Top page valence electron

Bohr model Neon structure.

New Bohr's molecular bonds.

- de Broglie "wavelength" and the maximum orbits.
- Four 1 × de Broglie wavelength orbits are impossible.
- New Argon, Krypton, Xenon, Radon structure.

*(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** × ( *1* × ) de Broglie wavelength orbit contains **two** ( *one* ) pairs of **±opposite** phases and **two** ( *one* ) midpoint lines.

These "**opposite**" wave phases **cancel** each other by *destructive* interference.

To avoid it, two orbits must cross **perpendicularly**.

*(Fig.2) Maximum orbits = midpoint lines + 2 (= two perpendicular orbits )*

As shown on this page, the number of *midpoint* lines ( related to **de Broglie** wavelength ) influences the number of **maximum** orbital number in the periodic table.

The **maxium** number of orbits in Ne becomes "**4**" (= 2 × *perpendicular* + 2 × **midlines** ).

4 × de Broglie wavelength contains **4** midlines, so the total orbital number of Kr becomes "**6**".

The **odd** numbers of "3", "5", "7" orbits are asymmetrical and **unstable**.

So the orbital numbers of "**Ar**" (= 3 × waveslength ), "*Xe*" (= 5 × waveslength ) remain the **same** as "**Ne**" and "*Kr*".

*(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) If Beryllium is composed of four 1 × de Broglie wavelength orbits ?*

As I said in Fig.2, the *maximum* orbital number in **1** × de Broglie wavelength is "**two**" of helium. It's **impossible** for atoms (= beryllium ) to contain **four** 1 × de Broglie wavelength orbits.

Here we explain this **Pauli** exclusion principle from the viewpoint of **de Broglie** wavelength.

*(Fig.6) 2 × two helium orbits perpendicular to each other = beryllium ?*

In helium-like atoms, e1 and e2 ( e3 and e4 ) orbits cross **perpendicularly**.

Here we try to combine these **four** orbits to make beryllium outer-electron orbits.

*(Fig.7) If beryllium outer electrons make "four" 1 × de Broglie wavelength orbits ..*

If the number of valence electrons of beryllium is "**four**", and all these orbits are **1** × de Broglie wavelength, their periodic motions become like Fig.7.

Due to **symmetry** and Coulomb repulsions among four electrons, "**tetrahedral**" distributions are formed in this process.

*(Fig.8) "e3" electron feels only "opposite" wave phases of "e1" and "e2" !*

But as shown in Fig.8, when "e1" and "e2" electrons pass the **upper** part of this atom, the **lower** part contains **only** "*opposite*" wave phases of "e1" and "e2" orbits, because they are **1** × de Broglie wavelength.

So "e3" ( and "e4" ) orbits feel *only* "**opposite**" wave phases of "e1" and "e2" orbits and be **disturbed** and **unstable**.

*(Fig.8') "e3" electron can pass the midpoints between e1 and e2, "safely"in Neon.*

On the other hand, in Neon consisting of four 2 × de Broglie wavelength orbits, each electron can move stably, **NOT** disturbed by the *opposite* phases of other orbits.

As shown in Fig.8', "e1" and "e2" orbits are just **symmetrical** with respect to "e3", when it passes the point which is the **same** distance from e1 and e2.

The wave phases of "*e1*" and "*e2*" are just **opposite** from each other at this point of "e3".

So their wave phases are **cancelled** out to be zero, which **doesn't** disturb "**e3**" de Broglie waves.

So, **different** from 1 × de Broglie wavelength, each electron in **2** × de Broglie wavelength orbit can move **safely** and stably, even when **four** orbits are included.

*(Fig.9) Both Ne and Ar forms stable "hexahedral" structure.*

Considering periodic table and **sudden** drop in the first *ionization* energy, it is quite natural to think **Neon** is **2** × de Broglie wavelength, and **Argon** is **3** × de Broglie wavelength in outer orbits.

Both Ne and Ar contain **eight** valence electrons, so each orbit contains **two** electrons in these atoms.

*(Fig.10) Orbits of Neon cross each other "perpendicularly".*

*(Fig.11) 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.12) 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.

( Strictly speaking, the positions of these opposite phases are a little different. But almost the same. )

"**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.13) Each Ar orbit contains two electrons and one hole.*

If Argon orbit is **3** × de Broglie wavelength, and each orbit contains **two** electrons, it must include one "**hole**". "Hole" is the **same** phase as the position of each electron, but it **doesn't** have an electron.

Also in this case of Argon, each two orbits ( ex. "e1" and "e2" ) can cross "**perpendicularly**", keeping *symmetrical* distribution.

*(Fig.14) Eight valence electrons of Ar can keep stable "hexahedral" structure.*

As shown in Fig.14, **eight** valence electrons of **Argon** can also move **stably** and keep *symmetrical* hexahedral distribution, like Neon, in which the **opposite** wave phases in two orbits *cross* **perpendicularly**.

*(Fig.15) When de Broglie wavelengths are different, their orbital sphapes are different.*

If Neon and Argon have the same numbers of valence electrons and orbits, their orbital **shapes** are a little different, because, 3 × wavelength orbit is a little **longer**.

As shown in Fig.15, the orbit of Argon is **closer** to "circular" (= **wider** in width ) than Neon.

( Of course, in these **multi**-electron atoms, each orbit **cannot** be genuine circular or elliptical ).

*(Fig.16) Ionization energies show us "where" de Broglie wavelengths change. *

In the same way, **sudden drop** in *inonization* energies show us "**Kr**" and "**Xe**" are the **last** atoms in **4** × and **5** × de Broglie wavelength orbits.

As seen in this and this, the **changes** of ionization energies from "4s" to "3d" orbitals are **too smooth**, meaning "4s" and "3d" are the **same** type of orbits. Quantum mechanical orbitals are too artificial.

As de Broglie wavelengths (= orbtial radius ) get bigger, the *spaces* between each orbits get **wider**.

So "**packing**" maximum electrons into one *single* orbit ( **5** electrons in **5** wavelength orbit ) causes strong **repulsive** Coulomb forces due to the **imbalance** in electron's distributions in all space.

In Kr and Xe, we consider each orbit contains **3** electrons and they form **three** pairs ( each pair is **two** orbits ), the total becomes **6** orbits ( 3 × 6 = 18 ).

*(Fig.17) 6 orbits of Kr outer electrons.*

Like Neon (= 4 orbits ), Krypton **six** orbits generate **three** pairs, each contains **two** *perpendicular* orbits with **opposite** phases and opposite *directions* ( e1-e1', e2-e2', e3-e3' ).

Fig.17 is **Krypton** orbits seen from the **upper** side.

**0.5** × de Broglie wavelength orbit is included between two closest **intersections**, like Neon and Argon.

*(Fig.18) Each orbit of Krypton's outer electrons.*

Considering the number of Kryoton valence electrons is "**18**", each **4** × de Broglie wavelength orbit contains **three** electrons and one hole (= 3 × 6 = 18 ).

When each orbit comes **closer** to another orbit, these two orbits tend to be *perpendicular* to each other.

( As shown in Biot-Savart law, the influence of de Broglie waves gets **stronger** in *inverse* proportion to the **distance** from de Broglie wave. )

*(Fig.19) e1 and e1' orbits form a "pair" crossing perpendicularly in opposite phases.*

Like Neon and Helium cases, each two orbits form a **pair**, in which these two orbits cross each other "**pependicularly**", and at these intersections, their wave phases are just opposite (= **±** ) to each other.

As a result, destructive intereference can be avoided, and wave phases at each intersection can be **neutralized** to be zero.

Basically the points at an **equal** distance between the two same phases (= + and + ) are just the opposite (= - ).
So considering Coulomb **repulsion**, other electrons tend to cross these **opposite**-phase points.

*(Fig.20) e1 and e2 orbits also cross perpendicularly in the "opposite" phases.*

As shown in Fig.17, when **0.5** de Broglie wavelength is included between two *intersections*, e1 and e2 ( next to e1 ) orbits with the "**opposite**" phases can also cross perpendicularly.

*(Fig.21) 4 × de Broglie wavelength orbits allow this crossing, too. ↓*

When each orbit is **4** ( or 5 ) × de Broglie wavelength, it includes **enough** wave "crest" and "trough" to make orbits *next to the next* (= e1 and e3 ) cross **perpendicularly** in the state of the **opposite** wave phases to each other.

As a result, e1 orbits can cross **all** other orbits ( e2, e3 and e1' - e3' ) perpendicularly in the states of the "**opposite**" wave phases to each other, when each orbit is **4** ( *NOT* 2 ! ) × de Broglie wavelength.

*(Fig.22) e1 and e1' cross perpendicularly + 4 × midpont lines = total 6 orbits !*

As shown in Fig.22, when e1 and e1' orbits cross **perpendicularly**, e2 and e3 orbits are just on the **different** *midpoint* lines (= zero phase ).

e2 and e3 have their own "**symmetrical**" *partner* orbits, e2' and e3', so the total is **four** midpoint lines are necessary in **6** orbits.
Because "perpendicular" e1 and e1' (= **2** ) + **4** midpoint lines become "**6**" orbits.

*(Fig.23) "e3" and "e1" orbits crash into each other in 2 × de Broglie wavelength.*

If we try to make **6** orbits using **2** × de Broglie wavelength orbits like Neon, we will find this is **impossible**, because the number of wave "*crest*" and "*trough*" is **too small** to **harmonize** all *6* orbital phases with each other.

First, if we try to do this, each orbital shape becomes "**unsymmetrical**" ( half wave phase is short, and another half is **long**, as shown in Fig.23 ).

When we suppose "e1" and "e2" (= next to e1 ) cross perpendicularly in the opposite phases, e1 and e3 (= next to the next ) electrons **crash** into each other !

Like Krypton, when each orbit crosses **all** other 5 orbits in the **opposite** wave phases *safely* (= **avoid** two electrons' **crash**, see Fig.19-21 ), their de Broglie wavelength must be **4** ( or 5 ), when **six** orbits are included.

*(Fig.24) To harmonize with all other orbits, 4 × de Broglie wavelength is indispensable.*

As shown in Krypton, when each orbit is **4** × de Broglie wavelength, each orbit can cross **any other** orbits in the **opposite** wave phases (= two same-phase electrons **don't** crash ), **harmonizing** with each other.

In Fig.24 left, *e1* crosses **e1'** orbit perpendicularly in the "**opposite**" phases.

In Fig.24 middle, *e1* crosses **e3** orbit perpendicularly in the "**opposite**" phases.

In Fig.24 right, *e1* crosses **e2** orbit perpendicularly in the "**opposite**" phases.

In case of **2** × wavelength like Neon, the number of wave "**nodes**" is **too small** to harmonize with *all other* 5 orbits in wave phases, which causes "**crash**" between two electrons somewhere.

*(Fig.25) To avoid strong Coulomb repulsion, "one hole" is necessary.*

Different from Neon and Argon case (= 8 electrons, hexahedron ), when the total number of outer electrons is "**18**", they **cannot** form regular polyhedron, so some **imbalance** is seen in their electrons' distiribtion.

For example, as shown in Fig.25, three electrons e1-e3 come closer to each other in the upper part of Krypton atom. In this state, "**e2**" electron has to enter **between** "**e1**" and "**e3**" in spite of their stronger Coulomb **repulsions**.

*(Fig.26) When 4 × de Broglie wavelength orbit contains only "3" electrons (= 1 hole ) *

If each 4 × de Broglie wavelength orbit contains only **3** electrons ( meaning the left one is empty "**hole**" ), these Coulomb repulsions can be **alleviated**.

For example, In Fig.26, "e2" electron can avoid entering the point between "e1" and "e3", thanks to its empty "**hole**". Alleviation of Coulomb repulsions happens **alternately** in different positions.

But if 4 × de Broglie wavelength orbits contains only 2 electrons (+ 2 holes ), its electron's number is **too small** to *maintain* this electrons' distribution and structure consisting of "**6**" orbits.

*(Fig.27) Xenon's orbit is longer than Krypton, so its shape gets "wider".*

As I said in Fig.2 and Fig.16, Xenon orbit is **5** × de Broglie wavelength.

So the total orbital number of Xenons becomes "**6**" like Krypton, because the *odd* "**7**" orbits **cannot** form symmetrical "*pairs*"

*(Fig.28) "e1" electron of Xenon crosses e1', e3 and e2 orbits in the "opposite" phases, safely.*

As shown in Fig.28, also when each orbit is **5** × de Broglie wavelength, its **number** of wave "crest" and "trough" is **enough** to make all six orbits "*harmonize*" and cross each other in the "**opposite**" phases.

In Fig.28 left, *e1* crosses **e1'** orbit perpendicularly in the "**opposite**" phases.

In Fig.28 middle, *e1* crosses **e3** orbit perpendicularly in the "**opposite**" phases.

In Fig.28 right, *e1* crosses **e2** orbit perpendicularly in the "**opposite**" phases.

*(Fig.29) *

Xenon's orbit is longer than Krypton, so the width of Xenon's orbit becomes **wider** than Krypton.

In any cases, **0.5** de Broglie wavelength is kept between two intersections.

*(Fig.30) Each orbit cannot be in "one plane" in multi-electrons.*

In these multi-electron atoms, each orbit **cannot** be in the *same one* plane like one-electron hydrogen.

So each orbital shape becomes more **complicated**, obeying the **total** effects of de Broglie waves ( ex perpendicular ) and Coulomb forces.

Of course, all six orbits must be **symmetrical** to each other in noble gases.

*(Fig.31) The total number of valence electrons of Rn is 32 (= 4 electrons × 8 orbits ). *

As **de Broglie** wavelengths increases, the **midpoint** lines (= zero phase ) increases in number, meaning they can contain **more** orbits *stably*.

In **6** × de Broglie wavelength orbit of Radon (= Rn ), the total **8** orbits can be included due to **increased** *midpoints*. So 4 electrons ( in each orbit ) × 8 orbits become **32** electrons in Rn atom.

*(Fig.32) Each Rn orbit contains 4 electrons and 2 holes.*

As I said in this section, as the number of electrons in one orbit becomes larger, **imbalance** in electron's distribution in all space becomes greater.

So in Radon case, we can think each orbit contains only **4** electrons ( + 2 holes ) to avoid stronger Coulomb repulsions and **imbalance** in its distribution.

Like Krypton case, each **two** orbits form a "**pair**", crossing perpendicularly and **safely** in the *opposite* wave phases.

e1-e4 ( or e1' - e4' ) electrons are moving in the **same** directions.

*(Fig.33) Radon outer electron's motion (= 6 × de Broglie wavelength ).*

Radon is **6** × de Broglie wavelength, so each orbit includes **enough** wave "*crest*" and "*trough*" to make all **8** orbits "**harmonize**" and cross each other in just the opposite phases.

In Fig.33 left, *e1* ( e2, e4 ) crosses **e3** ( e1, e1 ) orbit perpendicularly in the "**opposite**" phases.

In Fig.33 middle, *e1* crosses **e1'** orbit perpendicularly in the "**opposite**" phases.

In Fig.33 right, *e1* ( e3 ) crosses **e2** ( e1 ) orbit perpendicularly in the "**opposite**" phases.

As a result, we can prove **de Broglie** wavelength *determines* the **maximum** orbital number (= Pauli exclusion principle ), NOT depending on unrealistic spin, which magnetic moment is **too weak** to influence them.

In small atoms such as Helium and Neon, each electron can move **independently** in respect of electron's moving speed. But in large atoms such as Kr, Xe, Rn, their speed in the same orbit must be **equalized**.

Besically the *potential* energies of all electrons tend to be **equal** to each other, so all electrons' speed tend to be equal, too. When there are a little difference in their potential energies in the same orbit, it is thought faster electrons "push" slower electrons, equalizing their speed.

As shown in Compton effect, each **de Broglie** wave can exert some "**forces**", because they have "*momentum*"

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