Are Electrons Transferring Energy in Hybridisation Orbitals?

[robynmc]

Please help, very lost in khan academy!

1)The sp1, sp2 and sp3 hybridisation states look like dimensional i.e. a line, a plane and a xyz coordinate system, but I don't remember my resource explicitly stating this as the case. Is it a safe assumption/intuition? Or more objectively the case - not intuition, fact it's exactly like that?

2) are electrons transfering energy in some way in hybridisation orbitals? Concepts that average potential energy like formal charge and bond energy/length imply intermediate levels of energy, but potential energy of the orbital shells suggests any such changes must be instantaneous. Is the bond length an average, so a sigma bond is sort of vibrating? Or what's going on there

 

[DrClaude]

1)The sp1, sp2 and sp3 hybridisation states look like dimensional i.e. a line, a plane and a xyz coordinate system, but I don't remember my resource explicitly stating this as the case. Is it a safe assumption/intuition? Or more objectively the case - not intuition, fact it's exactly like that?

Yes, the two sp orbitals form a line, the three sp2 orbitals form a plane, and the four sp3 orbitals form a tetragonal pyramid. The bonds from these hybrid orbitals will lead to atoms being found in aline, in a place, or in a tetrahedral pyramid.

2) are electrons transfering energy in some way in hybridisation orbitals?

Not sure what you mean here. Hybrid orbitals (or even just plain orbitals for atoms other than hydrogen) are mathematical approximations. Hybridization is not a mechanism (orbitals transforming themselves), but rather a convenient approximate representations of what the electron eigenstates look like, to help explain molecular structure.

Concepts that average potential energy like formal charge and bond energy/length imply intermediate levels of energy, but potential energy of the orbital shells suggests any such changes must be instantaneous. Is the bond length an average, so a sigma bond is sort of vibrating? Or what's going on there

Like many things in quantum mechanics, bond lengths are not what you could think of as the distance between, say, two billiard balls. The most precise way to see it is as the expectation value of the distance between two atoms, but you can also view it as the average value of the distance between atoms in a classical (i.e., imperfect) classical representation of a bond as two atoms connected by a spring.

 

[robynmc]

Eigenstates will be eigenvectors and eigenvalues I suppose, which is more or less where I left off in maths many years ago. Would you recommend a flurry of revision and a bit of further digging into this? Can I get anywhere with fairly superficial linear algebra?

 

[DrClaude]

Eigenstates will be eigenvectors and eigenvalues I suppose, which is more or less where I left off in maths many years ago. Would you recommend a flurry of revision and a bit of further digging into this? Can I get anywhere with fairly superficial linear algebra?

I think you would need a better understanding of basic quantum mechanics. If you are coming more from the chemistry side, I would recommend Quantum Chemistry by McQuarrie or Molecular Quantum Mechanics by Atkins and Friedman.

 

[robynmc]

I think you would need a better understanding of basic quantum mechanics. If you are coming more from the chemistry side, I would recommend Quantum Chemistry by McQuarrie or Molecular Quantum Mechanics by Atkins and Friedman.

Thanks I'll see if I can find them. If hybridisation isn't a mechanism of energy exchange, can I think of hybridisation as the nuclear charges exchanging "influence" on... the electron cloud, or probability density, or whatever. So I'm looking at electron receptacles rather than the electrons themselves. Still seems odd to talk about the average though. Does that ever get used, I thought activation energies of reactions were determined experimentally

 

[DrClaude]

If hybridisation isn't a mechanism of energy exchange, can I think of hybridisation as the nuclear charges exchanging "influence" on... the electron cloud, or probability density, or whatever.

The nuclear charge is only there to provide the Coulomb potential. The perturbation comes from electron-electron interactions.

So I'm looking at electron receptacles rather than the electrons themselves. Still seems odd to talk about the average though. Does that ever get used, I thought activation energies of reactions were determined experimentally

I don't understand the link you are trying to make between bond lengths and activation energies. Do you mean as bond stretch during chemical reactions?

 

[DrDu]

2) are electrons transfering energy in some way in hybridisation orbitals? Concepts that average potential energy like formal charge and bond energy/length imply intermediate levels of energy, but potential energy of the orbital shells suggests any such changes must be instantaneous. Is the bond length an average, so a sigma bond is sort of vibrating? Or what's going on there

For many atoms, hybridization is not possible starting from the electronic ground state but is only possible for excited states. Forming hybrid orbitals is equivalent to applying an orthogonal transformation to the orbitals and for this to make sense, the orbitals in question have to be occupied with an equal number of electrons, each. Carbon in its ground state is $$\mathrm{s}^2\mathrm{p}^2$$, where the s orbital is doubly occupied and 2 of the p orbitals singly occupied, so a hybridization of s and p makes no sense.
However, starting from the excited state $$\mathrm{s}^1 \mathrm{p}^3$$, it makes sense to form hybrid orbitals. Pauling calls the energy necessary to go to the excited state "promotion energy", and promotion energy has to be smaller than the energy gain by forming more directed and stronger bonds due to hybridization. In oxygen, the energetic separation between the s and p orbitals is much larger than in carbon, whence hybridization is unimportant for the description of bonding in e. g. H2O.

 

[robynmc]

For many atoms, hybridization is not possible starting from the electronic ground state but is only possible for excited states. Forming hybrid orbitals is equivalent to applying an orthogonal transformation to the orbitals

So an endothermic process caused by electronegativity or something added, and as binding energy increases the configuration can change direction to absorb energy before reaching the ionisation energy? Configuration as available states before ionisation energy is reached?

and for this to make sense, the orbitals in question have to be occupied with an equal number of electrons, each. Carbon in its ground state is $$\mathrm{s}^2\mathrm{p}^2$$, where the s orbital is doubly occupied and 2 of the p orbitals singly occupied, so a hybridization of s and p makes no sense.
However, starting from the excited state $$\mathrm{s}^1 \mathrm{p}^3$$, it makes sense to form hybrid orbitals. Pauling calls the energy necessary to go to the excited state "promotion energy", and promotion energy has to be smaller than the energy gain by forming more directed and stronger bonds due to hybridization.

The promotion energy is immediately lost and returns to ground state in the absence of something to point the orthogonal transformation of the configuration at?

In oxygen, the energetic separation between the s and p orbitals is much larger than in carbon, whence hybridization is unimportant for the description of bonding in e. g. H2O.

This makes me think I'm still misunderstandng. The s orbital is not needed because there's a bunch of p orbitals reconfiguring. Sorry if I'm misusing terminology, I think I'm moving through the material too fast. Juggling to many new concepts!

 

[robynmc]

The nuclear charge is only there to provide the Coulomb potential. The perturbation comes from electron-electron interactions.I don't understand the link you are trying to make between bond lengths and activation energies. Do you mean as bond stretch during chemical reactions?

I think it's binding energies that have confused me. it looked like s orbital shouldn't be able to move into a p orbital, but it seems there's states available where it can

 

[DrJohn]

Hydribisation is a way to visualise the bonding and the geometry of molecules. It's simple, easy to work with and an easy way to get the geometry of a molecule, via VSEPR rules. It's possible you are reading too much into this topic and trying to relate things unnecessarily. Which might be the source of your confusion.

Hybridisation could perhaps be thought of as a simple substitute for molecular orbitals. Its big strength is it explains things in simple terms until you start digging into the fine details of the bonding. There are textbooks that do dig into the fine details, but in reality, unless you are good at the theory side of things, you don't need to study them.

Just accept the basic idea - atomic orbitals combine to make hybrid orbitals that then form the bonds in compounds and help you predict the shape of the molecule. It's almost a case of just accepting the idea and don't try to dig into it to relate it to the things you keep asking about. Molecular orbital theory however is a much better way to understand things like the number of unpaired electrons, spectra, aromatic bonding etc.

 

[DrDu]

Molecular orbital theory however is a much better way to understand things like the number of unpaired electrons, spectra, aromatic bonding etc.

Is it really? MO theory does not even yield a bond in rather simple molecules like F2. I would not like to discard the insights from either a VB nor MO point of view.

 

[DrJohn]

Paramagnetic O₂ - predicted / explained by MO theory, not by VB theory
Google for molecular orbitals in F₂ and look at the images section - clearly explains there are three bonding orbitals, and two antibonding orbitals, giving a net single bond.
Structure and bonding of benzene.

But let's not argue, as MO is usually only taught on an undergraduate course, while VB is taught at high school. One is easy for high school - VB, one is better for practising chemists - MO. Both work in many cases, but MO works in the cases that VB can't explain.

PS I never really needed MO for my research, VB was simple enough. Except I was working with compounds with planar N , such as PF₂NH₂ derivatives, and PF₂NHPF₄, my favourite new compound. And VB struggled with the idea of a planar NH₂X group. (Planar as shown by the electron diffraction molecular studies.)

 

[DrDu]

Paramagnetic O₂ - predicted / explained by MO theory, not by VB theory
Google for molecular orbitals in F₂ and look at the images section - clearly explains there are three bonding orbitals, and two antibonding orbitals, giving a net single bond.

Unpaired electrons and corresponding paramagnetism in O2 is also predicted by VB theory, two 3 electron bonds are simply stronger than one two electron bonds and one 4 electron bond.
This is at least known since 1937: https://pubs.rsc.org/en/content/articlelanding/1937/tf/tf9373301499/unauth
F2 is well known not to be stable in the Hartree-Fock limit: https://www.chem.pku.edu.cn/jianghgroup/docs/20190415114013847385.pdf
It is simply that the ionic structures in a MO wavefunction are wrong to describe molecules made up of highly electronegative elements like fluorine. A simple VB wavefunction (GVB in the paper) already yields a bound state.
This is standard undergrad knowledge by now: https://www.degruyter.com/document/doi/10.1515/9783110206852/html

 

[robynmc]

Hydribisation is a way to visualise the bonding and the geometry of molecules. It's simple, easy to work with and an easy way to get the geometry of a molecule, via VSEPR rules. It's possible you are reading too much into this topic and trying to relate things unnecessarily. Which might be the source of your confusion.

Hybridisation could perhaps be thought of as a simple substitute for molecular orbitals. Its big strength is it explains things in simple terms until you start digging into the fine details of the bonding. There are textbooks that do dig into the fine details, but in reality, unless you are good at the theory side of things, you don't need to study them.

Just accept the basic idea - atomic orbitals combine to make hybrid orbitals that then form the bonds in compounds and help you predict the shape of the molecule. It's almost a case of just accepting the idea and don't try to dig into it to relate it to the things you keep asking about. Molecular orbital theory however is a much better way to understand things like the number of unpaired electrons, spectra, aromatic bonding etc.

I was confusing map and territory I think. Valence bond theory for visualising structure, ok. Thanks!