[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 148 KB, 1404x824, orbits.jpg [View same] [iqdb] [saucenao] [google]
8192493 No.8192493 [Reply] [Original]

I'm trying to learn more basic physical chem and I've never been satisfied with the explanations given for pic-related representations of electron orbitals.

What does the 3-D surface represent? Is that some particular spatial probability density for finding an electron there, and if so what is that density? And what is meant by the opposite lobe of a p-orbital having this "negative" potential; how can it have negative probability density?

Also, e.g. for s-orbitals the density seems to peak at the nucleon location -- is it actually possible to find an electron right in the middle of the nucleus?

Finally how do energy levels modulate each orbital type? Are the lobes the exact same shape but expand farther out in space? Or are they somehow offset some distance from the nucleon and within that spherical shell, there can be no electron with that higher energy state? Or maybe still, they occupy exactly the same space but have higher momenta? The traditional model seems to imply that energy levels have "distinct" orbital regions, but from these model representations I can't see how that's possible unless they just overlap in space.

Thanks in advance

>> No.8192504 [DELETED] 

>>8192493
>that is some particular spatial probability for finding an electron there
fixed

Also, on the title, most of quantum, and in-turn, subatomic interactions are non-intuitive which is a difficult to grasp.

>> No.8192506

Just do the quantum. There is no intuitive explanation outside the math, which itself is blindingly obvious once you've worked through it.

>> No.8192507

>>8192493 (OP)
>that is some particular spatial probability for finding an electron there
fixed

Also, on the title, most of quantum, and in-turn, subatomic interactions are non-intuitive which is a difficult concept to grasp.

>> No.8192509

>>8192493
>is it actually possible to find an electron right in the middle of the nucleus?
Not a physicist, but I'm almost positive the answer is absolutely no.

>> No.8192514

>>8192506
Okay. But what is this model intended to even represent, in relation to the above questions? I would have thought these questions can be answered, they are pretty straightforward.

>> No.8192517
File: 3.00 MB, 480x360, Nothing But Trash.webm [View same] [iqdb] [saucenao] [google]
8192517

>>8192493
>electron orbital

Electrons don't "orbit."

>> No.8192523
File: 66 KB, 625x626, bait.png [View same] [iqdb] [saucenao] [google]
8192523

>>8192517

>> No.8192525

>>8192514
if you're asking questions like
>is it actually possible to find an electron right in the middle of the nucleus?
you should probably just start with wiki

>> No.8192527
File: 3.47 MB, 371x365, 1466872863192.gif [View same] [iqdb] [saucenao] [google]
8192527

>>8192493
>What does the 3-D surface represent?
They are isoprobability surfaces. The interpretation of the electron wavefunction (essentially what orbitals describe) is that the magnitude squared is equal to the probability density of finding an electron at a point in space. To find the probability of finding an electron within a volume, you integrate the probability density.

It is difficult to visualize 3D fields, so usually peeps pick a certain value and plot it's surface to get an idea of the shape of the field. That's what you see in your plots.

> And what is meant by the opposite lobe of a p-orbital having this "negative" potential
I think what people are trying to draw attention to is that the phase of the opposite sides of orbitals is opposite. When you square a single orbital, this "feature" cancels out and you get the shapes you see in your figure ignoring colors. However, when an electron's state is a combination of orbitals, the phase of the complex-valued wavefunction dictates how the orbitals interfere (be it constructive or destructive) with each other.

>Also, e.g. for s-orbitals the density seems to peak at the nucleon location -- is it actually possible to find an electron right in the middle of the nucleus?
Absolutely, although if you only integrate over a single point, the probability that you'll find an electron is 0.

In the world of quantum mechanics, the only thing not allowed for the family of ferimions (which electrons and nucleons are members of) is that no two fermions can share the same quantum state (read quantum numbers). Otherwise their is no notion of "solid" in the sense that things can occupy the same space. The quantum world is very ghostly in that respect. So how does an atom support itself? Reread the first sentence in this paragraph; for electrons this is called the Pauli exclusion principle, but it applies to fermions in general.

>Finally how do energy levels modulate each orbital type?
next post...

>> No.8192848

>>8192527
Thanks! Definitely interested in next post if you have time.

>> No.8192852

>>8192517
>>8192506
Fags.

>>8192525
>>8192509
Wrong.

>> No.8192968

>>8192493
Representative of probability I believe, where you can find electrons around the nucleus most of the time.

For s orbitals you can't find e at nucleus

For ie: Pi electrons the "negative" potential is just the sign of the wave function I believe. It's been a while since my phys chem, I do analytical

>> No.8192970

>>8192527
rip robopupper

>> No.8192974
File: 106 KB, 1101x725, hydrogen_orbitals___poster_by_darksilverflame-d5ev4l6.jpg [View same] [iqdb] [saucenao] [google]
8192974

>>8192493
The 3D surfaces basically represent the angular component of the wavefunction. The further out the surface is the greater the absolute value is at that angle. The probability density is the mofulus of the wavefunction squared so that's how the imaginary and negative values contribute.

It's hard to represent the full wavefunction in a 3D image because there are 4 things that need plotting, 3D location and the value of the wavefunction itself. This image shows a 2D slice of the orbitals, the blurriness shows the radial dependence that isn't shown in the other image.

>> No.8192998

>>8192974
>mofulus
*modulus

>> No.8193753

>>8192974

What do the (x,y,z) triplets mean?

>> No.8193754

>>8192998
I want to name an operator that it sounds incredibly cute.
>>8192493
If you haven't heard it by now, they're level surfaces of some 3d field of probabilities that the electron will be in that position or so.

>> No.8193762

>>8193753
They're not coordinates. They are three quantum numbers which give the "orbitals" a certain principle quantum number, angular momentum, and magnetic quantum number. They're dubbed 'n', 'l', and 'm'.

There are certain allowed values of each quantum number. For instance, n (the principle quantum number) must be greater than zero. Accepted values of n are 1, 2, 3, ..., and so on. Then we have l, which is our angular quantum number, which must can be any integer between 0 and n - 1. Let's say n is 2, if n = 2, then accepted values of l are either 0 or 1. Last but not least there is m, the magnetic quantum number, which usually does not receive much attention. It could be any integer between negative and positive l. Suppose l = 1, if l = 1, then accepted values of m are -1, 0, and 1.

l is probably the most noted quantum number because the number l takes is what "orbital" we get. I'll show an example below:

For s orbitals we have l = 0,
for p orbitals we have l = 1,
for d orbitals we have l = 2,
for f orbitals we have l = 3.

Supposedly, there exists an l = 4, which is called the "g" orbital. If they do exist, then there are nine of them, which are derived from the hydrogen wavefunction.

>> No.8193781

So certain m's are omitted in that diagram just because they are vertically (z-direction) oriented / are unsuitable for the 2-d cross sections, right?

Junior year chem is coming back to me. Thanks.

>> No.8193783

>>8193781
Meant for:
>>8193762

>> No.8193797

>>8193783
>m's are omitted in that diagram just because they are vertically (z-direction) oriented
This is pretty much the case.

Btw, you're welcome.