According to current theory, chemical bonds are formed by electrons. However, the theory is unable to explain either the geometry of the bonds or their strength. Nuclear physicists are not interested in this problem, because according to current theory, the electron shell of atoms, and therefore its geometry, is not directly determined by the nucleus. Which is a great pity, because they might realize that this is not the case. In fact, several parameters determine whether a chemical bond will form. Among them is the electronegativity of the elements involved, which is a parameter proportional to the size of the gap in the proton radial pulse. This determines both the position of the electrons in the bond, but also the location where the pulses from the bonding protons will hit each other. Furthermore, the geometry of the position of the individual protons participating in the bond. In the case of nuclear interactions, which significantly deform the symmetry of the pulsation of individual nucleons, in the case of chemical bonding, there is a small deformation of these pulses. Protons themselves are not able to maintain a strong chemical bond with each other due to their geometry. To do this, they need electrons, which have a dual task. On the one hand, they keep the system of bonding protons planar and on the other hand, they increase the strength of the bond with their pulses. How the pulsation of particles participating in a chemical bond changes is shown in the following example, for a chemical bond between a proton and deuterium.
Figure 42. Formation of the hydrogen-deuterium molecule.
The first image shows hydrogen and deuterium atoms, including the distribution of radial pulse intensities of all atomic particles. The second picture shows a hydrogen-deuterium molecule. Only partial changes in the pulses of all particles are shown. In the case of a hydrogen proton, the pulsation on the far side increases, which pushes the proton towards the deuterium atom. The same is true for a proton from a deuterium atom. The increased pulsation pushes the deuterium neutron further away, which increases the equilibrium distance between the proton and neutron. For both electrons, the pulsation increases in the direction, which pushes the two electrons towards each other. All the particles in a molecule also repel each other, which forces them to seek a new optimum. Each particle in the system tries to find a place where the sum of all the repulsive forces acting on it is the smallest. The third image shows a possible new optimized arrangement of the molecule.
Why is CO2 a gas while SiO2 is a solid?
Based on the previously explained principle of chemical bonding, in the next chapter I will try to analyze the double bonds of carbon or silicon with oxygen. According to the current classification, both elements belong to group 14 of the periodic table. Carbon belongs to the second period and silicon to the third period. Both elements have 4 electrons in their valence shells. So they should have similar chemical properties, yet their oxides could not possibly be more different. What is the cause?
According to the 3D model from the previous chapter, we know that oxygen has a total of 3 pairs of deuterium units. But only one of these pairs is active enough to form strong chemical bonds, as described in the figure below.
Figure 43. Oxygen a its proton pairs.
The figure below shows a chemical double bond between carbon and oxygen. This bond is formed by two protons on the carbon side, two protons on the oxygen side, and a total of four electrons that keep the whole system planar. The protons interact with each other with their radial pulses as indicated by the arrows. The radial pulses of the two neutrons from the carbon atom, as indicated by the blue dashed line, divide the space into two halves, in which separate bonding electron orbitals are located. As in the case of a single bond, in this case, the pulse activity of all four protons forming the bond is suppressed, allowing the two atoms to come closer together. At the same time, the pulse activity of these protons will increase on the opposite side, i.e. towards the center of both atoms. This increase in the intensity of the radial pulses is indicated in the image by the light red crescents around the bonding protons on the carbon atom. This also happens with the bonding protons, which are covered by neutrons in the picture, on the side of the oxygen atom. This increase in the radial and axial pulses of the bonding protons towards the center of the atom induces changes in the atom related to the chemical bond. This results in a change in the behavior of the atom itself, possibly also in the properties of other chemical bonds on the atom in question.
Figure 44. Double bond between carbon and oxygen.
Now we’re going to try to create the same double bond between silicon and oxygen. Silicon has valence nucleons that show the same arrangement as carbon, partially sandwiched between nucleons from the previous layer. It is this squeezing of protons on the side of the nucleus center, together with the interaction with paired neutrons, that increases the activity of proton pulsations in the direction out of the atom. This changes both the parameter of the radial proton pulse (i.e. electronegativity) and its intensity. The electronegativity of carbon is 2.5, respectively 1.5 in the case of silicon. At first glance, the situation around the double bond may appear to be the same as in the case of carbon. Difficulties arise if we want to increase the intensity of the coupling proton pulses on the side of the center of the atom. It is precisely the squeezing of protons by nuclei from the previous layer that prevents the axial pulsation activity from increasing. Increasing the intensity of the pulsation would be very energy-intensive. Such a bond would be relatively long and unstable. This is why silicon normally does not form double bonds.
Figure 45. Double bond between silicon and oxygen.
So what about the SiO2 molecule? In fact, it is a macromolecule that would be correctly written as SinO2n. Silicon normally forms single bonds that are oriented in a direction that allows the activity of the bonding protons to increase towards the center of the nucleus, as shown in the following figure.
Figure 46. Single bonds between silicon and two oxygens.
This is consistent with the structure as found for example here:
https://en.wikipedia.org/wiki/Silicon_dioxide#/media/File:Si-OCage.svg
Silicon-type protons have very limited bond angle options. They can form strong bonds, but these are fragile. They can crack when stressed. In contrast, protons in oxygen, carbon, and the like have a very wide range of bond angle possibilities. Therefore, bond angles measured in crystals containing these elements are very likely to differ from bond angles in a free molecule. While in crystals, bond angles, but also bond lengths, are optimized with respect to all bonds, including crystallization bonds, in a free molecule the bonds are optimized only with respect to the individual molecule. In the case of the SiO2 crystal, the bond angles are strongly determined by the silicon atoms.
The limited ability to form double bonds can be inferred analogously for other elements. If we add another deuterium unit to a silicon atom, we get a phosphorus atom, which has a similar arrangement of valence nucleons to nitrogen. There remains one pair of protons perpendicular to the 4He plateau, which is partially inserted into the nucleon from the previous layer. In addition, one unpaired proton and one pair of protons forming an acute angle with the 4He plateau. It is this pair of protons that is able to form a stable double bond with oxygen. The other protons, when phosphorus pentoxide is dissolved in water, form bonds with the -OH molecule. This is why phosphoric acid is a tribasic acid with the formula H3PO4, while nitric acid is a monobasic. Similarly, if we continue to the sulfur atom, we get the oxygen analog. Here we have two pairs of protons forming an acute angle with the 4He plateau, which are able to form a stable double bond with oxygen. The ring of neutrons from the previous layer and the 4He plateau itself are already stretched so much by the pressure of nucleons from the valence layer that the grip of the partially immersed proton pair is no longer so strong. A double bond with oxygen of this pair of protons is already possible, but it is unstable and energy-intensive.
This is because:
– the preferred product of sulfur combustion is sulfur dioxide,
– the dissolution of sulfur trioxide in water is exothermic
– the resulting acid is dibasic, because this unstable double bond immediately breaks down into two single bonds with the -OH group.
Single or double bond?
As mentioned in the previous chapter, the oxygen molecule is paramagnetic, meaning it is attracted to a magnetic field. The following image shows an O2 molecule. Unlike the previous double bonds, there is no neutron pulse here to divide the space between the atoms into two parts. The electrons are thus free to move throughout the entire space. The result is a single bonding superorbital.
Figure 47. Special bond in molecule O2.
Polarity of chemical bond
Depending on the position of the electrons in the chemical bond, they are divided into polar and nonpolar. In a nonpolar chemical bond, the bonding electrons are located in the middle. A nonpolar bond is usually formed by protons whose proton radial pulse characteristics, i.e. intensity and gap size, are identical. Pulses impinge on the electron from both sides, which have the same repulsive effect on the electron. In the case of a polar bond, the electron is located closer to the proton, whose pulse intensity is lower and the gap size is larger. Such a pulse has a weaker repulsive effect on the electron. The intensity of the pulse, and therefore its repulsive effect, weakens with distance from the source of the pulsation. Therefore, the electron in the binding takes a position where the repulsive effects of both pulses are balanced. Of course, we are talking about equilibrium positions around which electrons oscillate. An example of a nonpolar and polar chemical bond is shown in the following figure.
Figure 48. Nonpolar and polar chemical bonds
Ammonia
We will create an ammonia molecule from a nitrogen atom, according to its 3D model from the previous chapter, where we will create 3 bonds with hydrogen according to the rules given in the previous text. The ammonia molecule forms a trigonal pyramidal shape. The peculiarity of this molecule is that the free electron pair does not point symmetrically to the imaginary center of the triangle formed by the vertices of the hydrogen atoms. This arrangement would be expected according to the current quantum theory, which assumes that electrons orbit “freely” around the positively charged nucleus of the atom. In reality, the electron pair is more inclined towards one of the vertices of that triangle. Although there are several indications, such as this, that quantum theory may not be true, these indications are successfully ignored. The following 3D model shows an ammonia molecule, and below that is a representation showing the N-H bonds and the lone electron pair of nitrogen.
Figure 49. 3D model of ammonia
Figure 50. Model of ammonia