Top page ( correct Bohr model ).
New Bohr model, Neon, Argon.
(Fig.1) 2 × de Broglie wavelength orbits. Each is a pair of opposite phases.
One wavelength consists of a pair of "crest" (= + ) and "trough" (= - ) irrespective of transverse and longitudinal waves.
Here we suppose "+" phase contains an electron itself, and "-" phase is compressed by the electron's movement, --- which is " de Broglie wave ".
2 × de Broglie wavelength orbit contains two pairs of ±opposite phases and two midpoint lines.
(Fig.2) de Broglie "longitudinal" wave.
If we consider de Broglie wave as one of "longitudinal" waves, one de Broglie wavelength consists of a pair of "low" (= "+" phase ) and "high" (= "-" phase ) pressure regions.
(Fig.3) "Opposite" wave phases interfere with each other, destructively. → "perpendicular".
Helium atoms contains two 1 × de Broglie wavelength orbits ( n = 1 ).
So, two opposite wave phases cause "destructive" interference and instability.
To avoid this destructive interference, two orbits of helium have to cross each other "perpendicularly", because "perpendicular" means each wave phase can be independent from another.
(Fig.4) Two electrons have to pass each other on "midpoint" lines.
In one de Broglie wavelength orbit, a half of it is "opposite" wave phase.
So, "e1" electron of Fig.4 must go in the direction perpendicular to "e2" orbit to midpoint.
Two electrons in "perpendicular" orbits have to pass each other in the "parallel" direction at "midpoint", because "midpoint" line is zero phase, which has NO influence on another wave.
(Fig.5) Midpoint between ±opposite phases has NO influence on another wave.
"Midpoint" line between ±opposite phases contains "medium (= neutralized )" zero phase.
So this midpoint has NO influence on another wave phase (= NOT disturb other de Broglie waves ).
So, de Broglie wave can pass each other in the parallel direction, NOT disturbed on the midpoint lines.
(Fig.6) 2 × de Broglie wavelength orbit can naturally contain "two" electrons.
Different from one de Broglie wavelength orbits of helium, two de Broglie wavelength orbit can contain two electrons just on the opposite sides of the nucleus.
Because the wave phases at the opposite positions are the same (= both "+" phase ).
So two electrons can naturally enter the same orbit, NOT expelled by destructive interference.
(Fig.7) 2 perpendicular orbits + 2 midpoint lines = 4 orbits in Neon.
We can explain new Neon model using 4 orbits, each contains two electrons ( 2 × 4 = 8 valence electrons ). Each two orbits form a pair, when they cross perpendicularly.
When "two orbits" cross perpendicularly in each pair, each de Broglie wave pair must NOT be disturbed and obstructed by another pair of waves.
"NOT disturbed" means each pair ( consisting of two orbits ) must be on the midpoint lines of orbits in another pair. As a result, Neon can contain the maximum 4 orbits stably.
(Fig.8) Orbits of Neon cross each other "perpendicularly".
(Fig.8') Orbits of Neon cross each other "perpendicularly".
As shown on this page, we can show the appropriate new Neon model, in which orbits can cross each other "perpendicularly". "Perpendicular" crossing means they can avoid "destructive" interference.
(Fig.9) Orbits (= opposite wave phases ) always cross each other "perpendicularly".
Each electron can cross other electrons' orbits perpendicularly, when they are opposite wave phases to each other, because "perpendicular" crossing avoids destructive interference.
Furthermore. crossing of "opposite" wave phases can neutralize total wave phases at all points ( ±phases = zero ).
"Neutralization" of wave phases means any other electrons can be free from other wave ( bad ) influences, and move stably and independently, NOT to be disturbed.
(Fig.10) de Broglie "longitudinal" wave.
If we consider de Broglie wave as some longitudinal wave, each de Broglie wavelength consists of a pair of "low" and "high" pressure.
(Fig.11) "Low" + "low" pressures form "very low" pressure. - unbalanced.
When "low" and "low" pressure parts cross each other, it generates "very low" pressure.
"Very low" pressure in de Broglie wave field is bad for the pressure balance.
(Fig.12) "High" + "high" pressures form "very high" pressure.
When "high" and "high" pressure parts cross each other, it generates "very high" pressure.
"Very high" pressure in de Broglie wave field is bad for the pressure balance, too.
Of course, when both phases and their directons (= parallel ) agree with each other, it causes "constructive" interference, so enhances their displacement "synergistically".
But two de Broglie waves cross each other "perpendicularly", they are independent from each other. In this case, "field pressure" is very important factor to determine each orbit.
(Fig.13) "Low" and "high" pressure waves cross "perpendicularly" ← stable !
When "low" (= "+" phase ) and "high" (= "-" phase ) of de Broglie waves cross "perpendicularly", it generates "neutral" pressure in field. "Neutral" pressure is the most stable in balance.
Furthermore, "perpendicular crossing" can keep each electron's de Broglie wave phase independently.
(Fig.14) Opposite "e2" and "e1" phases form midpoint phase at e4,e3 crossing point.
In new Neon model, when two orbits cross perpendicularly, they are NOT disturbed by other orbits.
In Fig.14, "e4" electron cross "e3" opposite phase wave perpendicularly.
At this crossing point ( of "e3" and "e4" ), the influences from other electrons (= e1, e2 ) must be cancelled out and zero (= midpoint ) to keep the crossing waves stable.
As you see Fig.14, from the viewpoint of e3,e4 intersection, "e2" (= upper ) and "e1" (= lower ) orbits are just symmetrical and "opposite" wave phases !
This is the reason why the 4 even-number orbits of Neon can be stable.
When 2 orbits cross perpendicularly, other 2 symmetrical orbits cancel each other at another intersection.
(Fig.15) H2 molecule. two "perpendicular" orbits.
Hydrogen molecule (= H2 ) contains two hydrogen atomic orbits with 1 × de Broglie wavelength.
Condidering Coulomb interaction, when one electron (= e2 ) approaches another nucleus (= n1 ), these two orbits must cross "perpendicularly" to avoid destructive interference.
Two orbital de Broglie waves in the opposite direction cancel each other at a distance, so magnetic field loop is disrupted and its magnetic moment created in H2 is almost zero.
(Fig.16) Two orbits are too close → repulsion between electrons become stronger.
When internuclear distance of H2 molecule is shorter than the experimental value (= 0.7414 Å ), each electron is more attracted to another nucleus.
So electrons tend to stay longer between two nuclei.
Furthermore, the repulsive Coulomb force between two closer nuclei becomes stronger.
These increasing repulsive Coulomb forces between electrons and nuclei prevent two H2 nuclei from approaching each other.
(Fig.17) Most stable H2 configuration.
On this page, we can explain the experimental internuclear distance of H2 by Virial theorem.
"Average" positions of each hydrogen orbit is as shown in Fig.17.
As the two nuclei come closer to each other, each electron can interact with two positive nuclei.
So, the bond energy becomes bigger, which means stable H2 molecule is formed.
(Fig.18) Repulsive force of two electrons.
But when the two nuclei come closer to each other than the experimental value of 0.7414 Å, the repulsive force between two electrons are stronger.
Each hydrogen orbit must keep 1 × de Broglie wavelength, which prevents the orbit from being smaller. Stronger repulsion between two electrons reduces "effective" central positive charge of each hydrogen.
So under the common de Broglie wave's condition (= 1 × de Broglie wavelength ), each hydrogen radius "r" is larger, and bond energy is smaller than the experimental value in Fig.18.
(Fig.19) Carbon = 4 valence electrons + 4 holes = "tetrahedral".
Carbon has four valence electrons ( n = 2 ), so easily forms "tetrahedral" shape due to Coulomb balance. Regular "tetrahedron" is a part of regular "hexahedron".
So Carbon atom can be thought to consist of 4 tetrahedral electrons + 4 holes like Neon.
(Fig.20) Hydrogen ±electron's wave phases synchronize with carbon. →"hexahedron".
As I said in Fig.6, 2 × de Broglie wavelength orbit can contain two electrons.
Carbon tetrahedral-shape orbit includes four electrons and four holes.
When electrons of hydrogen atom approach and synchronize with the same phase hole, they can fuse and form stable hexahedral structure with 8 electrons like Neon.
When opposite (= "-" ) wave phase approaches carbon, hydrogen orbit synchronize with the opposite ( = "-" ) wave phase ( in this case, "e4" wave ), which is perpendicular to e1 orbit.
Because the same phase waves interfere "constructively" and enhance each other, and "opposite" phases repel each other due to their different phase motions.
If we consider field oscillations in three directions (= x, y, z ), hydrogen orbit can oscillate periodically in concert with another same wave phase.
(Fig.21) In C-C bond, an electron synchronize with another hole.
Also in carbon-carbon ( C-C ) bond, electons are thought to aim at stable hexahedral structure like Neon by electron-hole pairs.
As shown in Fig.21, an electron of one carbon can always approach a hole of another carbon, which is stable in Coulomb energy.
As shown on this page, the average carbon radius is about 0.64 Å, considering its ionization energies and 2 × de Broglie wavelength.
This bond length is almost kept in different carbon-atomic bonds ( C-H, C-C .. )
(Fig.22) In C=C bond, two electron-hole pairs are formed.
In C=C double bond, two electron-hole pairs are formed between two carbons.
In Fig.22, "e1" and "e3" electrons of carbon 2 approach correspondent holes of carbon 1, and make Neon-like hexahedral structure.
Comparing Fig.21 and Fig.22, you will easily find C=C bond is shorter than C-C bond.
(Fig.23) "e1" ( and "e2" ) electrons pass "midpoint" between two nuclei.
As two carbon nuclei are closer to each other, their repulsive Coulomb force becomes stronger.
To cancel this nuclear repulsion, each electron must enter between C = C nuclei.
Considering actual orbit of each electron, you find "e1" and "e2" electrons pass the midpoint between two nuclei periodically, which can cancel stronger internuclear repulsion.
(Fig.24) Alternating between "recipient" and "donor" carbons makes "plane".
Basically, each carbon's valence electrons aim at stable Neon-like hexahedral structrue, absorbing other 4 electrons. Carbon receiving these electrons is "recipient", and another carbon is "donor".
Due to Coulomb symmetry, each carbon is alternating between these "recipient" and "donor" roles.
In Fig.24, four holes of carbon 1 absorb 4 other electrons from two hydrogens and carbon 2.
On the other hand, 4 electrons of carbon 2 (= donor ) become the closest to surrounding atoms.
As you see, in this case, "e2" and "e4" electrons in both carbons are bonded to hydrogens, which form "plane-like" ethene ( H2C=CH2 ).
(Fig.25) 2 × de Broglie wavelength orbits "aim" at four orbits like Neon.
As I said, the orbit of two de Broglie wavelength contains two symmetrical midpoint lines with zero phase. So the total four orbits ( 2 × 2 perpendicular orbits ) become the most stable.
Because as shown in Neon, when four orbits are included, they are completely symmetrical with respect to both Coulomb and de Broglie waves.
(Fig.26) Oxygen bonds, H2O.
Oxygen atom contains six valence electrons, so regular octahedral shape is most stable with respect to Coulomb repulsion. But as shown in Fig.25, four orbits with eight electrons are symmetrical and stable as "de Broglie waves".
So oxygen also aims at Neon-like hexahedral structure to be stable in both Coulomb and de Broglie waves .
As a result, the bond angle of H2O (= 104.5o ) is between octahedron (= 90o ) and tetrahedron (= 109.5o )
(Fig.27) "Complete" circular and spherical orbital shapes are impossibe.
When three circular ( NOT elliptical ! ) orbits cross perpendicularly like Fig.27, it can obey de Broglie wave rule.
But thinking commonsensically, this "ideal" complete sphere is impossible.
Because six ( NOT one ! ) valence electrons are always moving and repelling each other, each orbit cannot be kept complete circular.
So the condition of complete "circle" and "sphere" is too severe in multi-electron atomic orbits.
(Fig.28) 2 × de Broglie wavelength orbit can contain two electrons.
As I said, 2 × de Broglie wavelength orbit consists of two pairs of ±phases (= two electrons ) and two midpoint lines.
(Fig.29) When two orbits of 2 × de Broglie wavelength cross perpendicularly.
In Fig.29, two orbits of 2 × de Broglie wavelength cross each other perpendicularly.
But different from helium case, Fig.29 orbits are unstable, when including only two orbits.
(Fig.30) When an electron moves 1/8 of its orbit, it becomes "free" and unstable.
Different from 1 × de Broglie wavelength helium, in this case, the same "+" phase of e2 electron easily approaches e1 electron.
When e1 moves 1/8 of its orbit (= Fig.30 left ), the wave phase of "e2" becomes neutral (= midpoint ).
So from 1/8 to 1/4 orbital region (= Fig.30 middle ), e1 electron feels the same "+" phase (= low pressure ) of "e2" eletron.
As a result, this e1 electron can fly somewhere else freely due to NO restriction of e2 opposite phase, which makes this structure unstable and changeable.
(Fig.31) When an electron moves 1/8 of its orbit, it needs "other restrictions" to be stable.
To keep these two perpendicular orbits, other orbits must help them.
In Fig.31, e4 opposite wave phase crosses e1 electron to keep these orbital shapes stable.
So only two orbits in 2 × de Broglie wavelength is NOT enough to keep orbital shape.
At least, four orbits like Neon is necessary for stability.
(Fig.32) "Average" wave phase is neutral (= zero phase ).
In Fig.30 right, electrons in two "perpendicular" orbits pass each other in parallel directions.
In this case, the wave influence of another electron (= e2 ) must be zero (= neutral ) like helium case.
As shown in Fig,32, this region contains the same amounts of ± wave phases, so the total is zero.
(Fig.33) Electrons "cannot" pass each other stably.
In unstable oxygen structure, this region is different from Neon case of Fig.34.
In this unstable oxygen, e2 opposite wave phase also crosses this point at the same time.
In this case, all excessive wave pressure must be pushed away toward "perpendicular" (= e2 ) direction.
But "e3" wave phases try to obstruct and cancel this pushing process.
On the other hand, in Fig.32, the total wave phase is the lowest pressure (= complete "+" phase ), so the process of "pushing away" is unnecessary.
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