*(Fig.1) ↓ Periodicity of ionization energies shows true valence electron number*

How can we know how many valence (= outer ) electrons each atom has ? We should look at the periodicity of the first ionization energies of atoms.

In atoms just after noble (= inert ) gases, ionization energies suddenly **drop**, which indicates their electrons enter **outer** *new* different orbitals.

From this change, we can know **true** numbers of valence electrons in each atom ( He = 2, Ne, Ar = 8, Kr, Xe = 18, Rn = 32 electrons ).

*(Fig.2) 4d and 3s, 6s and 4f orbitals are reversed in quantum mechanics !?*

Quantum mechanical orbital are self-**contradictory** in periodic table. In one-electron hydrogen atom, energy levels are higher in larger quantum number ( 3s < 4d < 4f < 5d < 6s ).

But in other multi-electron atoms, their orders are reversed ( 4s < 3d, 6s < 4f < 5d ) ! Schrodinger equation has NO ablity to predict energy levels in multi-electron atoms.

They just choose fake trial functions giving their desired artificial values. So meaningless and **useless**.

For example, though magnesium and singly-ionized alminum have exactly the *same* valence electrons, their orbital orders are **chaotic**.

*(Fig.3) Maximum numbers of orbits in noble gases*

As shown on this page, **de Broglie** wavelength influences the number of **maximum** orbital number in the periodic table.

The **maxium** number of orbits in Ne with 2 × de Broglie wavelength becomes "**4**" (= each 2 electrons × 4 orbits = **8** valence electrons ).

Krypton (= Kr ) with *4* × de Broglie wavelength has the maximum "**6**" orbits. Radon (= Rn ) with *6* × de Broglie wavelength has the maximum "**8**" orbits.

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.4) Quantum orbit satisfies an integer times de Broglie wavelength.*

Only Coulomb force is insufficient to explain why atomic energy levels are quantized, and why an electron doesn't fall into nucleus.

Bohr model succeeded in getting actual atomic energies, proposing each orbit is an integer times de Broglie wavelength, as well as Schrodinger's hydrogen .

An **integer** multiple of *de Broglie* wavelength means an electron can **avoid** destructive interference and be stable.

Without this de Broglie wave, each electron can be attracted to positive nuclei, until they **stick** to each other and its energy is **unlimitedly** lower !

So the repulsive force by electron's de Broglie wave is **strong** enough to keep the electron away from the nucleus and cause Pauli exclusion force.

*(Fig.5) Opposite phases of two de Broglie waves cancel each other*

In old Bohr's helium, **two** electrons are moving on the **opposite** sides of the nucleus in the *same* circular orbit (= **one** de Broglie wavelength ).

Considering Davisson-Germer interference experiment, *two* electrons of old Bohr's helium are clearly **unstable**.

**1**-de Broglie wavelength orbit consists of a pair of the opposite wave phases (= ±ψ ), which **cancel** each other by *destructive* interference.

Due to Coulomb repulsion between two electrons, one is always on the opposite side of another where the opposite de Broglie wave phases cancel each other.

Actually, old Bohr's helium of Fig.5 gives wrong ground state energy of helium, when you calculate it.

Old helium gives the total energy of **-83.33** eV, which is a little lower than the actual value of **-79.005** eV (= 1st + 2nd ionization energy of this ).

*(Fig.6) Actual Helium must avoid "destructive interference".*

If two 1 × de Broglie wavelength orbits are in the same plane in old Bohr's helium model, their **opposite** wave phases cause **destructive** interference and vanish.

To **avoid** *vanishing* de Broglie's wave, two electron orbits in actual helium must be perpendicular to each other. Each orbit is **one**-de Broglie wavelength.

If the two orbits are **perpendicular** to each other, their wave phases are **independent** from each other and can be **stable**, not canceling each other.

This helium model considering actual de Broglie wave interference just agrees with experimental results of all atoms !

*(Fig.7) 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** in all atoms.

*(Fig.8) All opposite wave phases just cross each other in Neon*

Neon contains **eight** valence electrons. Neon belongs to the 2nd line in periodic table, meaning its orbit is *2 × de Broglie* wavelength.

When just two 2 × de Broglie wavelenght orbits cross each other (= Fig.8 upper ), some electrons are **free** from their opposite wave phases.

With **four** orbits, **all** electrons and their opposite wave phases cross each other, forming stable noble gas Neon with no more room.

*(Fig.9) Each orbit contains 2 electrons and "1 hole" in Argon*

Argon (= Ar ) has **3** × de Broglie wavelength orbits which can contain up to **3** electrons.
But Argon has **8** valence (= outer ) electrons.

It means each orbit of Argon contains **2** electrons and *1 hole* ( 2 × **4 orbits** = total 8 electrons ).

When the number of Argon valence electron is 8, it makes symmetrical hexahedral structure.

*(Fig.10) 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.11) Four 3 × de Broglie wavelength orbits in Argon*

When each Argon orbit is 3 × de Broglie wavelength, how many orbits can Argon have ? Argon can have maximum **4** orbits.

When there are 4 orbits of 3 × de Broglie wavelength, all *opposite* wave phases just cross each other perpendicularly and become **stable** like helium.

Different from Neon with 2 × de Broglie wavelength orbit, Argon includes **4 holes**.

*(Fig.12) Periodic motion of Argon 8 valence elecrons*

This is periodic motion of Argon 8 valence electrons. Each electron ( or hole ) always crosses its opposite wave phase perpendicularly.

*(Fig.13) Total number of Ar valence electrons becomes 12 ( not 8 )*

*Argon* has 3 × de Broglie wavelength orbits, so each Ar orbit can contain up to **3** electrons. So total maximum electron number is 3 × 4 orbits is **12** ?

If Ar each orbit is fully loaded with 3 electrons, Coulomb *repulsions* among electrons are asymmetrically too **strong** ( all three layers include 4 electrons in Fig.13 upper ).

To **weaken** Coulomb repulsion, electrons ( each 2 ) have to be **removed** from upper and lower layers (= Fig.13 lower ). So when total electron umber of Ar is **8**, it is stable.

*(Fig.14) Two holes in oxygen form H2O molecular bond*

Oxygen (= O ) with **2** × de Broglie wavlength orbits has **4** orbits ( 2 electrons × 2 orbits + 1 electrons × 2 orbits = **6** valence electrons ). So oxygen has 2 holes.

Four 2 × de Broglie wavelength means even this oxygen structure must be close to tetrahedron like carbon.

Due to the asymetrical number "six", tetrahedral structure is a little distorted by Coulomb repulsion also in oxygen.

So its H2O bond angle 104.5^{o} is slightly smaller than pure tetrahedral 109^{o}

*(Fig.15) Sulfur (= S ) has 6 electrons and 6 holes. → H2SO4 is formed*

Sulfur (= S ) has **6** valence electrons, which means sulfur can bind to up to 2 electrons to form Argon-like structure ?

But as shown in sulfuric acid (= H2SO4 ), sulfur can bind to up to **4 oxygens** to form tetrahedral structure.

Because the sulfur with 3 × de Broglie wavelength orbit has **6 holes** enough to accept other atomic electrons and form molecular bond with 4 oxygens.

*(Fig.16) Because Oxygen has only "2 holes", and sulfur has "6 holes"*

Oxygen (= 6 ) also has 6 valence electrons like sulfur. But oxygen **cannot** bind to 4 oxygens to form molecules like tetrahedral H2SO4. Why ?

Because oxygen with **2** × de Broglie wavelength orbits has only **2 holes**, so there is **NO** room to accept valence electrons of 4 oxygens in order to form molecular bonds like sulfur.

Ozone hydrogen peroxide (= H2O5 ) has **nothing** to do with H2SO4

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

In atoms larger than Neon (= 4 orbits ), Krypton with 4 × de Broglie wavelength has **6** orbits.

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.*

Krypton with six 4 × de Broglie wavelength orbits has 18 valence 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.

*(Fig.19) All opposite wave phases just cross each other*

Krypton is in the fourth row of periodic table, which means it has **4** × de Broglie wavelength orbit.

If there are **6 orbits** with 4 × de Broglie wavelength, **all** their electron's opposite wave phases always cross
each other, and **cancel** their magnetic property

*(Fig.20) When Krypton contains 24 electrons, their repulsion is too strong*

Seeing the fourth row in periodic table including 18 elements, we can think noble gas Krypton has the maximum **18** valence electrons.

When Krypton with **4** × de Broglie wavelength orbits has **6** orbits, it can have the maximum **24** electrons (= 4 × 6 ) ? Unfortunately **not**.

Kryton orbitals consists of **four** *layers* in vertical direction, with each having six electrons, as seen in Fig.20 upper.

But in this case, each six electrons included in the **top** and **bottom** layers show **stronger** Coulomb repulsion than other two middle layers.

So to be *symmetric* in Coulomb repulsion, the top and bottom orbital layers must decrease electrons in their most stable state ( Fig.20 below ).

Like swing of a pendulum, electrons moving back and forth between unstable and **stable** states with maximum **18** electrons.

*(Fig.21) Each Rn orbit contains 4 electrons and 2 holes.*

Radon (= Rn ) with eight 6 × de Broglie wavelength orbits has 32 valence electrons.

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.

*(Fig.22) 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.22 left, *e1* ( e2, e4 ) crosses **e3** ( e1, e1 ) orbit perpendicularly in the "**opposite**" phases.

In Fig.22 middle, *e1* crosses **e1'** orbit perpendicularly in the "**opposite**" phases.

In Fig.22 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.

*(Fig.23) If Rn contains 6 electrons × 8 orbits = 48 electrons ..*

Each orbit of Radon (= Rn ) is 6 × de Broglie wavelength, so each Rn orbit can contain up to **6** electrons. Radon has **8** orbits, then total is 6 × 8 orbits = **48** electrons ?

But the number of Radon valence electron is only 32. Why ? Radon consists of *6 layers* like Fig.23 right ( Sorry, this picture of Rn is a little thinner than actual one ).

If each orbit contains maximum **6** electrons, especially in upper and lower layers, Coulomb repulsions among electrons become too **strong**.

So 48 valence electrons is **impossible** in Radon.

*(Fig.24) Each orbit of Radon contains 4 electrons + 2 holes*

When each orbit of Radon contains up to **4** electrons and *2 holes*, Coulomb repulsion among electrons can be weakened, then stable.

As shown in Fig.24 right, interelectronic repulsion especially upper and lower 2 layers is asymmetrically too **strong**, compared to middle layers.

If we decrease 6 fully loaded electrons to **4** in upper and lower 2 layers, the whole Rn electron distribution becomes *symmetric* and normal in their Coulomb repulsion.

This is the reason why Radon has **32** (= 4 × 8 orbits ) valence electrons instead of 48 electrons.

2016/9/23 updated. Feel free to link to this site.