• Phosphorus Valence Electrons Number
• Phosphorus Valence Electrons Configuration
• Phosphorus Valence Electrons And Ion

By Prof. L. Kaliambos (Natural philosopher in New Energy)

October 30, 2015

The total number of electrons present in the valence shell of an atom are called valence electrons, and there are a total of five electrons present in the valence shell of phosphorus (3s²3p³). Thus, phosphorus has five valence electrons. Valency of Phosphorus (P). Phosphorus is an element which is part of Group 15A neutral Phosphorus Atom has five valence electrons. These are contained in the third energy level of the atom. Due to the number of valence electrons, Phosphorus is capable of forming three bonds with other elements. The Valence Electrons are found in different types of orbitals. Electron Configurations and Orbital Diagrams KEY Draw orbital diagrams for the following elements: 1. Phosphorus ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 4p. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. Whatever his categories, Brandt’s phosphorus was a spectacular sight. Artist Joseph Wright of Derby immortalized this moment a century later in his painting The Alchymist. Within 50 years of its discovery phosphorus was being produced and sold to apothecaries, natural philosophers, and showmen, who made the element the centerpiece of. A way to find valence electrons without the periodic table is using the atomic number and drawing a diagram. The atomic number is how many protons and electrons the atom has. That means an atomic number of 8 (oxygen), has 8 protons and 8 electrons.

A phosphorus atom is an atomof the chemical element phosphorus with symbol P and atomic number 15. However unlike for hydrogen, aclosed-form solution to the Schrödinger equation for the many-electron atomslike the phosphorus atom has not been found. So, various approximations, suchas the Hartree–Fock method, could be used to estimate the ground state energies.Under these difficulties I published my paper “ Spin-spininteractions of electrons and also of nucleons create atomic molecular andnuclear structures” (2008) by analyzing carefully the electromagnetic interactionsof two spinning electrons of opposite spin. Under this condition we may use this correct image of phosphorus with the following electron configuration:

1s².2s².2p_x².2p_y².2p_z².3s².3p_x¹.3p_y¹.3p_z¹

According to the “Ionization energies of the elements-WIKIPEDIA”the ionization energies (in eV) of the element phosphorus are the following: E₁ =10.48669 , E₂ = 19.7694 , E₃ =30.2027 , E₄ = 51.4439 , E₅ = 65.0251, E₆ = 220.421, E₇ = 263.57 , E₈ =309.60, E₉ = 372.13 , E₁₀ = 424.4, E₁₁ = 479.46 . E₁₂= 560.8. E₁₃ =611.74, E₁₄ = 2816.91, and E₁₅ = 3069.842 .

Here the -( E₁ +E₂ + E₃ ) is equal to the ground state energyE( 3p_x¹ + 3p_y¹ + 3p_z¹)of the three outer electrons. Then the - ( E₄ + E₅)is equal to the ground state energy E(3s²). Also, the -(E₆ + E₇ + E₈ +E₉ + E₁₀ + E₁₁ ) isequal to the ground state energy E( 2p_x² + 2p_y² +2p_z²) . On the other hand the -( E₁₂ + E₁₃ )is equal to the ground state energy E(2s²) , while the -( E₁₄ +E₁₅) equals the ground state energy E(1s²). See also my papers about the explanation ofionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS based on my paper of 2008.

For understanding theionization energies E₁, E₂, E₃ , E₄ ,E₅ , E₆ , E₇ , E₈ , E₉ ,E₁₀ , E₁₁ , E₁₂, and E₁₃,which givethe ground state energies of (2s².2p_x².2p_y².2p_z².3s².3p_x¹.3p_y¹.3p_z¹)you can see my “EXPLANATION OF PHOSPHORUS IONIZATIONS”.

In the same way for theexplanationof the E₁₄ and E₁₅ which give the ground stateof the 1s² electrons one must apply the basic formula of mypaper of 2008.

EXPLANATION OF -( Ε₁₄ + E₁₅ ) = - 5886.752 = E( 1s²), WHICH GIVES THE GROUND STATE ENERGY OF THE TWO 1s² ELECTRONS

Asin the case of helium the binding energy E(1s²) is due to thetwo remaining electrons of 1s² with n = 1. Thus we maycalculate the binding energy by applying my formula of 2008 for Z =15 as

E(1s²) = [(-27.21))15² (+ 16.95)15 - 4.1 ] /1² =- 5872.1

Howeverthe experiments of ionizations give - (E₁₄ + E₁₅ )= - 5886.752 . In other words one sees here that after the ionizations myformula of 2008 gives the value of 5872.1 eV which is smaller than theexperimental value of 5886.752 eV. Under this condition of ionizations Isuggest that n = 1 becomes n < 1 due to the fact that theionizations reduce the electron charges and now the nuclear charge is muchgreater than the electron charge of the two remaining electrons. So for Z =15 wedetermine the n by writing

(E₁₄ +E₁₅) = 5886.752 eV = - E(1s²) = -[(-27.21) )15² + ( 16.95)15 - 4.1 ] /n²

Thensolving for n we get n = 0.9988.

However in the absence of adetailed knowledge about the electromagnetic force between the two spinningelectrons of opposite spin physicist today using wrong theories cannot explainthe ground state energy of the electrons 1s². For example underwrong theories based on qualitative approaches many physicists believeincorrectly that the second electron of the 1s² shell is less tightlybound because it could be interpreted as a shielding effect; the other electronpartly shields the second electron from the full charge of the nucleus. Anotherwrong way to view the energy is to say that the repulsion of the electronscontributes a positive potential energy which partially offsets the negativepotential energy contributed by the attractive electric force of the nuclearcharge.

Under such false ideas Ipublished my paper of 2008 . You can see the paper in “User Kaliambos”.

Historically, despite theenormous success of the Bohr model and the quantum mechanics of the Schrodingerequation based on the well-established laws of electromagnetism in explainingthe principal features of the hydrogen spectrum and of other one-electronatomic systems, so far, under the abandonment of natural laws neither was ableto provide a satisfactory explanation of the two-electron atoms. In atomicphysics a two-electron atom is a quantum mechanical system consisting of onenucleus with a charge Ze and just two electrons. This is the first case ofmany-electron systems. The first few two-electron atoms are:

Z =1 : H^- hydrogenanion. Z = 2 : He helium atom. Z = 3 : Li⁺ lithiumatom anion. Z = 4 : Be²⁺ beryllium ion.

Prior to the development ofquantum mechanics, an atom with many electrons was portrayed like the solarsystem, with the electrons representing the planets circulating about thenuclear “sun”. In the solar system, the gravitational interaction betweenplanets is quite small compared with that between any planet and the verymassive sun; interplanetary interactions can, therefore, be treated as smallperturbations.

However, In the helium atomwith two electrons, the interaction energy between the two spinning electronsand between an electron and the nucleus are almost of the same magnitude, and aperturbation approach is inapplicable.

In 1925 the two young Dutchphysicists Uhlenbeck and Goudsmit discovered the electron spin according towhich the peripheral velocity of a spinning electron is greater than the speedof light. Since this discovery invalidates Einstein’s relativity it met muchopposition by physicists including Pauli. Under the influence of Einstein’sinvalid relativity physicists believed that in nature cannot existvelocities faster than the speed of light.(See my FASTER THAN LIGHT).

So, great physicists likePauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism infavor of wrong theories including qualitative approaches under an idea ofsymmetry properties between the two electrons of opposite spin which lead tomany complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen aclosed form solution to the Schrodinger equation for the helium atom has notbeen found. However various approximations such as the Hartree-Fock method ,canbe used to estimate the ground state energy and wave function of atoms”.

It is of interest to notethat in 1993 in Olympia of Greece I presented at the international conference“Frontiers of fundamental physics” my paper “Impact of Maxwell’s equation of displacement current on electromagnetic laws and comparison of the Maxwellian waves with our model of dipolic particles '. In that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY .At the same time I tried to find not only the nuclear force and structurebut also the coupling of two electrons under the application of the abandonedelectromagnetic laws. For example in the photoelectric effect the absorption oflight contributed not only to the increase of the electron energy but also tothe increase of the electron mass, because the particles of light have mass m =hν/c^2. (See my paper 'DISCOVERY OF PHOTON MASS' ).

However the electron spinwhich gives a peripheral velocity greater than the speed of light cannot beaffected by the photon absorption. Thus after 10 years I published my paper 'Nuclear structure..electromagnetism' (2003), in which I showed not onlymy DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that theperipheral velocity (u >> c) of two spinning electrons with opposite spingives an attractive magnetic force (F_m) stronger than the electricrepulsion (F_e) when the two electrons of mass m and charge (-e) areat a very short separation (r < 578.8 /10¹⁵ m). Because ofthe antiparallel spin along the radial direction the interaction of theelectron charges gives an electromagnetic force

F_em = F_e - F_m .

Therefore in my researchthe integration for calculating the mutual F_em led to thefollowing relation:

F_em = F_e - F_m = Ke²/r² - (Ke²/r⁴)(9h²/16π²m²c²)

Of course for F_e =F_m one gets the equilibrium separation r_o =3h/4πmc = 578.8/10¹⁵ m.

That is, for aninterelectron separation r < 578.8/10¹⁵ m the two electronsof opposite spin exert an attractive electromagnetic force, because theattractive F_m is stronger than the repulsive F_e . Here F_m is a spin-dependent force of short range. As aconsequence this situation provides the physical basis for understanding thepairing of two electrons described qualitatively by the Pauli principle, whichcannot be applied in the simplest case of the deuteron in nuclear physics,because the binding energy between the two spinning nucleons occurs when thespin is not opposite (S=0) but parallel (S=1). According to the experiments inthe case of two electrons with antiparallel spin the presence of a very strongexternal magnetic field gives parallel spin (S=1) with electric andmagnetic repulsions given by

F_em = F_e + F_m

So, according to thewell-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/10¹⁵ m a motional EMF produces vibrationsof paired electrons.

Unfortunately today manyphysicists in the absence of a detailed knowledge believe that the twoelectrons of two-electron atoms under the Coulomb repulsion between theelectrons move not together as one particle but as separated particlespossessing the two opposite points of the diameter of the orbit around thenucleus. In fact, the two electrons of opposite spin behave like oneparticle circulating about the nucleus under the rules of quantum mechanicsforming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV dependson the Ze charge of nucleus as

Ev = 16.95Z -4.1

Of course in the absence ofsuch a vibration energy Ev it is well-known that the ground state energyE described in eV for two orbiting electrons could be given by the Bohr modelas

E = (-27.21) Z².

So the combination of theenergies of the Bohr model and the vibration energies due to the opposite spinof two electrons led to my discovery of the ground state energy of two-electronatoms given by

-E = (-27.21) Z² +(16.95 )Z - 4.1

For example the laboratorymeasurement of the ionization energy of H^- yields an energy ofthe ground state -E = - 14.35 eV. In this case since Z = 1 weget

-E = -27.21 +16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z =2 we get

-E = - 108.8 + 32.9 - 4.1 =-79.0 eV

The discovery of thissimple formula based on the well-established laws of electromagnetism was thefirst fundamental equation for understanding the energies of many-electronatoms, while various theories based on qualitative symmetry properties lead tocomplications.

Answer: A neutral Phosphorus Atom has five valence electrons. These are contained in the third energy level of the atom.

The Phosphorus Element as it would be represented on a modern periodic table. Source

Phosphorus is an element which is part of Group 15 (formally known as the Pnictogen group) and is directly below the nitrogen atom. As was mentioned before, a neutral Phosphorus Atom contains five valence electrons in the third shell. You can see a Bohr Model of Phosphorus below this paragraph. It also supports the original answer of five valence electrons.

Bohr Model of a Phosphorus Atom.

Due to the number of valence electrons, Phosphorus is capable of forming three bonds with other elements. Yet since it has a third-energy level, it can form more than three bonds if it is favorable to do so. Phosphorus tends to form an anion even though it has a low electronegativity. Due to its high reactivity, phosphorus is never found a pure element on earth.

Electron Configuration of Phosphorus with a Lewis Diagram on the side as well. Source

The Valence Electrons are found in different types of orbitals. Two are in the s-orbital (a circular orbital) while the other 3 are in p-orbitals (bell-shaped orbitals). There are a total of 4 orbitals in the valence shell to minimize electron-electron repulsion. Since their is less electron-electron repulsion, this formation is actually more stable than one with an extra electron. Therefore Phosphorus has a higher ionization than its Group-16 (Chalcogen) neighbors Sulfur.

Phosphorus Valence Electrons Number

Phosphorus as it moves in a biochemical process. Source

Phosphorus is necessary for some functions in life and is an essential component of fertile soil. It moves as a compound (not as a pure element) through what is called the Phosphorus cycle from the soil to plants and then back into the air before returning to the soil. Phosphorus is also essential component of matches. It was discovered in 1669 by Hening Brand.
Sources:

Phosphorus Valence Electrons Configuration

http://thechemistryguru.com/phosphorus/

Phosphorus Valence Electrons And Ion