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Author Topic: hybridisation  (Read 2337 times)

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overlord16

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hybridisation
« on: May 30, 2017, 02:16:43 AM »

'hybridization is an operation carried out not actually on orbitals but on the mathematical functions that define them'. Does this means that hybridization does not alter the actual shape of orbitals and no actual mixing of orbitals takes place?
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Burner

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Re: hybridisation
« Reply #1 on: June 13, 2017, 05:13:54 AM »

It is better to say that hybridization is a 'mathematical' theory used to explain molecular geometries. Chemists introduces the concept of hybridization of atomic orbitals to explain the shape of molecules in valence bond theory. It is only one of the many theories that generally works in predicting/explaining molecular shapes, others including molecular orbitals theory, VSEPR...

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Enthalpy

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Re: hybridisation
« Reply #2 on: June 13, 2017, 12:57:31 PM »

'hybridization is an operation carried out not actually on orbitals but on the mathematical functions that define them'.
This sentence is meaningless, so don't bother to understand it. If the maths differ from the observation, it means that the model is wrong and ripe for the trash bin. Who wrote that, a scientist?
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Irlanur

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Re: hybridisation
« Reply #3 on: June 13, 2017, 09:52:14 PM »

Quote
    'hybridization is an operation carried out not actually on orbitals but on the mathematical functions that define them'.

This sentence is meaningless, so don't bother to understand it. If the maths differ from the observation, it means that the model is wrong and ripe for the trash bin. Who wrote that, a scientist?

I disagree a bit. The sentence is meaningless in a way that orbitals are mathematical objects only anyway (apart from the Hydrogen atom maybe). But it is correct in the way that Hybridization is by no means a physical process. it is a unitary transformation of mathematical objects (the orbitals).
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overlord16

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Re: hybridisation
« Reply #4 on: June 14, 2017, 08:46:55 AM »

'hybridization is an operation carried out not actually on orbitals but on the mathematical functions that define them'.
This sentence is meaningless, so don't bother to understand it. If the maths differ from the observation, it means that the model is wrong and ripe for the trash bin. Who wrote that, a scientist?

It was written in Peter Sykes.
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Enthalpy

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Re: hybridisation
« Reply #5 on: June 15, 2017, 09:11:20 AM »

And so he meant that useful mathematical operations don't need to reproduce the observations?
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overlord16

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Re: hybridisation
« Reply #6 on: June 17, 2017, 12:43:26 AM »

And so he meant that useful mathematical operations don't need to reproduce the observations?

That is what I didn't understand. The above sentence is quoted as it is from the book.
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Irlanur

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Re: hybridisation
« Reply #7 on: June 18, 2017, 09:11:30 PM »

Quote
And so he meant that useful mathematical operations don't need to reproduce the observations?

I can't read this into the original sentence...
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Enthalpy

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Re: hybridisation
« Reply #8 on: June 22, 2017, 08:35:26 AM »

"In maths but not for real" is just the opposite of a successful model. Either they do hybridize in the observations and the corresponding maths are useful, or both are wrong.

Or in intermediate cases, which is the case for hybridization, the model is known to be wrong, is outdated, but is still useful for it simplicity. Then the math operations shall please give at least some results that resemble some observations.

I wouldn't put that "orbitals are mathematical objects only". Wave functions are observable and observed, precisely in the case of orbitals. They define, when they're written as functions of xyzt, from which locations the particle acts simultaneously and with which amplitude and phase. Few things could be less real, even if the observation isn't always trivial, and even if we have no simple mathematical expression for them most often.

And more generally, I hate commenting wording details about QM. That's the wrong approach.
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Irlanur

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Re: hybridisation
« Reply #9 on: June 25, 2017, 08:45:32 PM »

Quote
I wouldn't put that "orbitals are mathematical objects only". Wave functions are observable and observed, precisely in the case of orbitals.

Orbitals are only wave functions in the case of atomic hydrogen. In any other case, they are mathematical functions only, with little or no meaning on its own. We write an approximation of the real wave function in terms of these orbitals. Hybridization then is not a physical process. not at all. it is a unitary transformation of the orbital basis only. That's what I think the author of the original sentence meant. And I think he is right. And I think this should be emphasized much more, given the amount of similar questions in chemistry forums.
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Enthalpy

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Re: hybridisation
« Reply #10 on: June 25, 2017, 09:55:15 PM »

It must be a matter of wording then.

I understand orbitals as wave functions of electrons trapped in atoms or molecules or ions, whether we know an algebraic solution or not. For instance, we say Homo and Lumo despite no exact expression is known.
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Enthalpy

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Re: hybridisation
« Reply #11 on: June 27, 2017, 07:50:35 PM »

And I've forgotten that orbitals are only the stationary solutions.
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Irlanur

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Re: hybridisation
« Reply #12 on: June 29, 2017, 10:52:40 PM »

Maybe I just quote wiki for once:

Quote
Bonding orbitals formed from hybrid atomic orbitals may be considered as localised molecular orbitals, which can be formed from the delocalised orbitals of molecular orbital theory by an appropriate mathematical transformation [a unitary transformation (Irlanur)]. For molecules with a closed electron shell in the ground state, this transformation of the orbitals leaves the total many-electron wave function unchanged. The hybrid orbital description of the ground state is therefore equivalent to the delocalised orbital description for ground state total energy and electron density, as well as the molecular geometry that corresponds to the minimum total energy value.

I think that mathematical transformations that leave observables unchanged do not have a physical meaning. And since the individual orbitals are in this sense arbitrary (in contrast to the actual wavefunction) I would say that they do not have a physical meaning themselves. Nevertheless, in certain situations, there might be some heuristics which allow one to draw conclusions from orbitals in a certain basis. (e.g. photoelectron spectra from Koopmann's theorem).

To come back to the original question. I think the sentence makes somewhat sense. The wording is strange because orbitals are just mathematical functions, so I would not know what 'orbitals themselves' are.
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Enthalpy

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Re: hybridisation
« Reply #13 on: July 03, 2017, 12:24:24 AM »

If the mathematical transformation is correct, then the solution must exist physically too, unless the theory has a flaw. True, you pointed out "effect not observable".

The problem I see with hybridization is that carbon's 2s and 2p have different energies. While any linear combination of these orbitals still is a valid wave function for a trapped electron, it is not a stationary solution, not an orbital. Explanations I've seen up to now combine the amplitudes of 2s and 2p, which are incomplete expressions of the wave functions, and forget the energy term. This is not a valid transformation - unless I missed something.

So to my understanding, the math is wrong and experiments refute the hybridization model too (notably the energy spectrum of methane). So I find pretty futile to say "maths but not real" or any variant when both are wrong.
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Irlanur

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Re: hybridisation
« Reply #14 on: July 03, 2017, 04:55:53 AM »

I suspect a problem in your understanding (and I feel this is a rather bold statement).

Orbitals themselves are one-electron functions. They are never stationary solutions of a many-electron problem. In the usual formalism, the total wavefunction is a slater-determinant which combines products of orbitals in an anti-symmetric way. A unitary transformation among the orbitals DOES Change the orbitals, but it does not change the slater-determinant (i.e. the wavefunction), and it also doesn't change the total energy.

In a way, hybridization is a coordinate transformation (a coordinate transformation is also a unitary transformation). A coordinate transformation is a mathematical process which does not change the physics of the system. BUT: some coordinate systems are more intuitive and also mathematically easier to handle as others.
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