Top page ( correct Bohr model ).

New Bohr model, Neon, Argon.

- de Broglie wavelength, ±opposite phases.
- Reason why Neon contains "4 orbits" stably.
- Various molecular bonds ( H2, C-C, C=C ).

*(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, C=O bonds, 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.

In the same way, C=O double bonds can be explained by this realistic atomic models.

The point is two empty holes of oxygen atom with six valence electrons are always **occupied** by two electrons from the other carbon atom, as seen in Fig.22 lower.

So **No** more electrons of other atoms can approach or form new molecular bonds with this oxygen atom of C=O bond, which is the realistic mechanism of Pauli explucion principle.

*(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 ).

When an electron of H atom belonging to carbon-1 (= recipient carbon in the upper figure ) approaches the the carbon-1 nucleus, an electron of H atom belonging to carbon-2 (= donor carbon ) moves farther away from the carbon-2 whose electron approaches the H-atomic nucleus, instead.

**Repulsive** forces between two hydrogen atomic electrons **prevent** all electrons of both hydrogens bound to two carbons 1 and 2 from approaching the carbon nuclei at the same time, which is why ethene's two hydrogen atoms exist in the same one plane, as shown in the upper figure.

*(Fig.24') ↓ Linear and circular motions. *

As shown in the Broglie wave interference of an electron in a linear motion, the influence of each de Broglie wave reaches very distant area = 5~6 atomic distance.

But when an electron moves around a nucleus in a **circular** motion, the area the de Broglie wave reaches is **smaller**.

Because when an electron moves with a velocity of v around an nucleus in a circular orbit with the radius r, in the point which is 3r distant from the nucleus, de Broglie wave phase must move three times faster (= 3v ) to keep up with the inner electron moving at a speed of v, which is difficult.

Considering Pauli repulsion (= caused by destructive interference between de Brogle wave phases ) between two hydrogen atoms bound to different atoms ( the non-convalent distance between two hydrogen atoms H-H is about 1.3 Å ), we can estimate the influence of each electron's de Broglie wave reaches as far as the point which is less than 3r distant from the nucleus.

*(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 (= 104.5^{o} ) is **between** *octahedron* (= 90^{o} ) and *tetrahedron* (= 109.5^{o} )

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