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Author Topic: hybridisation  (Read 586 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: Yesterday at 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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