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Old 04-18-2015, 01:40 PM   #6461
JamSessionFreak
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No, the probability density of finding the electron inside the atom or rather inside the atomic orbital is always 95%, that's how the orbital is defined. It's not 100% because that would include the entire universe due to the nature of waves.

Probability density is this special kind of integral where you integrate the square of the wave function of a particle. It's purely mathematical and it is related to where the electron wave is localised I think. I'm a bit wary of going further into this because it will force me to make analogies which I won't be completely sure about.

I put 'finding' in quotes because to observe an electron you have to bounce a photon off of it - this is scattering and due to the wavelength of the photons used I think nodes don't really have much effect on it. Again, going further into this is forcing me to make certain 'classical' visualisations which I'm not perfectly comfortable making due to my limited knowledge of the subject.

Exact opposite with the nodes. More nodes, higher energy state, less stable. In guitar terms, higher pitch means more energy and less stability. The string is most stable when unfretted so in it's base state.

This is a picture of all possible molecular orbitals of a benzene ring where the p-orbital electrons are delocalized in a cloud above and beneath the molecular plane. The bottom-most picture, where all of the blue orbital halfs are above the plane and all of the red ones are beneath it has no node - no changes switches between red and blue (remember, visualisation) - therefore it is the most stable one. The uppermost picture where the red and blue colours constantly switch has 6 nodes - 6 colour changes - and is the least stable. This state of 6 p-orbitals is actually less stable than six lone p-orbitals so if electrons would forced to 'inhabit' this state it would actually make the entire system less stable.

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Old 04-18-2015, 01:43 PM   #6462
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I'm sorry if I'm becoming more and more vague but this is the part of quantum physics where you start to rely heavily on mathematics. I can't describe it anymore like I'd describe say an orange. The orange is now becoming more and more abstract, defined not by its appearance and shape but by certain equations that govern how it behaves as an object.
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Old 04-18-2015, 01:48 PM   #6463
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I'm not sure how much they teach you in chemistry but s, p, d, f, ... orbitals are atomic orbitals of different energy states. As units of volume they look sort of like this. Notice how the s-orbital, the one in the lowest energy state, has no nodes and is just a ball. As you go to p and d orbitals the number of nodes increases which is indicative of higher energy states.



The nodes are the parts of empty space, where you are very unlikely to find an electron. So the entire orbital always has a 95% probability density for an electron - if you sum up the probability density of the entire orbital (this means integrating the square of the electron wave function across the entire volume of the orbital) you get 95%. But certain parts that make up the orbital have varying probability densities. Nodes have a probability density of practically zero so if you'd say integrate the function square across almost all of the orbital except the nodes, you'd still get 95% or maybe like 94,99999999% or some shit like that.
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Old 04-18-2015, 01:53 PM   #6464
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lol trust me, thats not vague for me

Learned pretttttty much nothing from 11th grade Chem. And the pi thing is kind of giving me some flashbacks to calc 1 and the class' s clock that had pi/2 as 6pm, etc...


Infinite energy for changing the bottom was it, right? So the top would not require much energy (if thats even possible to add electrons to), which would have the electromagnetic field more stable environment?


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Old 04-18-2015, 02:02 PM   #6465
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Oh, please don't connect this to potential wells and the energy needed to get an electron out of said well. That's going in the wrong direction or rather would require a shitton of text to explain every nuance.

Look at it like this - if an electron would inhabit the non-bonding molecular (not atomic) orbitals, it would have to vibrate at a higher frequency than if it would just be in a lone atomic orbital. So you would actually need to give the electron energy to inhabit that orbital. Since the electron is always looking to be in the lowest possible energy state, it would rather not bond at all than to inhabit that orbital, which is why it is called a non-bonding orbital.

However, if the electron inhabits the bonding molecular orbital it inhabits an energy state with a lower energy than the energy of a lone atomic orbital. Because of this the electron actually loses energy by bonding and that gives it a hard on.

To get two atoms to bond, you have to combine their atomic orbitals into an molecular orbital in such a way that their electrons lose energy when the atoms bond. The bonded state has a lower energy than the two nonbonded atoms so the electrons can get their perv on in the bonded state.

This basically means that the energy of the molecular orbital is lower than the sum of the energies of the two separate atomic orbitals.

And this is the theory of molecular orbitals which is one of the models we use to explain bonding and predict the thermodynamics of molecules. The other one is valence bond theory. One works better for some bonds and molecules, the other for other bonds and molecules but overall both are good enough that we don't throw them out the window.
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Old 04-18-2015, 02:52 PM   #6466
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After checking out some old textbooks I want to make a correction. Probability density is the unintegrated square of the wave function, by integrating it you get actual probability. As for the physical, measurable implications of said probability, I'd rather not comment on those because of some problems with 'collapsing' an electron down to a particle. Something about the observer effect I'm not completely sure about.

Maybe we're able to make some measurements I thought were impossible due to instrumental limitations.
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Old 04-20-2015, 12:28 AM   #6467
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Quote:
Originally Posted by JamSessionFreak
Look at it like this - if an electron would inhabit the non-bonding molecular (not atomic) orbitals, it would have to vibrate at a higher frequency than if it would just be in a lone atomic orbital. So you would actually need to give the electron energy to inhabit that orbital. Since the electron is always looking to be in the lowest possible energy state, it would rather not bond at all than to inhabit that orbital, which is why it is called a non-bonding orbital.

However, if the electron inhabits the bonding molecular orbital it inhabits an energy state with a lower energy than the energy of a lone atomic orbital. Because of this the electron actually loses energy by bonding and that gives it a hard on.

To get two atoms to bond, you have to combine their atomic orbitals into an molecular orbital in such a way that their electrons lose energy when the atoms bond. The bonded state has a lower energy than the two nonbonded atoms so the electrons can get their perv on in the bonded state.

This basically means that the energy of the molecular orbital is lower than the sum of the energies of the two separate atomic orbitals.


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Old 04-20-2015, 12:36 AM   #6468
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Originally Posted by jobbye6
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The most stable state
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