nucleus than the hydrogen electrons 8,9,10. (Fig.3 upper).
(So the +X and +Y coordinates of ele 11 need to be positive.)
But if we try to make +X of ele 11 minus, the "nuc" value becomes shorter, which makes the "average" nuc much shorter than 4500 MM.
So we need to make +X and +Y of ele 11 positive.

Bohr model Ethylene (H2C=CH2)

Breaking all bonds ( 4 × C-H + C=C ) of ethylene (=ethene) requires 2253.1 kJ/mol (= 23.35 eV)
So the ground state energy of ethylene (8 carbon valence electrons + 4 hydrogen atoms) becomes -23.35 -148.024 × 2 -13.606 × 4 = -373.822 eV
According to the Virial theorem, the total potential energy (tV) becomes 2 × -373.822 = -747.644 eV.
The C=C and C-H bond lengths are 13390 MM, and 10870 MM, respectively. (1 MM = 10-14 meter).
And C=C-H angle is 121.3 degrees.

Fig. 6. Ethylene's electrons (H2C=CH2)
ethylene

The ethylene molecule also consists of a pair of the 1st and 2nd versions of methane.
Between two carbon nuclei, the four electrons are attracted by another nucleus avoiding each other.
There is one problem in ethylene.
In case of ethylene, C=C-H angle of 121.3 degrees is bigger than ethane C-C-H angle of 111.17 degrees.
So we can not neglect the influence of the hydrogen electrons on the carbon nuclei.
The eight carbon valence electrons and two hydrogen nuclei are arranged "symmetrically" both in the 1st and 2nd versions of methane.
But the hydrogen electrons (e8-e11) are distributed "unsymmetrically".
(e10,11 of the 2nd version of methane is closer to the carbon nucleus than e8,9 of the 1st version methane like Fig.2.)
So the two carbon nuclei are attracted in the direction of the 2nd version of methane alternately.
As a result, the average forces of the hydrogen electrons acting on the carbon nuclei are zero.

Actually, in Fig.6, the orbits of e0,1 (1st version) is longer than e4,5 (2nd version) orbits like Fig.3. So the influences of hydrogen's electrons are balanced.
But here we use the Virial theorem. so we suppose that the long and short sides of elliptical orbits are the same, and we can not discriminate it.

To average these forces of hydrogen electrons, we "temporarily" use 1st × 1st versions of methane as shown in Fig.6.
The total energy of each methane is the same, but we need to correct the interaction between hydrogen nuclei later.
(The original ethylene is a little bigger (0.681 eV) than Fig.6 in the total potential energy.)

Here we try to express the ethylene molecule using the next sample JAVA program.
About the detailed methods, see this page.

Sample JAVA program of ethylene (Virial theorem)
(If you copy and paste the program source code inside into a text editor, you can easily compile and run this.)

Fig. 7. Ethylene's electrons (H2C=CH2)
distribution

Table 6. Parameters of ethylene (C=C 13390 MM), control.
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -3869 0 5249 6520 2044 -37 0 -22 1.998
ele 1 -3869 0 -5249 6520 2044 -37 0 22 1.998
ele 2 3500 5294 0 6346 1396 0 0 0 2.006
ele 4 3750 -5260 0 6459 2094 41 24 0 1.997
ele 5 3750 5260 0 6459 2094 41 -24 0 1.997
ele 6 -3568 0 -5290 6380 1347 11 0 -8 2.003
ele 8 560 -1100 4270 4444 1156 -183 299 0 1.000
ele 9 560 1100 -4270 4444 1156 -183 -299 0 1.000
ele 10 -560 -4280 -1100 4454 1147 184 0 290 1.000

In Table 6, the total potential energy (tV) is -747.43 eV, which is almost same as the experimental value (-747.644 eV).
Next we choose "15000" in the scrollbar, and click "C=C (MM)" button.

Table 7. Parameters of ethylene (C=C 15000 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -3910 0 5260 6554 2033 -16 0 -10 2.003
ele 1 -3910 0 -5260 6554 2033 -16 0 10 2.003
ele 2 3490 5210 0 6270 1490 -6 4 0 1.994
ele 4 3770 -5270 0 6479 2095 22 14 0 2.003
ele 5 3770 5270 0 6479 2095 22 -14 0 2.003
ele 6 -3590 0 -5197 6316 1431 21 0 -16 1.990
ele 8 600 -1100 4290 4469 1152 -189 301 0 1.001
ele 9 600 1100 -4290 4469 1152 -189 -301 0 1.001
ele 10 -610 -4290 -1100 4470 1146 193 0 293 0.999

When C=C length is longer (Table 7), the repulsive forces of the carbon nuclei becomes weaker, so the balance of the forces breaks down.
(It is similar to the ethane case of Fig. 5 upper panel.)
The "tetrahedral structure" is more broken to make the carbon nuclei stable (= FX, FY, FZ of C0,1 are small)
The maximum error is 0.010, which is bigger than the control of Table 6.
And in Table 7, the total potential energy is -717.74 eV, which is higher than the experimental value (= -747.644 eV).
So the binding force is weaker in Table 7.
Next we choose "12500" in the scrollbar, and click "C=C (MM)" button.

Table 8. Parameters of ethylene (C=C 12500 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 -3850 0 5200 6470 2078 -53 0 -30 1.988
ele 1 -3850 0 -5200 6470 2078 -53 0 30 1.988
ele 2 3500 5350 0 6393 1331 5 -3 0 2.015
ele 4 3750 -5240 0 6443 2106 57 31 0 1.995
ele 5 3750 5240 0 6443 2106 57 -31 0 1.995
ele 6 -3620 0 -5350 6459 1249 13 0 -10 2.015
ele 8 560 -1150 4265 4452 1145 -185 308 0 1.000
ele 9 560 1150 -4265 4452 1145 -185 -308 0 1.000
ele 10 -560 -4280 -1100 4454 1138 183 0 286 1.001

When C=C length is shorter (Table 8), the repulsive forces of the carbon nuclei becomes stronger.
(It is similar to the ethane case of Fig. 5 lower panel.)
The "tetrahedral structure" is more broken to make the carbon nuclei stable (= FX, FY, FZ of C0,1 are small)
The maximum error is 0.015, which is bigger than the control of Table 6.

These results show that the 13390 MM of C=C length is the balanced value.
About the acetylene (HC ≡ CH), see this page.
In the acetylene, the carbon electrons tend to be more attracted to another carbon nucleus, so "tetrahedral" structure of carbon is more broken than the ethane.
So the acetylene is more unstable and reactive.

Bohr model Oxygen molecule (O=O)

Breaking the O=O bond of the oxygen molecule requires 498 kJ/mol (= 5.16 eV)
So the ground state energy of oxygen molecule (12 oxygen valence electrons ) becomes -5.16 -433.103 × 2 = -871.366 eV
(433.103 eV is the sum of the 1-6th ionization energies of the oxygen atom.)
According to the Virial theorem, the total potential energy (tV) becomes 2 × -871.366 = -1742.732 eV. And O=O length is 12074 MM.

Fig. 8. Oxygen molecule electrons (O=O)
O2

In the oxygen molecule, it is thought that two "octahedral" oxygen atoms are bound to each other by the Coulomb force
(One is rotated by 90 degrees as shown in Fig.8. The oxygen atom has only one version of "octahedron", because it already has electrons which are on the opposite sides of the nucleus. )
We use the next sample JAVA program to express this oxygen molecule.

Sample JAVA program of O2 (Virial theorem)
(If you copy and paste the program source code inside into a text editor, you can easily compile and run this.)
In this program, the central charge (+6e = +8e -2e) is supposed to be +6.271, so de Broglie's waves of single oxygen atom becomes just 2.000.

Fig. 9. Oxygen molecule electrons (O=O)
distribution

Table 9. Parameters of Oxygen molecule (O=O 12074 MM), control.
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 3295 0 -3165 4568 5832 0 0 0 1.985
ele 1 3295 0 3165 4568 5832 0 0 0 1.985
ele 2 -3200 0 3260 4568 6004 -4 0 -4 2.002
ele 4 10 -4580 0 4580 5958 -18 0 0 1.998
ele 6 -3295 -3165 0 4568 5832 0 0 0 1.985
ele 8 3200 3260 0 4568 6004 4 -4 0 2.002
ele 9 3200 -3260 0 4568 6004 4 4 0 2.001
ele 10 -10 0 -4580 4580 5958 18 0 0 1.998
ele 11 -10 0 4580 4580 5958 18 0 0 1.998

In Table 9, the total potential energy (tV) is -1741.3 eV, which is almost same as the experimental value (-1742.732 eV).
The oxygen nuclear charge is bigger than carbon, so the de Broglie's waves is a little more unstable than the carbon case.
Next we choose "13000" in the scrollbar, and click "O=O (MM)" button.

Table 10. Parameters of Oxygen molecule (O=O 13000 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 3295 0 -3178 4577 5848 -6 0 -6 1.988
ele 1 3295 0 3178 4577 5848 -6 0 6 1.988
ele 2 -3200 0 3263 4570 5997 -5 0 -5 2.001
ele 4 5 -4584 0 4584 5943 -13 0 0 1.998
ele 6 -3295 -3178 0 4577 5848 6 -6 0 1.988
ele 8 3200 3263 0 4570 5997 5 -5 0 2.001
ele 9 3200 -3263 0 4570 5997 5 5 0 2.001
ele 10 -5 0 -4584 4584 5943 13 0 0 1.998
ele 11 -5 0 4584 4584 5943 13 0 0 1.998

When O=O length is longer (Table 10), the repulsive forces of the oxygen nuclei becomes weaker
But in Table 10, the total potential energy is -1738.93 eV, which is higher than the experimental value (= -1742.732 eV ).
So the binding force is weaker in Table 10.
Next we choose "11000" in the scrollbar, and click "O=O (MM)" button.

Table 11. Parameters of Oxygen molecule (O=O 11000 MM).
eNo X (MM) Y (MM) Z (MM) nuc (MM) CF fx fy fz Waves(wn)
ele 0 3295 0 -3154 4561 5777 20 0 19 1.981
ele 1 3295 0 3154 4561 5777 20 0 -19 1.981
ele 2 -3195 0 3269 4571 5999 -7 0 -7 2.004
ele 4 15 -4582 0 4582 5960 -27 0 0 1.998
ele 6 -3295 -3154 0 4561 5777 -20 19 0 1.981
ele 8 3195 3269 0 4571 5999 7 -7 0 2.004
ele 9 3195 -3269 0 4571 5999 7 7 0 2.004
ele 10 -15 0 -4582 4582 5960 27 0 0 1.998
ele 11 -15 0 4582 4582 5960 27 0 0 1.998

When O=O length is shorter (Table 11), the repulsive forces of the oxygen nuclei becomes stronger.
So the "octahedral structure" of oxygen is more broken to make the oxygen nuclei stable (= FX, FY, FZ of O0,1 are small)
The maximum error is 0.019, which is bigger than the control of Table 9.

The oxygen molecule is more unstable and reactive than H2O.
It is thought that this property of O2 can be explained by the unstable wave's values of Table 9.

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2010/12/6 updated. Feel free to link to this site.