How Chemistry Narrowly Avoids Negating Quantum-Mechanics

According to Quantum-Mechanics, the ultimate solution to the question, of Wave-Particle Duality, no matter how deeply this solution is buried, lies in the idea, that Particles cause Waves. Hence, the particles are more-ultimately real, and waves are not. In certain cases such as phonons, this even extends beyond waves-in-a-vacuum, to sound waves, that can be modeled as quasi-particles.

One rule which this evokes is the notion, that if (A) causes (B) with certainty, then it cannot be true that (B) causes (A). And to my mind, this has presented the greatest challenge with Chemistry.

The way Chemistry is understood to work today, the electrons that were loosely stated to be orbiting the nucleus, are actually occupying Quantum-Mechanical states around the nucleus, thus merely being attached to the nucleus, and they occupy shells, which are subdivided into orbitals. Further, these orbitals have known wave-functions, that follow from QM. Hence, the s2 -orbitals are spherical, the p6 -orbitals are perpendicular, and the d10 and f14 -orbitals have the more-complex geometries, which are possible modes of resonance. If all the orbitals belonging to a shell are filled, then indeed the shell becomes spherical itself, and this is best exhibited with inert gasses, which therefore also have ideal cancellation of the nuclear charge at close distance, and which therefore also lack electronegativity. (:1)

The main point of confusion which is possible here, is in the fact that these orbitals and their wave-functions seemingly define the chemical and physical properties of the element, except for anything related to its mass. The suggestion follows, that since the electrical field of the nucleus is strong enough to manipulate the wave-functions, it can also end up displacing where the particle ultimately occurs. In so doing, this action on the orbital would seem to suggest that the wave-function can also be said to change the particle-parameters, thereby creating a contradiction with the way in which QM is currently taught.

There is a specific observation which we can make about this subject, which causes Chemistry to avoid contradicting QM by the width of a hair.

These s, p, d and f -orbital geometries are only thought to exist, if their electrons are unpaired. Each orbital is capable of holding up to 2 electrons, and an orbital which only holds 1 electron is said to be “half-filled”. It has these formally-defined properties when half-filled.

There has never been a precedence in Chemistry, in which a half-filled orbital can be shared by two atoms. But some sort of entity needs to be shared between 2 or 3 atoms, in order actually to form a bond, and in order to change position around either atom. (:2)

When orbitals are filled by 2 electrons each, these two electrons perform a dance which electrons are already famous for, in which both their spin-vector and their magnetic dipole moment pair up, to cancel out. This is also known as “spin-spin decoupling”, and causes the electron to resemble a Fermion less, resulting in some quasi-particle that resembles a fluid more – i.e. a massive Bose particle.

The same affinity causes electron-pairs to form Cooper Pairs, which ultimately result in superconductivity. But in Chemistry, it forms charge-droplets, which are able to change position on an atom or molecule, and which can be shared between 2 or 3 atoms, thus forming either the sigma-bond or the pi-bond known.

The important fact to understand, is that This quasi-particle does not represent a wave-function, and so its mutability also does not represent the mutability of a wave-function. This charge-droplet has mass.

What is typical about this subject in a negative way however, is that Chemistry teaches us that the single-atom state is also the original state of an element. This is not true in practice. When most elements are gained on an industrial scale, they are dug out of the ground already-bound to other elements, and chemical reactions are used to purify them. Even when they do form the only element in a quantity, the atoms are still usually bound to others of the same element, thus forming O2, N2, Cl2, etc.. (:3)

Only, in an abstract way, it is still the desire of the atom from its single-atom state, to complete its orbitals, thus completing an octet ideally, that forms the best first-approximation of how it will react. Many other behaviors are known, in which some of the electrons remain unpaired, thus resulting in paramagnetism, or in which the completed valence shell ends up with a higher number of completed orbitals, than 4. And these frequent exceptions make Chemistry harder to study, than ‘Chem 101′ would have taught it.


1:) Unlike reactive elements, the inert gasses commonly do occur in the single-atom state.

2:) The way I was taught the meaning of the phrase ‘Multi-Centered Bonds’, 3-center bonds are already extremely rare, and the main examples I know are certain hydrides – where the hydrogen atom can also be related to an unshielded nucleus, such as in Beryllium Hydride. In this example, each hydrogen atom shares a bond with 2 beryllium atoms, resulting in a chain.

Another example I know, involves the Pi bond from a double-bond, where double or triple bonds exist as Sigma bonds – which form an axis between two atoms – plus 1 or 2 Pi bonds, which stand out from the Sigma bond as mirror-images at 90⁰ angles in 3D. This involvement of Pi bonds is also what makes quadruple bonds between two atoms impossible. Well, the Pi bond can become attracted to a 3rd atom, acting as a cation or an oxidizer, thus forming another kind of 3-center bond.

By default, a covalent bond is an electron-pair shared by only 2 nuclei, while electron-pairs attached to a single nucleus do not form a bond, but nevertheless exist, and contribute to satiating the electronegativity of the atom in question.

Further, those single-centered electron-pairs are often available, for ‘coordinate covalency’ to form, starting from an atom that is not coordinate yet, and that so far has a formal charge of zero. Once coordinate, its valence will have changed, and it will acquire the corresponding formal charge.

3:) It can happen that when a quantity of atoms belongs to the same element, each atom is bound to more than one other atom. Hence, covalent solids can form, out of one element, as opposed to the molecular solids. In this way, Sulfur can consist of the molecular solid S8. But when we heat it, shortly after melting, it will want to form the polymer Sn. And in other examples, the 3-dimensional structure of the solid is too complex to be stated fully in this notation, so that notations like Al(m) and B(s) are often written, to refer to the metal or the solid form, without being more specific.

This generally does happen, if in the abstract single-atom representation, an element has fewer than 4 valence electrons. Hence, this will happen to most of the elements in the periodic table – when pure – even though practical chemistry often tends to focus on the atypical, well-behaved elements, towards the upper-right corner, that are also the basis for life-forms.

(Edited 12/16/2017 : )


The only way metallic beryllium could complete its octet – and thereby, its valence shell – would be by forming a crystal with as many electron-pairs, as there are beryllium atoms, to balance the charge. But I think that in such crystals, as soon as 1 electron pair would be shared by 4 or more atoms, this may no longer be called a covalent bond.

In practice, metallic beryllium has a “close-packed hexagonal structure“.

(Edit 12/19/2017 : )

When spheres are hexagonally close-packed, as the WiKi-article states, the order in which their layers fill empty positions, one layer ‘up’ or ‘down’, each new sphere should be positioned within one out of two triangular cusps, that formed between 3 spheres belonging to the previous layer, so that each sphere of the first layer, is also positioned at a cusp, between 3 spheres of the new layers. But there are actually different reasons for which a hexagonal structure could form. One such structure misses an atom at the center of each hexagon. And in case they are close-packed, there are actually 2 available ways in which this can progress to each new layer. When this progresses to yet another layer, only 1 out of the 2 available arrangements actually corresponds to that of the first layer.


And so to distinguish what the exact, hexagonal configuration of an unknown substance is, in general, the ordering of layers should be determined, as possibly being AA, or ABA, or ABCBA, or just plain random… But the only other element I was aware of, which actually does this, would be helium in its solid state.

Because helium is completely non-reactive, it only solidifies when liquid and put under approximately 10 atmospheres of pressure, and the vulgarization would then be correct, to say that ‘Helium atoms only stay packed together, because they have been forced to do so, due to high pressure and very low temperature.’ And in that case, helium neither prefers the ABA nor the ABCBA order, of the close-packed configuration… Its ordering is random.

If that situation was also to describe metallic beryllium, then the result would be, that beryllium has no tensile strength ! But we know factually, that real beryllium has good tensile strength, which suggests to the contrary, that it shares electron-pairs in fixed positions. Based on that observation, it might be a bit safer to speculate, that the ordering of the hexagonal close-packed structure of beryllium, is actually ‘ABA’.

In that case, the atoms no longer truly correspond to spheres either.


And if this is true, then 3-dimensionally, a configuration would be plausible, in which each of its electron-pairs would be shared by 5 beryllium atoms, and each beryllium atom would conversely be connected to 5 electron-pairs, that correspond simultaneously to the 3 atoms in the hexagonal layers above and below, plus corresponding to the 2 positions directly above and below the current layer’s atoms.

If the (stressed) ‘vertical’ bond in such a scenario, from one layer to the next, did not belong to the atoms’ valence shell, then the physical properties of the solid would more-closely resemble that of layered materials, such as clays or micas. But we also know that true beryllium has approximately as much strength to hold its successive layers to each other, as it does to hold each layer together.

And this would also be why, I find the WiKiPedia diagram of the hexagonal structure, in the article just about beryllium, so hard to understand.

This would give beryllium in its pure form, a so-called ‘extended octet’ (of 10 electrons). It would only be possible, due to the highly electrophilic nature of beryllium ions, and would be spectacular.



But alternatively, an electron-configuration would be plausible, in which pure beryllium only has ‘a regular octet’ of 8 electrons. This configuration would be easiest to visualize, as if it was a “Wurtzite crystal“, tetrahedral. The difference would be in the fact that true Wurtzite crystals are binary compounds, and that in visualizing beryllium, we’d need to substitute the shared electron-pairs, as corresponding to 1 of the 2 elements in a Wurtzite crystal.

This scenario would suggest that alas, each electron-pair is shared by 4 beryllium atoms, and that each atom shares 4 electron-pairs.

After much contemplation, I’ve concluded, that I should believe in the electron-configuration, which breaks the fewest rules.



Periodic Table:

One of the facts which I’ve learned, is that we still cannot predict the physical properties of unknown substances, entirely by following simple rules about Physics and Chemistry. We actually need to measure the physical properties of the elements at least, to deduce their structure.

Is the tensile strength of beryllium, indeed weaker along one axis, from what it is along the other 2 axes?’

A case in point is the simple question of the order, in which orbitals half-fill. In Chem 101 we’re taught, that the order in which they half-fill is:

  • First, The s2 orbital, then
  • One p6 orbital, then
  • One d10 orbital, if there are any f14 orbitals, then
  • The f14 orbitals, completely, if there are any, then
  • The remaining d10 orbitals, completely, if there are any, then
  • The remaining p6 orbitals.

Only after the d and f -orbitals are completely-filled, and all the p-orbitals half-filled, would the s, and lastly the p -orbitals completely fill.

In other words, even though the series of rare-earth elements follows lanthanum, lanthanum itself has no even-partially-filled f -orbitals. And the same is true for the actinides. The fact that lanthanum must receive one electron into its d -orbital, before the filling of f -orbitals begins, is also a pattern begun with the alkaline-earths, which must receive 1 electron into their p -orbitals, before the filling of d -orbitals begins. This is why there are generally two similar elements in each period, before the filling of d -orbitals leads to the transition elements.

According to that, the electron-configuration of beryllium should be:

[He] 1s2 1p6

Further, the reason why elements generally try to complete an octet, is the fact that their s and their p -orbitals, together, form their ‘valence shell’, and that together, these two orbital-types have space for 8 electrons, if completely filled. The effects which the d and the f -orbitals have on valence can also be strong, but are just harder to characterize in a systematic way.

Hydrogen forms an exception, because it doesn’t even have any p -orbitals.

In Physical Chemistry we learned, that as we progress further into the periodic table, even when just encountering the transition-elements, the order in which the d and f -orbitals fill completely, differs from this basic rule, due to the energy of the fields and repulsion between the electrons.



Slicing a Hexagonal Close-Packing in 3D:

A hexagonal close-packing may have as desirable feature, that if sliced along a plane inclined by 60⁰ in 3D to the original plane, doing so may reveal an equivalent plane, in which the spheres are packed efficiently, in 2D again, and with the same scaling. If this was fulfilled, then it would indicate ‘good 3D symmetry’, even though some elements and compounds form crystals, which lack this symmetry.

(Edit 12/20/2017 :

In order for this to be possible, the layering-order must be ABC in each chosen plane.


One limitation of the above diagram is, that it assumes that the view has just been inclined 60⁰. In reality, we’d need to consider what happens, when the view is inclined 60⁰ and rotated 30⁰. Yet, this does not change the fact that the ordering of the spheres would need to repeat itself exactly, every 2nd layer, as seen in 2D. Otherwise, we’d see a row of spheres that’s shifted with respect to the newly-defined plane. Every second row of spheres would need to be, as if shifted in alternating directions.


Therefore, If solidified helium exhibits poor adherence to ABC ordering in one plane, then it will also exhibit poor 2D packing in another, inclined plane. But, because the helium atoms themselves do not exhibit any preferred directional properties, this would also mean, poor 2D packing in whatever plane the description started in. )

And this should mean in practice, that any success to get helium to solidify in a cryogenic lab, can at best be a partial success.


Print Friendly, PDF & Email

2 thoughts on “How Chemistry Narrowly Avoids Negating Quantum-Mechanics”

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>

This site uses Akismet to reduce spam. Learn how your comment data is processed.