# B How and why do electron quantum numbers affect bonding?

1. Sep 10, 2017

### Hallucinogen

Hi, I just have a few questions I'm struggling to find straightforward answers to online.

The 4 quantum numbers of an electron in an atom describe the energy level, shape and suborbital of the orbital, and the fourth assigns a value to the electron's spin. Question 1) why is it in lone atoms that having two electrons of opposite spin is more energetically desirable? Why is there this "push" towards having all suborbitals filled with 2 electrons with opposite spins?
2) why do covalent bonds, for example sigma bonds, form due to two electrons also having opposite spins? What's so great about having a bond with two electrons in it of opposite spin compared with having two separate atoms each with a lone electron of either spin?
Question 3) I understand that some covalent bonds take in energy while most release energy, but I don't really understand why. Does the uniting of two electrons from two orbitals into one orbital... just... release latent energy due to opposite spins sharing one orbital?
4) Does anyone know why orbitals change shape when formed? For example, an sp3 orbital joins an s orbital, to form a sigma orbital. I know orbitals are wavefunctions, so why does the wavefunction change form when two orbitals become one? Does it go back to the quantum numbers?

2. Sep 10, 2017

### blue_leaf77

1) From Pauli exclusion principle, an orbital cannot be occupied by more than two electrons and if it is fully occupied the spin must be opposite in order for the total wavefunction to be antisymmetric.
2) Again from Pauli exclusion principle, if the spin part of the wavefunction is antisymmetric (which occur when the two electrons have opposite spin direction), then the spatial part should be symmetric. Symmetric spatial part means that the two electrons have a larger likelihood to be localized in a certain region hence providing h¥the necessary bonding force for the molecule.
3) It comes from the solution of the TISE, it turns out that some pair of atoms each of which being in certain state is more stable if the form a molecule because the energy of the resulting molecule lies lower than the energy of the separated atoms. A typical bonding energy curve for a diatomic molecule looks like this. As you can see the there is a global minimum in this curve which corresponds to the stable geometry of the newly formed molecule. Obviously a chemical reaction following this curve necessitates the release of energy. Some other curve however has a slightly more involved shape in that they may have a barrier to the right of the minimum which means in order to form such molecule, certain amount of energy must be supplied.

Last edited: Sep 11, 2017
3. Sep 11, 2017

### DrDu

I know, this is old-fashioned, but somewhere in dusty shelves there still exist some books on the topic.

To your question 3: The point is not that pairing of the spins would lower energy in any way. Rather the lowering of the energy already occurs for 1 electron. The reason lies in Heisenberg's uncertainty relation: In an atom, the electron tries to get as near to the nucleus as possible to reduce potential energy. However, this localization to a distance $\Delta l$ near the nucleus will lead to an inrease of the electrons momentum and correspondingly its kinetic energy. So the electron cannot fall completely into the nucleus but remains on average on a finite distance (Bohr's radius in the case of hydrogen). Now if you have two atoms instead, the electron can be more delocalized as the potential through gets larger. Hence it can get nearer to the nuclei without it's kinetic energy increasing as much as would be the case in an isolated atom. In the end the total energy of the electron can get lower in the molecule than in the isolated atom. This is the main principle of chemical bonding. The prototypic example with just one electron is the ion $\mathrm{H}_2^+$.
Now, adding an additional electron to the same orbital is only possible if the second electron has opposite spin as compared to the first one. While the electrons will still repell each other electrostatically, there will be still a net energetic gain from filling an orbital with two electrons, i.e. $\mathrm{H}_2$ will be still energetically lower than 2 H, although the binding energy will be less than two times the binding energy in $\mathrm{H}_2^+$.

Question 4: Orbitals aren't real entities but more mathematical constructs to form an approximate wavefunction for the molecule. From the previous answer to question 3 it should be clear that the energetic gain in bond formation will be the larger the more the orbitals allow for a hopping between the two atoms. Mathematically this corresponds to maximizing the overlapp of the orbitals on the two atoms. For hybrid orbitals this overlapp is often larger than for unhybridized atomic orbitals.

4. Sep 13, 2017

### Hallucinogen

Okay. But there isn't much of a reason why an orbital cannot be occupied by more than two electrons other than the fact that the 4 quantum numbers are the simplest and best explanation of particle acceleration data, right? Am I correct in thinking that this is also the main reason why the Pauli exclusion principle is inferred? Or can it be justified by other means?
I read this to understand what you meant about antisymmetric wavefunctions. So the more fundamental reason why two electrons must be of opposite spin in a sub-orbital is because they must obey a wavefunction that is antisymmetric whilst their spacial wavefunction is symmetric, because they're identical particles? And if that were not the case, they wouldn't be identical particles? Am I understanding correctly?
It also says that this reasoning only applies to non-interacting particles, but two electrons in a sub-orbital are interacting, right?
I think I understand your answer to 2) and 3), thank you.

5. Sep 13, 2017

### Hallucinogen

Okay, but why will it release energy?
Okay, so the shape of the wavefunctions/orbitals merely reflect where 90% of the maximum probability overlap is, between the 2 nuclei?

6. Sep 13, 2017

### Staff: Mentor

Three of the four quantum numbers come from solving Schrodinger's equation for an electron in the Coulomb potential of the nucleus. The fourth, spin, was originally discovered experimentally but turned out to be an integral part of quantum field theory.
The exclusion principle follows from the antisymmetry of the wave function when the two particles are exchanged. If the two particles are indistinguishable then the state after exchange will be the same as the state before exchange; but antisymmetry means that the state after exchange will be the negative of the state before exchange. This is possible only if both are equal to zero, and that's the exclusion principle.

7. Sep 13, 2017

### blue_leaf77

The reasoning that the electrons in an atom can be thought to be occupying a set of well-defined orbitals is an approximate picture. It's only exact if there is only one electron such as H-like atoms. In reality, the eigenfunction of a multi-electron atom is formed from an infinite linear combination of the so-called Slater determinant. Each Slater determinant describes the electron configuration for a given set of orbitals by assuming non-interacting picture. The interaction of all electrons can be accounted for by forming a linear combination of any possible Slater determinant, hence of any possible set of orbitals.

Last edited: Sep 14, 2017
8. Sep 17, 2017

### Hallucinogen

Thanks Nugatory and blue leaf for your replies.
So, when two orbitals become one and maximise their overlap between the two nuclei, each electron can maximise its proximity to the nucleus : kinetic energy ratio, where kinetic energy must be lowest and proximity to the nucleus highest?
Another question: Does the release of energy upon bond formation have anything to do with con/destructive waves of the new wavefunction within the new orbital formed?
Also it was stated that there is no strict shape or boundary for the electron orbital distributions (shapes). Elsewhere I read that the contour of what represents the 90% probability distribution of the orbitals is determined by 2 or more forces varying with each other, producing a Lissajous geometry. I suppose in the case of electron orbitals this would be the electric charges of the electrons and protons? Or is it the linear combination of Slater determinants again?
As a kind of summary, I'd like to ask if it's correct to say that, the driving force for bonds to form is energetic, but the rules of the road are the quantum numbers and the Pauli exclusion principle?

9. Sep 17, 2017

### Hallucinogen

I'm browsing this: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4032414/
"covalent bonding is driven by the attenuation of the kinetic energy that results from the delocalization of the electronic wave function"
I don't understand this definition of delocalization. I learned that delocalization means electrons aren't associated with a single atom and is the reason metals are so hard. Is it also used to refer to electrons forming a new wavefunction between nuclei in a covalent bond?

10. Dec 23, 2017

### Hallucinogen

A few more questions. For one electron in between two nuclei, it can delocalise / move its wavefunction to the other nucleus within the same potential energy (http://www.chem1.com/acad/webtut/bonding/TunnelBond.html). Whereas, when there are two electrons in this situation, the wavefunction of the electron's spatial overlap between the atoms is maximised as they undergo bond formation, and one can have opposite spin so that they can increase the probability density of their orbitals within the same potential energy allowance.
Is it correct to say that the Pauli exclusion principle isn't directly causative here, it's just a law that allows these observations to happen?
Also, why is potential energy so important to minimise, as opposed to kinetic energy? Why is having a high kinetic energy close to the nucleus no problem?
So tunnelling does not affect the kinetic energy nor potential energy, but delocalisation decreases kinetic energy because it increases the volume occupied (spatial aspect of the probability wavefunction)?
It also seems like tunnelling between two nuclei acts against the maximisation of orbital overlap between nuclei? Since it moves the probability wavefunction close to the nuclei of both atoms at the expense of between them?

11. Dec 24, 2017

### DrDu

That's an excelent source! Yes, delocalisation is here meant also as delocalisation over the two atoms being bonded.

12. Dec 24, 2017

### DrDu

There is a theorem, called the virial theorem, which states that $V=2E$ and $T=-E$.
The principal picture is the following: The possibility of the electronic wavefunction to delocalise between two atoms increases the spatial uncertainty, and, thus, by the Heisenberg uncertainty principle, decreases the uncertainty of the momentum and therefore also decreases the average kinetic energy. The actual picture is more complicated, though, as in reaction the atomic orbitals become smaller in the directions perpendicular to the bond direction. This increases the kinetic energy of the electrons a bit but decreases even more the potential energy of the electrons.
Confer also the following article by Kutzelnigg:
The Physical Mechanism of the Chemical Bond
http://onlinelibrary.wiley.com/doi/10.1002/anie.197305461/full

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted