Top page (correct Bohr model including the two-electron atoms).

Strange "Spin" is NOT a real thing.

Special relativity is wrong.

- Quantum mechanical atomic radius is NOT useful.
- Radius of carbon-like atoms ( C, Si ).
- Radius of oxygen-like atoms ( O, S ).
- Radius of boron-like atoms ( B, Al ).
- Radius of nitrogen- and fluorine-like atoms ( N, F, P, Cl ).
- Electronagetivity can be explained by true radius.

*(Fig.1) Orbital radius by quantum memcanical hydrogen ?*

As shown in Fig.1, if we calculate average orbital radius of Schrodinger'd hydrogen, its value becomes about 1.5 × Bohr radius.

This result is **different** from that gotten from **Virial** theorem.

*(Fig.2) Virial theorem of hydrogen → Bohr radius.*

As shown in this page, Schrodinger's hydrogen also **satisfies** Virial theorem, and its radius is just **Bohr radius** like classical Bohr model.

This means what we call atomic radius of Hartree-Fock orbital is **inconsistent** with Virial theorem.

This inconsistency originates in **unrealistic** probability **spreading** of quantum mechanical model.

Also in 1s hydrogen wavefunction, the probability density at Bohr radius becomes maximum.

But their average value is different from Bohr radius.

*(Fig.3) Difference between two atomic radii.*

When two atoms of the same kind are bonded through a **single** bond in a neutral molecule, one **half** of that bond length is referred to as the covalent atomic radius.

For example, the internuclear distance of H2 molecule is 0.74 angstrom, so hydrogen atomic radius becomes 0.74 × 0.5 = **0.37** angstrom.

These covalent atomic radii are gotten from experiment, and have **nothing** to do with orbital radii by quantum mechanical methods.

Actually, as shown in Fig.3, orbital radius by quantum mechanical approach is completely **different** from the actual atomic radius.

*(Fig.4) Atomic radius of hydrogen and helium.*

When we think about the relations between covalent atomic radius and experimental molecular bonds, the **actual configurations** of molecules are very important.

For example, in hydrogen molecule (H2), two orbits of hydrogen atom need to be almost **parallel** to each other, considering their energetically **stable** states.

So 2 × atomic radius is not equal to bond length in H2 molecule.

As shown in this page, we can get the correct energy states using classical orbits based on **Coulomb** and **de Broglie** relations in H2 molecule.

On the other hand, the quantum mechanical models **cannot** give clear and realistic atomic pictures at all.

In helium atom, the atomic orbital radius using new Bohr's helium (= 0.30 Å ) is almost **same** as experimental covalent atomic radius (= 0.32 Å ).

This agreement shows actual helium configuration.

*(Fig.5) Ethane molecular model. *

Single C-C bond length is 1.54 angstrom, so covalent atomic radius becomes 1.54 × 0.5 = 7.7 angstrom.

But as shown this page or this page, real rotating radius of carbon valence electron becomes **0.64** angstrom (= 0.6400 × 10^{-10} meter ), when we consider Coulomb and **de Broglie** wavelength.

This value 0.64 is a little smaller than 7.7.

So covalent atomic radius (= 0.77 ) itself does not mean real orbital radius (= 0.64 ).

In this page, we show Bohr orbit ( de Broglie wavelength ) based methods can give **excellent** atomic radius, which **agrees** with experimental results such as electronegativity.

*(Fig.A-1) σ bond = two electrons are going back and forth between two H atoms ?*

According to the quantum mechanics and Pauli exclusion principle, when two electrons of opposite spins enter one orbital, it becomes stable, they insist.

So in σ covalent bond, two electrons should belong to **closed** 1s orbitals in left or right H atoms, **as one** .

This means two electrons are always going **back and forth** as a pair between two H atoms ?

Unfortunately, each H nucleus has **NO** power to attract **two** nagetive electrons **at the same time**.

So this idea based on Pauli exclusion principle and closed orbital is **unrealistic**, and **neglects** basic Coulomb interaction.

*(Fig.A-2) C-C bond = two electrons are going back and forth ?*

C-C bond is very long (= 1.5400 angstrom ).

In spite of this **long** bond length, two electrons **must** always form **a pair** to satisfy Pauli excluion principle ?

Considering **realistic** Coulomb interaction, it is quite natural that each electron belongs to **each** Carbon nucleus, **not** forming a pair. ( See Fig.7. )

The important point is that quantum mechanics **cannot** tell **where** electrons exist inside molecular bond.

They are like "**black box**".

So quantum mechanical molecular bonds, which cannot tell any **concrete** informations, are completely **useless**, when we think about larger molecules.

*(Fig.A-3) An electron is spreading as sphere in QM ?*

According to quantum mechanics, we **cannot** designate a **single** electron in hydrogen atom.

So a single elecctron of quantum mechanics is always spreading as wavefunction.

For example, when a single electron is spreading as **sphere**, we can consider that electron exists at the **center** of that sphere, from the viewpoint of potential energy.

If the electron is at the center, it just **overlaps** +e nucleus, and H atom completely becomes **neutral**.

So that H atom **cannot** interact with other atoms at all.

So an electron in H atom needs to be **localized** at some place like Bohr model, when it can interact with other atoms.

*(Fig.A-4) H-H bond = two electrons are avoiding each other.*

In hydrogen molecules, two electrons tend to be attracted to other nuclei.

But if we try to express this state using wavefunction sphere, they are repelling each other, as shown in Fig.A-3 lower figure.

On the other hand, if we use classical orbits, we can express H molecule **naturally**.

In this natural model, two electrons are avoiding each other (= **up** and **down** ), attracted to other positive nuclei, which can form covalent bond.

In conclusion, we need to use concrete **localized** electron in various molecules to express covalent bonds.

**Electronegativity** ( Pauling ) is the power of an atom in a molecule to **attract** nagative electrons to itself.

This electronegativity by Pauling is the average value gotten by comparing various **experimental** bond energies.

The electronegativity of hydrogen atom (H) is **2.20**, and that of carbon (C) is **2.55**.

Attracting powers of H and C atoms are almost **same**.

*(Fig.6) "Force 1" acting on valence electron (= e0 ) in methane.*

As shown in this page and this page (Fig.22), the rotating radius of four valence electrons of carbon is about **0.6400** × 10^{-10} meter, when we consider the total 1-4 ionization energies and "**2**" de Broglie wavelength orbit.

As a result, when valence electron e0 is closest to hydrogen nucleus, their distance becomes about **0.4500** angstrom, by **1.0900** (= C-H ) - 0.6400 .

At this point, ( Coulomb ) "**force 1**" is acting between a valence electron **e0** of carbon and H0 nucleus.

*(Fig.7) Periodic motion of ethane ( CH3-CH3 ).*

Four valence electrons of carbon are arranged as **tetrahedral** structure considering their symmetric distribution.

Each electron is actually **moving** around each centeral carbon nucleus, so Fig.7 upper and lower patterns are repeated periodically in ethane, if we try to know "**real**" motions in ethane. ( See also this page. )

*(Fig.8) "Force 2" acting on valence electron (= e0 ) in ethane.*

Fig.8 is the x-z cross-sectional view of ethane in Fig.7.

C-C bond length of ethane is about **1.5400** angstrom, and distance between valence electron e0 and C0 nucleus is about 0.6400 angstrom, when we consider "**2**" de Broglie wavelength orbit like methane.

As a result, the distance between e0 and another carbon nucleus C1 is about **0.9000** angstrom (= 1.5400 - 0.6400 ).

And central carbon nucleus (= C1 ) has about **+4e** positive charge ( +6e - 2e ), when two 1s negative charges are subtracted from +6e carbon nucleus.

"**Force 2**" is the force acting between this **e0** electron and C1 nucleus.

*(Fig.9) Coulomb forces 1, 2 in methane and ethane are almost same.*

From Fig.6 (= methane ) and Fig.8 (= ethane ), we can know the forces acting between e0 valence electron and another nucleus ( H0 or C1 nucleus ).

Using Coulomb inverse square law, "force 1" in methane proves to be almost **same** as "force 2" in ethane !

This result agrees with almost **same electronegativity** of H and C atoms.

And the result of Fig.9 proves that both in methane and ethane, the idea of Coulomb relation and de Broglie wavelength are very **important**.

In this section, we review the methods of obtaining real atomic radius of **carbon**-like atoms (= C, Si .. ) based on **Coulomb** and **de Broglie** relations.

*(Fig.10) Tetrahedral structure of carbon-like atoms.*

Considering symmetric structures of **four** valence electrons of carbon-like atoms, it is natural that those four electrons are distributed **tetrahedrally** around nucleus.

Actually, methane (= CH4 ), tetrafluoromethane (= CF4 ), silane (= SiH4 ), and siilicon tetrafluoride (= SiF4 ) are known to have **tetrahedral** structure.

This means the concepts of "s" and "p" ortbitals in quantum mechanics are completely **useless**, and **NOT** real.

They try to introduce the new concept of "sp3" orbitals, but this manipulation is very **artificial**, I think.

These unrealistic electron distributions such as "s", "p" and "d" **prevent** us from expressing various molecular bonds by real and clear picture.

*(Fig.11) Valence electrons of carbon-like atoms.*

Here, using valence electrons' distribution of Fig.11, we compute "real" atomic radius and central positive charge Z.

The length of one side in tetrahedron ( Fig.11 ) is supposed to be"**2r**".

The total potential energy V is the sum of attractive V among +Ze nucleus and electrons, and repulsive V among four outer electrons.

*(Eq.1) Total potenial energy V.*

The first term of Eq.1 is potential energies between Ze nucleus and each electron.

And the second term of Eq.1 is the repulsive potential energies among four valence electrons.

Tetrahedron has six sides, so there are **six** interactions among electrons.

*(Eq.2) Total kinetic energy T.*

Total kinetic energy is the sum of kinetic energies in four valence electrons.

Using Virial theorem, this T can be gotten from total V. ( T = -1/2 V = -E. )

From the relation of Eq.2, we obtain the **velocity** "v" and de Broglie **wavelength** (= h/mv ) of each electron.

*(Eq.3) Force F acting on each electron.*

Eq.3 means force F acts on each electron.

The first term of Eq.3 is the attractive force by nucleus.

And the second term is repulsive forces from three other electrons.

Using the above velocity (v) ( of T ) and force F, we suppose the centrifugal force is equal to the force (F), as follows,

*(Eq.4)*

This r_{1} is "*temporary*" radius, which appears by this relation.

Using this r_{1}, we can calculate the number of **de Broglie's waves** (= *wn* ) contained in one orbit, as follows,

*(Eq.5)*

Sample JAVA program ( atomic radius ).

C language program ( atomic radius ).

In this program, first we choose the kind of atoms.

If you compute carbon like atoms (= C, Si ), input the valiue of "0", first.

Next input the central charge Z and total 1-4 ionization energies of all valence electrons.

From these inputted values, this program computes potential energies and force of tetrahedral carbon using Eq1 - Eq.5.

And using Virial theorem ( E = -T = 1/2 V ), this program calculate **de Broglie wave** in one orbit

In the screen, this de Broglie wave's number (= almost 2.0 ), and the distance between each valence electron and the central nucleus are shown.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Carbon (C) | 11.260 | 24.383 | 47.888 | 64.494 | - | - | - |

Silicon (Si) | 8.151 | 16.346 | 33.493 | 45.142 | - | - | - |

For example, the sum of 1-4 ionization energies in carbon is **148.025** eV.

When Z = *+4.0*, this wn becomes 1.87 (almost *1.9*) and rotating radius is almost 6000 MM. ( 1 MM = 10^{-14} meter. )

And when Z = **+4.22**, this wn becomes just **2.00** and rorating radius is almost **6400** MM.

This little difference between 4.0 and 4.22 is caused by the fact that two inner 1s electrons are **apart** from +6e carbon nucleus.

And this result agrees with those obtained from the program below, in which four valence electrons' motions are actually computed.

Sample JAVA program -- Visualization of carbon four valence electrons. ( central charge +4.22. )

After running the program above, and choose "1.0" or "2.0" de Broglie wave in the scrollbar, and click "orbit" button, you can understand when Z = **4.22**, one orbit is just **2.0** de Broglie wavelength, and its rotating radius (= nuc ) is about **6400** MM (= 0.6400 × 10^{-10} meter. )

See also this page.

*(Fig.12) Real atomic radius of carbon ( C ).*

So we can define the **real** atomic radius of carbon as 0.6400 angstrom and the central positive charge as **+4.22**.

As I say later, these results **agree** with experimental results of bond length and electronegativity very well.

Next, the sum of 1-4 ionization energies in silicon ( Si ) is **103.132** eV.

Outer electrons of silicon is in n =3 ( **three** de Broglie wavelength ) orbits, so this value is smaller than n=2 carbon (= 148.025 ).

After running program above, choose "0" ( = carbon-like atoms ), in the same way.

And when Z = **+5.05**, de Broglie waves in one orbit becomes just **3.00** and rorating radius is almost **1.1500** angstrom.

*(Fig.13) Real atomic radius of silicon ( Si ).*

The atomic radius of silicon (Si) is bigger than that of carbon due to "**3**" de Broglie wavelength.

Actually, Si-H bond length is **1.4800** angstrom is much longer than C-H bond length (= 1.0900 angstrom ).

And due to **wide** distribution of 8 × n=2 middle electrons of silicon, central positive charge (= +5.05 ) is bigger than +4.

*(Fig.14) Long atomic radius of silicon ( Si ) → unstable.*

In **silane** (= SiH4 ), the hydrogen atoms have a partial **negative** charge and the silicon has a **positive** charge due to the difference between their **electronegativity**.

This property can be easily understood using new atomic radius based on de Broglie wavelength.

The distance between Si and its valence electron e0 is **longer** than that between e0 and H nucleus, which causes **negative** hydrogen atom.

And due to long distance of Si-H bond and the weak attracting power of silicon, silane (= SiH4 ) is much **unstable** than methane (= CH4 ).

Silane gas is easily **pyrophoric**, and it undergoes spontaneously combustion in air without the need for external ignition.

*(Fig.15) New atomic radius can explain experimental Si-Si bond length.*

The experimental value of single Si-Si bond length is **2.3300** angstrom.

New atomic radius of silicon based on de Broglei wavelegnth is 1.1500 angstrom.

2 × 1.1500 = **2.3000** is near the experimental bond length of 2.3300 angstrom.

So these results almost agree with the actual values.

*(Fig.16) Oxygen-like atom + hydrogen. *

Oxygen -like atoms (= O, S, Se, Te ) has **six** valence electrons.

So it is natural that these six electrons are distributed **octahedrally** considering Coulomb **repulsive** forces among them.

Actually, *H-S-H, H-Se-H*, and *H-Te-H* angles are **92.1, 91.0**, and **89.5 degrees**, respectively. ( = almost **90** degrees).

In the case of oxygen, the hydrogen atom is **closer** to the oxygen atom than other oxygen-like atoms.

So, the **repulsive** force by hydrogen nuclei is so strong that the water (H2O) can **not** keep "octahedral" structure to make O nucleus stable.

*(Fig.17) Sulfur hexafluoride, SF _{6} = octahedron. *

According to the quantum mechanics, sulfur (S) has **six** valence electrons like oxygen atom.

( 3s × 2 + 3p × 4 = 6. )

And stable **sulfur hexafluoride** is known to have just octahedral structure, in which F-S-F angle is just **90** degree.

This result clearly indicates, 3s^{2} 3p^{4} state claimed by quantum mechanics is **wrong** and useless, and symmetric repulsive **Coulomb** force among six electrons is right.

*(Fig.18) Oxygen-like atomic structrure.*

Here we compute **real** atomic radius of oxygen-like atoms using octahedral structure of Fig.18.

Considering Coulomb interactions among electrons and nucleus, total potenial energy V becomes

*(Eq.6) Total potential energy.*

In Eq.6, the first term is attractive potential energy between +Ze nucleus and six electrons in Fig.18.

And the second and third terms are repulsive V among six valence electrons.

*(Eq.7) Total kinetic energies.*

Using Virial theorem ( T = -1/2 V ), we can know total kinetic energy, each electron's velocity "v" and de Broglie wavelength (= h/mv ).

Total kinetic energy is the sum of six electrons.

*(Eq.8) Force F acting on each electron.*

F of Eq.8 is the total attracting force acting on the electron "**e1**".

The first term of Eq.8 is the attracting force by +Ze nucleus.

The second term is repulsive force by e0, e2, e4 and e5, and the third term is the force by e3, as shown in Eq.8.

In the same way of carbon-like atoms, using Eq.4 and Eq.5, we obtain de Broglie waves in one orbit.

Sample JAVA program ( atomic radius ).

C language program ( atomic radius ).

If you compute oxygen like atoms (= O, S ), input the valiue of "1", first.

Next input the central charge Z and total 1-6 ionization energies of all valence electrons.

From these inputted values, this program compute potential energies and force of octahedral oxygen using Eq.6 - Eq.8 and Eq.4 - Eq.5.

On the screen, this de Broglie wave's number, and the distance between electron and nucleus are shown.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Oxygen (O) | 13.618 | 35.121 | 54.936 | 77.413 | 113.899 | 138.120 | - |

Sulfur (S) | 10.360 | 23.338 | 34.790 | 47.222 | 72.595 | 88.053 | - |

The sum of 1-6 ionization energies in oxygen is **433.107** eV.

When Z = *+6.0*, this wn becomes 1.88 (almost *1.9*) and rotating radius is almost 4300 MM. ( 1 MM = 10^{-14} meter. )

And when Z = **+6.273**, this wn becomes just **2.00** and rorating radius is almost **4600** MM.

*(Fig.19) Real oxygen atomic radius.*

This small difference between 6.0 and 6.273 is caused by the fact two inner 1s electrons are apart from +8e oxygen nucleus.

( This influece is prominent especially in 5th and 6th ionization energies. )

These values agree with experimental phenomena.

*(Fig.20) Coulomb forces 1, 2 in O-H and O-C bonds.*

As is gotten above, real atomic radius of valence electron of oxygen is **0.4596** angstrom, when we consider 1-6 total ionization energies and their **two de Broglie** wavelength orbits.

**O-H** bond length of water is **0.9584** angstrom, and the average single **O-C** bond length is known to be **1.4300** angstrom.

Subtracting the atomic radius of 0.4596 from these bond lengths, we can know Coulomb forces 1,2 between electron and another nucleus.

*(Fig.21) Coulomb forces 1, 2 in O-H and O-C bonds are almost same.*

H nucleus is **+e** and, C nucleus is almost **+4e** (= +6e - 2e ), subtracting two inner 1s electrons from +6e nucleus.

As a result, Coulomb forces 1 and 2 between electron and another nucleus are almost **same** in O-H and O-C bonds.

Because the electronegativity of H (= **2.20** ) and C (= **2.55** ) atoms are almost same.

But due to strong attracting power of O atom, e0 of H2O is more attracted toward O nucleus than O-C bond in Fig.20. ( See Fig.21. )

These results demonstrate that valence electrons of oxygen in O-H and O-C bonds are moving obeying the **common** physical principle, and
**Coulomb** and **de Broglie** relations are **effective** also in these cases.

*(Fig.W-1) Electron distribution change in H2O *

As shown in Fig.W-1, if H-O-H angle (= 104.45 degree ) is **bigger** than 90 degree of octahedral structure, e0 and e1 move **outward**, attracted to hydrogen nuclei.

As a result, e0 - O - e1 angle becomes **wider** than 90 degree.

**Following** this formal change, e4 - O -e5 angle becomes **smaller** than original 180 degree.

As a result, **repulsive** force by hydrogen nuclei is **cancelled** out by these configuration change.
( See this page. )

Because e4 and e5 **pull** O nucleus toward H atoms.

*(Fig.W-2 ) Electron configuration change is blocked, when H-O-H is 90 degree.*

On the other hand, if H-O-H angle is **90** degree, electron e0 and e1 **cannot** move outward, so e4 and e5 **cannot** move, either.

Because, O-e0 and O-el lines are on the **same** lines of O-H0 and O-H1 from the **beginning**.

So configuration change to cancel repulsive force of H nuclei does **NOT** happen.

Furthermore, this repulsive force by H nuclei becomes **stronger** when 90 degree than that of 104.45 degree, due to its sharp angle.

So the **structual** change to cancel these repulsive force of H nuclei is the reason of this angle.

*(Fig.W-3) Sulfur tetrafluoride ( SF4 ).*

Similar configuration change is seen in sulfur tetrafluoride (= SF4 ).

Sulfur (= S ) has six valence electrons like oxygen.

Due to the strong electronegativity of fluorine, sulfur's electrons are attracted toward fluorines and one side of electron density of S becomes weaker, which causes configuration change.

This mechanism is a little different from H2O above, but their configuration change is similar.

( The important difference between fluorine and hydrogen is seven valence electrons of F tend to **repel** electrons of other atoms, as the number of valence electrons in other atoms increases, while H tends to be attracted to them. )

*(Fig.22) Real atomic radius of sulfur ( S ).*

Next, the sum of 1-6 ionization energies in sulfur ( S ) is **276.358** eV.

Outer electrons of sulfur is in n =3 ( three de Broglie wavelength ) orbits, so this value is smaller than n=2 carbon (= 433.107 eV ).

After running program above, choose "1" ( = oxygen-like atoms ), in the same way.

And when Z = **+7.18**, de Broglie waves in one orbit becomes just **3.00** and rorating radius is almost **0.8622** angstrom.

As I said above, middle n=2 eight electrons are spreading more **widely** than n=1 electrons.

So effective central charge (= + 7.18 ) is bigger than +6.

*(Fig.23) C-C and C-S bond length ( experimental values ).*

The exprimental values of C-C and C-S bond length are **1.5400** and **1.8200** angstrom, respectively.

The **electronegativities** ( Pauling ) of carbon (= **2.55** ) and sulfur (= **2.58** ) are almost **same**.

The electronegativity represents the power of attracting an electron of another nucleus.

So we investigate the forces acting on e0 (= carbon valence electron ), because this e0 is the **target** electron of ionization.

*(Fig.24) Attracting powers of C and S are almost same.*

We compute the Coulomb forces ( f1 and f2 ) between e0 and another nucleus ( C or S ) in Fig.23 and Fig.24.

The results shows the powers of attracting electron are almost **same** in carbon and sulfur.

This means these classical orbits agree with almost **same electronegativities** in C and S atoms ( 2.55 and 2.58 ).

*(Fig.25) BF3 = triangular.*

Boron (= B ) has **three** valence electrons.

If classical Coulomb force is always effective, the compounds of boron must be **triangular**.

Actually, boron trifluoride (= BF3 ) is known to be **trigonal planar**.

So the original "s" and "p" orbitals **cannot** be used also in boron-like atoms.

They try to introduce the new concept of sp2 hybrid orbitals temporarily, but this means original wavefunctions are completely **unuseful** in various molecules.

*(Eq.9) Total potential energy V of boron - like atoms..*

We suppose three valence electrons of boron-like atoms are distributed **triangularly**.

Considering Coulomb interactions among electrons and nucleus, total potenial energy V becomes like Eq.9.

In Eq.9, the first term is attractive potential energy between +Ze nucleus and three electrons.

And the second term is repulsive V among three valence electrons.

*(Eq.10) Total kinetic energies.*

Using Virial theorem ( T = -1/2 V ), we can know the total kinetic energy and each electron's velocity.

Total kinetic energy is the sum of three electrons.

*(Eq.11) Force F acting on each electron.*

F of Eq.11 is the total attracting force acting on an electron.

The first term of Eq.11 is the attracting force by +Ze nucleus.

The second term is repulsive force by other two electrons.

In the same way of other atoms, using Eq.4 and Eq.5, we obtain de Broglie waves in one orbit.

Sample JAVA program ( atomic radius ).

C language program ( atomic radius ).

If you compute boron like atoms (= B, Al ), input the valiue of "2", first.

Next input the central charge Z and total 1-3 ionization energies of all valence electrons.

From these inputted values, this program compute potential energies and force of triangular boron using Eq.9 - Eq.11 and Eq.4 - Eq.5.

On the screen, this de Broglie wave's number, and the distance between electron and nucleus are shown.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Boron (B) | 8.298 | 25.155 | 37.931 | - | - | - | - |

Aluminum (Al) | 5.986 | 18.828 | 28.448 | - | - | - | - |

The sum of 1-3 ionization energies in boron is **71.384** eV.

When Z = **+3.22**, this wn becomes 2.00 and rotating radius is **7996** MM. ( 1 MM = 10^{-14} meter. )

*(Fig.26) Real boron atomic radius.*

As shown in Fig.26, rotating radius of boron is very **long** (= about **0.8000** angstrom ) for its small central charge (= +3.22 ).

This very long atomic radius is related to **small** electronegativity of boron (= **2.04** ).

*(Fig.27) Borane (= BH3 ) is very unstable. *

The borane is composed of one boron and three hydrogens.

But this borane molecule is very **unstable**.

Because as shown in Fig.27, valence electron "e0" of boron tend to be **attracted** to hydrogen nucleus due to its long radius.

On the other hand, B-H bond length is **long** and boron nucleus has **weak** attracting power ( Z = 3.0 ), hydrogen electron and nucleus **cannot** come close to boron nucleus.

In this process, hydrogen electron ( accompanied by its nucleus ) is **repelled** by e0 electron, and they are **separated** from each other.

*(Fig.28) Unusual bond of Diborane (= B2H6 ). *

Diborane (= B2H6 ) is colorless gas, and ignite spontaneously in moist air at room temperature.

But this diborane is more stable than borane.

Boron tends to form unusual bonds such as electron-**deficient**.

In Fig.28, green ball is boron valence electrons, and red is hydrogen electron.

Boron's electrons tend to be attracted to two positive H nuclei of center, outer hydrogen's electrons also tend to be **attracted** toward ( slightly positive ) boron, and their bond strength becomes a little stronger.

( Unfortunately, quantum mechanics **cannot** give concrete mechanism of diborane. )

*(Fig.29) Real atomic radius of aluminum ( Al ).*

Next, the sum of 1-3 ionization energies in aluminum ( Al ) is **53.262** eV.

Outer electrons of aluminum is in n =3 ( three de Broglie wavelength ) orbits, so this value is smaller than n=2 boron (= 71.384 eV ).

After running program above, choose "2" ( = boron-like atoms ), in the same way.

And when Z = **+4.0**, de Broglie waves in one orbit becomes just **3.00**, and rorating radius is almost **1.3880** angstrom.

As I said above, middle n=2 eight electrons are spreading more **widely** than n=1 electrons.

So effective central charge (= + 4.0 ) is bigger than +3.

*(Fig.30) Weak electronegativity → aluminium hydride = crystal.*

Aluminum (Al) has very **long** rotating radius (= 1.3880 angstrom ).

So its valence electrons easily **leave** aluminum, which means aluminium easily becomes **ionized** to be stable.

For example, in aluminium hydride, the distance between Al nucleus and its valence electron is much longer than that between H nucleus.

So alminum hydride becomes **crystal** and their valence electrons are **moving around** to make crystal's bonds strong.

( The electronegativity of alminum is **1.61**, which is smaller than 2.20 of hydrogen. )

It can be said that diborane of Fig.28 is **incomplete** version of crystal.

*(Fig.31) Distribution of five valence electrons in nitrogen (central charge = +5.25e)*

In the nitrogen atom, it is *much more difficult* to visualize the electrons' motions than carbon atom.

As the nitrogen has *five* valence electrons, it **cannot** be expressed as a *regular polyhedrons* like carbon (tetrahedron), oxygen (octahedron).

**Phosphorus pentafluoride** (PF5) is known to have structure like Fig.31.

( Phosohorus has five valence electrons like nitrogen. )

So we suppose five valence electrons of nitrogen are distributed like F of PF5.

Sample JAVA program (distribution of N electrons)

Here we use sample program above to seek symmetric distribution of nitrogen.

Also in the nitrogen atom (N), the two 1S electrons are a little *apart* from the +7e nucleus of N.

So the "effective" central charge of this figure is a little bigger than the +5e ( = +7e -2e ).

First, we use **+5.25e**, which is *0.25e+* bigger like the carbon case.

( This central charge Z can ne changed freely, input arbitrary value into the designated textbox. )

*(Fig.31') Distribution of five valence electrons in nitrogen (central charge = +5.25e)*

In this program, the nitrogen nucleus is gray.

Here we use the new units, ( 1 MM = 10^{-14} meter).

Each coordinate of electrons (+X (MM), +Y (MM), +Z (MM)) in the text box means "relative" positions from this nucleus.

"nuc (MM)" means the distance between the nucleus and each electron.

You can change "nuc" by entering the values into the designated text box, and press the Enter key.

*(Fig.32) Forces acting on electron and nucleus.*

V (eV) and T (eV) mean the potential and kinetic energies of each electron.

CF means the force toward the nucleus acting on each electron.

(fx, fy, fz) mean the each component of the "remaining" force other than CF.

So when five valence electrons of nitrogen is **symmetrically** distributed around nucleus, these (fx, fy, fz) should be close to zero.

( FX, FY, FZ ) mean each component of the force acting on the nucleus.

For the stability, these FX, FY, FZ should be close to zero, too.

*(Fig.33) Force = 1000.*

The unit of force: *1000* = the force between +e nucleus and an -e electron which are Bohr radius apart from each other.

"tV" is the total potential energy (eV).

*Waves (wn)* means the number of de Broglie's waves contained in one orbit of each electron.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Nitrogen (N) | 14.534 | 29.601 | 47.449 | 77.474 | 97.890 | - | - |

According to the *Virial theorem*, "average" total potential (V) and kinetic (T) energies satisfy the relation,

V = 2 × E = -2 × T

The total energy (E) of the five valence electrons of nitrogen is *-266.948 eV* (= the sum of the 1-5th ionization energies), so this potential energy (V) becomes 2 × -266.948 = **-533.89 eV**.

When the valence electrons are distributed "**symmetrically**" around the nucleus, the distances between electrons and nucleus (nuc) need to be the same in all valence electrons.

And the forces acting on electrons *need to be toward the nucleus* in the equlibrium state.

Based on total force (F) acting on each electron, we define "*temporary*" radius (r_{1}) like Eq.4 and Eq.5, as follows,

*(Eq.12)*

This means the centrifugal force is supposed to be equal to F.

"v" is the electron's velocity which value can be gotten from each kinetic energy.

And based on these values, we can find the values (*wn*) of de Broglie's waves, which is contained in one orbit of each electron, as follows,

*(Eq.13)*

As shown in Table. 5, when the total potential energy (*tV*) of nitrogen valence electrons is ** - 534.05 eV**, this wn (Waves) of each electron becomes almost **2.0**.

eNo. | +X (MM) | +Y (MM) | +Z (MM) | nuc (MM) | CF | fx | fy | fz | Waves |
---|---|---|---|---|---|---|---|---|---|

ele 0 | 2666 | -4617 | 0 | 5331 | 3904 | 0 | 0 | 0 | 1.998 |

ele 1 | 2666 | 4617 | 0 | 5331 | 3904 | 0 | 0 | 0 | 1.998 |

ele 2 | 0 | 0 | 5332 | 5332 | 3878 | 0 | 0 | 0 | 1.992 |

ele 3 | -5332 | 0 | 0 | 5332 | 3904 | 0 | 0 | 0 | 1.998 |

ele 4 | 0 | 0 | -5332 | 5332 | 3878 | 0 | 0 | 0 | 1.992 |

nuc FX | nuc FY | nuc FZ | total V (eV) | Charge Z | Distance (MM) |
---|---|---|---|---|---|

0 | 0 | 0 | -534.05 | Z = 5.25 | nuc= 5332 |

You can change the coordinates (+X, +Y, +Z), nuc (MM) and charge Z freely.

Table 5 shows that the all five valence electrons of nitrogen are moving on the orbits of "**2**" de Broglie's wavelength.

In this case, each component of the force acting on the nucleus becomes ( FX, FY, FZ ) = ( 0, 0, 0 ), which means *stable* nucleus.

Of cource, (fx, fy, fz) of each electron needs to be very small (= zero ) in the equilibrium state.

But nitogen with five electrons **cannot** be completely symmetric.

( In Fig.31 model, bond angles of **90** and **120** degree are mixed. )

*(Fig.34) Real atomic radius of nitrogen.*

As a result, real atomic ( rotating ) radius of nitrogen valence electrons become **0.5332** angstrom.

And the central charge is **+5.25**.

( The small difference 0.25 is due to 1s electrons, which are apart from +7e nucleus. )

Next we think about the phosphorus (P) atom, which also has **five** valence electrons in n=3 orbital.

We can use the sample program above, also in this case.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Phosphorus (P) | 10.487 | 19.770 | 30.203 | 51.444 | 65.025 | - | - |

The sum of 1-5 ionization energy of phosphorus is **176.929 eV**.

According to Virial theorem, the total potential energy V becomes 2 × 176.9 = **-353.8 eV**.

Input "**6.12**" into charge Z textbox, and "**9825**" into nuc textbox after running the program of nitrogen.

eNo. | +X (MM) | +Y (MM) | +Z (MM) | nuc (MM) | CF | fx | fy | fz | Waves |
---|---|---|---|---|---|---|---|---|---|

ele 0 | 4912 | -8508 | 0 | 9825 | 1402 | 0 | 0 | 0 | 2.996 |

ele 1 | 4912 | 8508 | 0 | 9825 | 1402 | 0 | 0 | 0 | 2.996 |

ele 2 | 0 | 0 | 9825 | 9825 | 1394 | 0 | 0 | 0 | 2.988 |

ele 3 | -9825 | 0 | 0 | 9825 | 1402 | 0 | 0 | 0 | 2.996 |

ele 4 | 0 | 0 | -9825 | 9825 | 1394 | 0 | 0 | 0 | 2.988 |

nuc FX | nuc FY | nuc FZ | total V (eV) | Charge Z | Distance (MM) |
---|---|---|---|---|---|

0 | 0 | 0 | -353.58 | Z = 6.12 | nuc= 9825 |

When Z = 6.12 and the distance between each electron and P nucleus is **0.9825** angstrom ( 1 MM = 10^{-14} meter ), each electron orbit becomes almost "**3**" de Broglie wavelegth.

*(Fig.35) Real atomic radius of phosphorus (P).*

The radius of phosporus is longer than nitrogen.

So the electronegativity of phosphorus is weak (= 2.19 ).

And due to the widely spreading n=2 electrons, its effective central charge Z is +6.12, which is bigger than +5.

*(Fig.36) Single P-P bond length.*

Single P-P bond length is known to be **2.2100** angtrom.

The atomic radius (= **0.9825** angstrom ) based on de Broglie wavelength is about an **half** of this single bond length.

So also in phosphorus atom, de Broglie and Coulomb relations prove to be effective.

*(Fig.N-1) What angle between valence electrons ? *

Methane molecule (= CH4 ) is just **tetrahedral** structure, so H-C-H bond angle is just **109.5** degree.

And H-O-H angle of water is **104.45** degree, which is a little different from 90.0 degree of octahedral structure.

H-N-H angle of ammonia (= NH3 ) is **106.7** degree, then what is the **real** angle between 5 valence electrons of nitrogen in molecules ?

*(Fig.N-2) Angles between 5 and 6 valence electrons. *

To know the real angles between valence electrons, we should investigate **larger** atoms of the **same** number of valence electrons.

Because as shown in this page, due to the **short** bond length (= 0.96 Å ) of H2O and the strong repulsive force by two H nuclei, the water **cannot** keep octahedral structure.

On the other hand, H-S bond length of SH2 is longer (= 1.34 Å ), so we can know more exact angle between 6 valence electrons, which is close to 90 degree (= **92.1** degree ).

And H-P bond length (= 1.42 Å ) is longer than H-N bond length (= 1.02 Å ).

So H-P-H bond angle of PH3 (= **93.4** degree ) is closer to the **real** angle between 5 valence electrons .

Actually the average angle among 5 valence electrons in Fig.31 nitorogen model is between 90 and 120 degree.

Also in the fluorine atom, it is **difficult** to visualize the electrons' motions.

As the fluorine has *seven* valence electrons, it **cannot** be expressed as a *regular polyhedrons* like carbon and oxygen.

We try to find as much "*symmetrical*" distribution as possible using the following sample program.

Sample JAVA program (distribution of F electrons)

*(Fig.37) Distribution of seven valence electrons in fluorine.*

In this program, we start from the fluorine configuration of Fig.37 left.

Electrons e0, e1, e2 and e3 are symmetric to each other, and e5 and e6 are symmetric.

So we can change only coordinates of e0, e4, and e5, which determine other positions.

( In this starting positions, e0, e1, e2 and e3 are on x-y plane, and e4, e5 and e6 are on x-z plane. )

And we seek the distribution of seven valence electrons in which its potential energy becomes the lowest (= experimental value ).

*(Fig.37') Distribution of seven valence electrons in fluorine (central charge = +7.3e)*

In this program, the fluorine nucleus is gray.

Here we use the new units, ( 1 MM = 10^{-14} meter).

Other manipurations and calculation methods are the same as the upper nitrogen.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Fluorine (F) | 17.423 | 34.971 | 62.708 | 87.140 | 114.243 | 157.165 | 185.186 |

According to the *Virial theorem*, "average" total potential (V) and kinetic (T) energies satisfy the relation,

V = -2 × T = 2 × E.

The total energy (E) of the seven valence electrons of fluorine is **-658.836 eV** (= the sum of 1-7 the ionization energies), so this potential energy (V) becomes 2 × -658.836 = **-1317.67 eV**.

As shown in Table 9, when the total potential energy (*tV*) of fluorine valence electrons is ** - 1317.71 eV**, this wn (Waves) of each electron becomes almost **2.0**.

eNo. | +X (MM) | +Y (MM) | +Z (MM) | nuc (MM) | CF | fx | fy | fz | Waves |
---|---|---|---|---|---|---|---|---|---|

ele 0 | 2537 | -3045 | 563 | 4003 | 9147 | 9 | 22 | 78 | 1.990 |

ele 1 | 2537 | 3045 | 563 | 4003 | 9147 | 9 | -22 | 78 | 1.990 |

ele 2 | -2537 | 3045 | 563 | 4003 | 9147 | -9 | -22 | 78 | 1.990 |

ele 3 | -2537 | -3045 | 563 | 4003 | 9162 | 0 | 0 | 0 | 1.991 |

ele 4 | 0 | 0 | 4003 | 4003 | 9162 | 0 | 0 | 0 | 1.991 |

ele 5 | -2500 | 0 | -3127 | 4003 | 9119 | 162 | 0 | -130 | 1.987 |

ele 6 | 2500 | 0 | -3127 | 4003 | 9119 | -162 | 0 | -130 | 1.987 |

nuc FX | nuc FY | nuc FZ | total V (eV) | Charge Z | Distance (MM) |
---|---|---|---|---|---|

0 | 0 | 9 | -1317.71 | Z = 7.3 | nuc= 4003 |

You can change the coordinates (+X, +Y, +Z), nuc (MM) and charge Z freely.

If you try changing these values, you will know the de Broglie's waves ( *Waves* ) become almost **2.0**, when the total potential energy is about *-1317.71 eV*.

This result shows that all seven valence electrons of fluorine are moving on the orbits of "**2**" de Broglie's wavelength.

In this case, each component of the force acting on the nucleus becomes ( FX, FY, FZ ) = ( 0, 0, 9 ), which values are almost *zero* and means stable nucleus.

Of cource, ( fx, fy, fz ) of each electron needs to be small in the equilibrium state.

( As I said, these values cannot be completely zero. )

*(Fig.38) Real atomic radius of fluorine (= F ).*

As shown in Fig.38, the atomic ( rotating ) radius of seven valence electrons of fluorine becomes **0.4003** angstrom (= 4003 MM ), which is very **small**.

And the central charge Z is **7.3**.

*(Fig.39) Strong electronegativity of F.*

Actually, F-H bond length (= 0.9200 angstrom ) is shorter than C-H bond length (= 1.0900 angstrom ).

Furthermore, the distance between F nucleus and its valence electron e0 (= 0.4003 angstrom ) is shorter than that between H nucleus (= 0.5200 angstrom ), though its central charge Z is **larger** (= +7.0e ) than hydrogen (= +e ).

These structures appear as strong **electronegativity** of fluorine.

The electronegativities are F (= **3.98** ), H (= 2.20 ), and C (= 2.55).

So this result shows Coulomb and de Broglie relations are **valid** also in fluorine.

Next we think about chlorine (= Cl ) atom, which also has **seven** velence electrons in **n = 3** energy levels.

Atomic Name | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
---|---|---|---|---|---|---|---|

Chlorine (Cl) | 12.968 | 23.813 | 39.610 | 53.465 | 67.800 | 97.030 | 114.196 |

The total ionization energies of seven valence electrons of chlorine is **408.88 eV**.

So the total potential energy V becomes 2 × -408.88 = **-817.76 eV**.

After running the same program as fluorine, and input each coordinate of Table 9 (= fluorine ).

Then, when we input charge Z = **8.26** and nuc = **7640** into the designated textboxes, we can get the following values.

eNo. | +X (MM) | +Y (MM) | +Z (MM) | nuc (MM) | CF | fx | fy | fz | Waves |
---|---|---|---|---|---|---|---|---|---|

ele 0 | 4841 | -5811 | 1074 | 7640 | 2972 | 2 | 6 | 21 | 2.990 |

ele 1 | 4841 | 5811 | 1074 | 7640 | 2972 | 2 | -6 | 21 | 2.990 |

ele 2 | -4841 | 5811 | 1074 | 7640 | 2972 | -2 | -6 | 21 | 2.990 |

ele 3 | -4841 | -5811 | 1074 | 7640 | 2972 | -2 | 6 | 21 | 2.990 |

ele 4 | 0 | 0 | 7640 | 7640 | 2975 | 0 | 0 | 0 | 2.993 |

ele 5 | -4770 | 0 | -5967 | 7640 | 2965 | 44 | 0 | -35 | 2.986 |

ele 6 | 4770 | 0 | -5967 | 7640 | 2965 | -44 | 0 | -35 | 2.986 |

nuc FX | nuc FY | nuc FZ | total V (eV) | Charge Z | Distance (MM) |
---|---|---|---|---|---|

0 | 0 | 0 | -817.19 | Z = 8.26 | nuc= 7640 |

As shonw in Table 11, when the distance between each electron and Cl nucleus is **0.7640** angstrom (= 7640 MM ), and its central charge Z = **8.26**, each orbit becomes almost "**3**" de Broglie wavelength.

( The total V of -817.19 is almost same as experimental value of -817.76 eV. )

Due to widely spreading of n = 2 middle electrons, effective charge Z (= +8.26 ) is bigger than +7

*(Fig.40) Real atomic radius of chlorine (= Cl ).*

Comparing Fig.40 and Fig.38, the distribution of seven electrons are larger in chlorine than fluorine.

Actually the electronegativity of chlorine (= 3.26 ) is **smaller** than that of fluorine (= 3.98 ).

And the difference of spreading areas in Cl and F is related to moleculr bond angle.

*(Fig.41) Fluoroform (= HCF3 ) and chloroform (= HCCl3 ) .*

As shown in Fig.41, F-C-F angle (= **108.5** degree ) of fluoroform is a little smaller than basic tetrahedron (= 109.5 degree ).

Because the valence electrons around C nucleus tend to be **attracted** toward F, and electron density of F sides becomes lower .
But Cl-C-Cl angle of chloroform is a little bigger (= **110.4** degree ), though Cl has stronger electronegativity than carbon.

Because seven valence electrons of chlorine spread over **larger** area (= 0.7640 angstrom ) than fluorine (= 0.4003 angstrom ).

These **repulsive** interactions becomes slightly stronger in chloroform.

( Due to electron's wide distribution, positive chlorine nucleus exerts stronger repulstive force. )

*(Fig.42) Electronegativity.*

The electronegativity ( Pauling ) was gotten comparing various bond energies of various molecules composed of different pairs.

As the difference of eletronegativity between atoms becomes stronger, its bond includes more "**ion**" effects, which make that bond stronger.

Electronegativity represents the power of **attracting** an electron in each atom.

The hydrogen and carbon atoms have almost same electronegativity ( H = **2.20**, C = **2.55** ).

The positive charge of C nucleus is bigger than H, but the electron orbit of H atom is **one** de Broglie wavelength, which is **shorter**.

So H atom can come close to another nucleus, **keeping** its own electron. ( It's related to why C-H bond of methane cannot be shorter than experimental value. )

As a result, they are almost same.

*(Fig.43) "Force 1" acting on valence electron (= e0 ) in methane.*

As is explained above, the rotating radius of four valence electrons of carbon is about **0.6400** × 10^{-10} meter, when we consider the total 1-4 ionization energies and "**2**" de Broglie wavelength orbit.

As a result, when valence electron e0 is closest to hydrogen nucleus, their distance becomes about **0.4500** angstrom, by **1.0900** (= C-H ) - 0.6400 .

At this point, ( Coulomb ) "**force 1**" is acting between a valence electron **e0** of carbon and H0 nucleus.

*(Fig.44) "Force 2" acting on valence electron (= e0 ) in ethane.*

C-C bond length of ethane is about **1.5400** angstrom, and distance between valence electron e0 and C0 nucleus is about 0.6400 angstrom, when we consider "**2**" de Broglie wavelength orbit like methane.

As a result, the distance between e0 and another carbon nucleus C1 is about **0.9000** angstrom (= 1.5400 - 0.6400 ).

And central carbon nucleus (= C1 ) has about **+4e** positive charge ( +6e - 2e ), when two 1s negative charges are subtracted from +6e carbon nucleus.

"**Force 2**" is the force acting between this **e0** electron and C1 nucleus.

*(Fig.45) Coulomb forces 1, 2 in methane and ethane are almost same.*

From Fig.43 (= methane ) and Fig.44 (= ethane ), we can know the forces acting between e0 valence electron and another nucleus ( H0 or C1 nucleus ).

Using Coulomb inverse square law, "force 1" in methane proves to be almost **same** as "force 2" in ethane.

This result agrees with almost **same electronegativity** of H and C atoms.

*(Fig.46) Coulomb forces 1, 2 in O-H and O-C bonds.*

As is gotten above, real atomic radius of valence electron of oxygen is **0.4596** angstrom, when we consider 1-6 total ionization energies and their **two de Broglie** wavelength orbits.

**O-H** bond length of water is **0.9584** angstrom, and the average single **O-C** bond length is known to be **1.4300** angstrom.

Subtracting the atomic radius of 0.4596 from these bond lengths, we can know Coulomb forces 1,2 between electron and another nucleus.

*(Fig.47) Coulomb forces 1, 2 in O-H and O-C bonds are almost same.*

As shown in Fig.47, Coulomb forces 1 and 2 between electron and another nucleus are almost **same** in O-H and O-C bonds.

So also in bonds with oxygen, the electronegativity of H and C atoms are same using **real** atomic radius.

*(Fig.48) Coulomb forces 1, 2 in N-H and N-C bonds.*

As is gotten above, real atomic radius of valence electron of nitrogen is **0.5330** angstrom, when we consider 1-5 total ionization energies and their **two de Broglie** wavelength orbits.

**N-H** bond length is **1.0100** angstrom, and the average single **N-C** bond length is known to be **1.4700** angstrom.

Subtracting the atomic radius of 0.5330 from these bond lengths, we can know Coulomb forces 1,2 between electron and another nucleus.

*(Fig.49) Coulomb forces 1, 2 in N-H and N-C bonds are almost same.*

As shown in Fig.49, also in bonds with nitrogen, the electronegativity of H and C atoms are same using **real** atomic radius.

Basically the target electron of ionization is one **closest** to another nucleus (= "e0" ).

Hydrogen atom is composed of one nucleus and one electron.

When another outer electron "e0" is closest to H nucleus, H electron moves away from this "e0".

( This gap between H nucleus and electron causes attracting power. )

So a pair of one nucleus + one electron is one unit of ionization power.

C atom can be considered as having four pairs of a nucleus + an electron, so its power is four times stronger than H atom when their distances are the same.

Atoms (= A ) | A-C bond ( Å ) | A-H bond ( Å ) | A-e0 ( Å ) | e0 - C nuc | e0 - H nuc | e0-C / e0-H |
---|---|---|---|---|---|---|

Carbon (C) | C-C 1.5400 | C-H 1.0900 | 0.6400 | 0.9000 / | 0.4500 | = 2.000 |

Oxygen (O) | O-C 1.4300 | O-H 0.9584 | 0.4596 | 0.9700 / | 0.4980 | = 1.948 |

Nitrogen (N) | N-C 1.4700 | N-H 1.0100 | 0.5330 | 0.9400 / | 0.4800 | = 1.958 |

Boron (B) | B-C 1.5500 | B-H 1.1900 | 0.8000 | 0.7500 / | 0.3900 | = 1.923 |

Fluorine (F) | F-C 1.3500 | F-H 0.9200 | 0.4000 | 0.9500 / | 0.5200 | = 1.827 |

Silicon (Si) | Si-C 1.8500 | Si-H 1.4800 | 1.1500 | 0.7000 / | 0.3300 | = 2.121 |

Phosphorus (P) | P-C 1.8400 | P-H 1.4400 | 0.9800 | 0.8600 / | 0.4600 | = 1.870 |

Sulfur (S) | S-C 1.8200 | P-H 1.3400 | 0.8600 | 0.9600 / | 0.4800 | = 2.000 |

Chlorine (Cl) | Cl-C 1.7700 | Cl-H 1.2700 | 0.7640 | 1.0060 / | 0.5060 | = 1.988 |

*(Fig.50) Coulomb forces 1, 2 in A-H and A-C bonds are almost same.*

The electronegativity of H (= 2.20 ) and C (= 2.55 ) are almost same.

This means the **powers** of them attracting another electron are almost same.

In Fig.50, when the distance ratio of e0 - C nuc / e0 - H nuc is near "**2.00**", their attracting powers are equal, considering their nuclear charges.

As shown in Table 12, the power of H and C are almost **same** in **various** atoms, because e0 - C nuc / e0 - H nuc is near "**2.00**".

To be precise, electronegativity of carbon is a little stronger than hydrogen, so this ratio tends to be a little smaller than 2.00.

*(Fig.51) Attracting force in H, C, B atoms.*

Next we investigate the relations between the electronegativity and attracting force of each atom.

Fig.51 shows forces of attracting "e0" electron in H, C, and B atoms.

( "e0" belongs to left carbon in all cases. )

Of course, rotating **radius** (= 0.6400 Å ) of carbon based on de Broglie wave is used here.

Attracting power is biggest in carbon (= C ), and weakest in boron (= B ).

*(Fig.52) Attracting force in H, C, B atoms.*

As shown in Fig.52, the electronegativities are strongest in carbon, and weakest in boron in the same way as force.

These results shows the **real** rotating radius based on Coulomb and de Broglie relations are **correct** also from the viewpoint of **electronegativity**.

Atoms (= A ) | charge Z | C-A bond ( Å ) | "e0"-A length | Force | Electronegativity |
---|---|---|---|---|---|

Silicon (Si) | 5.04 | C-Si 1.8500 | 1.2100 | 3.44 | 1.90 |

Boron (B) | 3.22 | C-B 1.5500 | 0.9100 | 3.88 | 2.04 |

Phosphorus (P) | 6.12 | C-P 1.8400 | 1.2000 | 4.25 | 2.19 |

Hydrogen (H) | 1.00 | C-H 1.0900 | 0.4500 | 4.94 | 2.20 |

Carbon (C) | 4.22 | C-C 1.5400 | 0.9000 | 5.21 | 2.55 |

Sulfur (S) | 7.18 | C-S 1.8200 | 1.1800 | 5.15 | 2.58 |

Nitrogen (N) | 5.25 | C-N 1.4700 | 0.8300 | 7.62 | 3.04 |

Chlorine (Cl) | 8.26 | C-Cl 1.7700 | 1.1300 | 6.47 | 3.16 |

Oxygen (O) | 6.27 | C-O 1.4300 | 0.7900 | 10.05 | 3.44 |

Fluorine (F) | 7.30 | C-F 1.3500 | 0.7100 | 14.48 | 3.98 |

As shown in Table 3, a close **correlation** is found between attracting **force** and *electronegativity* in each atom.

Generally, as the electronegativity is **bigger**, the force of attracting an electron is **bigger**.

Again, it is proved that de Broglie relation (= 1, 2, 3 wavelegnth ) and Coulomb force are **effective** in all atoms.

2013/3/28 updated. Feel free to link to this site.