Background
When an electron orbits an atomic nucleus, it has the probability of being in a given region that is described as an electron cloud. This cloud takes different shapes depending on the energy level of the electron, with a sequence described in the Periodic Table of Elements. These are the s, p, d and f blocks of atomic elements and their shapes:
Orbital Shapes (s, p, d and f)
Explanation
The proposed tetrahedral nucleus structure, along with rules for proton spin alignment that is the cause of the repelling force used to calculate orbital distances, can explain the shapes of the s, p, d and f orbitals. The electron is always attracted to the atomic nucleus at any angle. However, at certain angles, the alignment of opposite-spin protons causes a change in the repelling force. If this alignment did not exist, all orbitals would have a spherical shape like the s orbital. When the alignment exists, it cuts what would otherwise be a sphere. An animation is provided with a short introduction, but each orbital is unique, so they are explained in further detail below.
Orbital Shapes – Created from alignment of opposite-spin protons during rotation causing a change in the orbital force
S Orbital
The simplest shape is the spherical, s orbital, although there can be various orbitals of this shape in an atom due to quantum leaps of the electron. The explanation of the transition from 1s to 2s and other orbital jumps is described described in the quantum leap section. One of the causes is the alignment of same-spin protons in the atomic nucleus. Helium (Z=2) has two opposite-spin protons, but lithium (Z=3) is the first atomic element with two protons with the same spin. This causes one electron to be pushed out to the 2s subshell. Beryllium has two pairs of protons now with the same spin, thus two electrons are pushed out to the 2s subshell. The proposed nucleon structure for these elements are shown below.
The nucleon structure is also mapped to a VSEPR molecular geometry class, because it is possible that molecules get their shapes as an extension of how the nucleus itself is structured. The first four elements may be planar (2D) in structure given the stability of 7Li and 9Be which are proposed in symmetric arrangement. These proposed structures would match the known electron configurations in 1s and 2s and also the stability of the elements with these number of protons and neutrons.
Configuration of Protons and Neutrons for He, Li and Be
S-Orbital Shape
As the proton spins on three axes, it creates a spherical shape. H, He (Z=1, Z=2) have no protons with gluons that align (differing spins) and are confined to 1s. Li, Be (Z=3, Z=4) begin to have protons that align spin and gluons; this greater axial force pushes electrons to 2s.
Shape of the s orbital
P Orbital
Beginning with boron, a 3D tetrahedral structure begins to form. It is no longer planar (2D). There are six protons in the 2p subshell (B to Ne). This is the 3rd level of a tetrahedron. The side view of an atomic element, based on the axis of rotation, is shown below. The first four protons (H to Be) are now arranged as the first two layers of the tetrahedron.
Beginning with the 2p subshell, neutrons will be excluded from the view for simplicity to visualize the nucleus structure. However, neutrons are assumed to continue to fill the spaces between protons.
Protons forming in nucleus. The p orbital has six protons to complete the third level of a tetrahedral structure.
In the figure above, the dashed line is the focus for why the p orbital has a different shape than the s orbital. The p orbital appears as a dumbbell – a spherical shape like the s orbital cut in half. As the atomic nucleus spins, individual protons also spin. There are two times during a rotation that three protons align – 90° and 270° (below).
Two points in the proton’s spin rotation have an intersection where the axial force aligns for opposite spin protons
P-Orbital Shape
The p orbital is a dumbbell shape because the electron is pushed out twice during the rotation to the 3p subshell when an opposite-spin proton aligns gluons with two same-spin protons.
Dumbbell shape of p orbital due to two points in rotation where sum of forces is not at 2p distance
P-Orbital Proton Fill Order
Protons with spins aligned with the atomic nucleus’ spin will fill first as there is less energy required before a proton with opposite spin is filled in the nucleus structure. Protons also fill from the center then outwards for geometric stability. The figure below shows the fill order of atomic elements from boron (B) to neon (Ne) in both the side view of the nucleus and the bottom row (third row) which is being filled with protons.
Fill order of p orbital electrons (side view and bottom view shown)
D Orbital
The d orbital contains 10 electrons. This is the 4th level of the tetrahedron. This is illustrated in the figure below. Note that the 3s and 3p protons are not shown in this tetrahedral view, but are addressed in the section on nucleus structure.
Protons forming in nucleus. The d orbital has ten protons to complete a fourth level of a tetrahedral structure.
With three spin-aligned protons, it would have a spherical shape, yet four times during the rotation it will have gluons that align with a proton of the opposite spin to force an electron out to 4d.
Four points in the proton’s spin rotation have an intersection where the axial force aligns for opposite spin protons
D-Orbital Shape
The d orbital is a clover shape because the electron is pushed out four times during the rotation when an opposite spin proton aligns gluons with three spin-aligned protons.
Dumbbell shape of d orbital due to four points in rotation where sum of forces is not at 3d distance
D-Orbital Proton Fill Order
At Z=21, scandium (Sc) is the first element that begins a d orbital. As protons always build from the center then outwards for stability, the first proton is placed in the center (refer to the figure below). In a 4th row of a tetrahedron, this is the first time that a unit is in the center of axis of rotation. This creates a unique shape relative to other d orbital shapes (refer to shape highlighted in yellow).
Fill order of the 1st d orbital electron (bottom view shown)
The next three elements build outwards from the center, occupying the three sides of the triangle as shown below. These now have the clover shape as there are four points in the rotation where the repelling, axial force distance changes. These take place on the x-y, x-z and y-z planes of the tetrahedron while it spins. This is highlighted in yellow below.
Fill order of the 2nd to 4th d orbital electrons (bottom view shown)
The final spin up proton must be placed on one of the existing three sides. This is manganese (Mn). Since it shares a tetrahedral face with another spin up proton (x-y), its orbital will also be in this plane but will be shifted slightly based on the protons location as shown in the figure below. This is also highlighted in yellow below.
Fill order of the 5th d orbital electron (bottom view shown)
Finally, all of the five spin down protons complete the 4th row of a tetrahedron to complete the orbitals.
Fill order of the spin down orbital electrons (bottom view shown)
F Orbital
The sequence for the f block is unique. Beginning with lanthanum (Z=57) it starts a block that contains 15 elements. The 5th level of a tetrahedron has 15 units. There are 15 elements for the f block (Z=57 to 71), although an odd number affects the number of orbitals (14 / 2 = 7). It converts a proton to neutron in the next d block to compensate, beginning with the 5d block.
Protons forming in nucleus. The f orbital has 15 protons to complete a fifth level of a tetrahedral structure.
F-Orbital Shape
The f orbital is more complex, but follows the same rules based on proton alignment as the p and d orbitals. When completely full it is similar to the d orbital, but cut in half (eight lobes instead of four). It is based on the points in the nucleus rotation where the gluons of opposite spin protons align.
Shape of the f orbital due to eight points in rotation where sum of forces is not at 4f distance
F-Orbital Proton Fill Order
Similar to the d orbital, the first proton has a unique shape because it is in the center and does not have multiple protons in alignment on the tetrahedral edge. The 5th row of a tetrahedron has three center protons now (Z=57 to 59). As a result, these three elements have different shapes than the remaining spin up protons which will be placed on the triangle’s edge (tetrahedral face). These shapes are highlighted in yellow in the figure below.
Fill order of the 1st to 3rd f orbital electrons (bottom view shown)
Also, similar to the d orbital, protons continue to build outwards from the center. The next three protons (Z=60 to 63) occupy the space at the edge of the triangle and tetrahedral face. Now, with many protons in alignment on this face, it has the lobe shapes seen in the p and d orbitals, but they are cut in half due to an extra proton now matching spin during the nucleus rotation. These shapes are highlighted in yellow below.
Fill order of the 4th to 6th f orbital electrons (bottom view shown)
Once again, similar to the d orbital, the last spin up proton must be placed on one of the three triangular edges. It will share an edge with an existing x-y spin up proton, but the orbital is shifted on this plane because of the location of the proton.
Fill order of the 7th f orbital electron (bottom view shown)
Finally, seven spin down protons are added to the 5th row of a tetrahedral structure to complete the orbitals. There are now 7 spin up and 7 spin down protons. This matches the orbitals seen in the f series. However, there is one more space in the 5th row of a tetrahedron because it has 15 units. One last proton completes this row and it causes the next d block series to have 9 elements. This is confirmed in the Periodic Table of Elements as the 5d block (Z=72 to 80) contains 9 elements. In addition, the transition from 4f to 5d and again from 5f to 6d shows a mass increase that includes at least three neutrons before the next element. This means that the d block has a neutron take the position of a proton so that it can have 9 protons in a row, otherwise it requires 10 units to complete a row (two of the three neutrons are separation neutrons and the third occupies the proton’s position).
Fill order of the remaining f orbital electrons (bottom view shown)
Using the same rules that enabled the calculations of orbital distances, specifically that the proton is a pentaquark with gluons that align to cause a repelling force, the probability nature and shape of orbitals can be logically explained. The shapes match a nucleus structure that is based on a tetrahedral sequence. This nucleus structure is then further validated by the atomic element sequence seen in the Periodic Table of Elements, described in more detail in the next section.
Proof
Proof of the energy wave explanation for the orbital shapes is the explanation of:
- The probability cloud and orbital shapes of the s, p, d and f atomic orbitals (see above) using the same tetrahedral structure proposed in the nucleus structure and orbital distances.