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.
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
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.