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P orbitals

by Dual Op Amp
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Dual Op Amp
#1
Aug29-04, 10:26 AM
P: 151
How would three P orbitals, X,Y,Z, in the same subshell, give the electrons different quantum energies?
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marlon
#2
Aug29-04, 11:15 AM
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Quote Quote by Dual Op Amp
How would three P orbitals, X,Y,Z, in the same subshell, give the electrons different quantum energies?

Because of the spin-orbit-coupling. It is this coupling that gives rise to the 3 p-orbitals.

When you would apply an extern electric field, the three orbitals will position themselves in three different ways, each corresponding to a different energy-level.

regards
marlon
Dual Op Amp
#3
Oct11-04, 12:36 PM
P: 151
I don't understand this, I am talking about atoms that obey the octec rule. The second subshell of the second shell, it has three P orbital, I want to know why, there can't be an external electric field in everything that obeys the octec rule. I'm saying quantum physics states that no two electrons can have the same energy, and since the higher the shell, the more energy, no two electrons can exist in the same subshell. except for the fact that two electrons can spin in the opposite direction. Yet, there are a total of 3 P orbitals in an octec obeying atom, that's 6 electrons, not 2.

Pieter Kuiper
#4
Oct11-04, 01:21 PM
P: 143
P orbitals

Quote Quote by marlon
When you would apply an extern electric field, the three orbitals will position themselves in three different ways, each corresponding to a different energy-level.
This is correct. It is a kind of crystal field splitting.

But this is independent from spin-orbit splitting, which in itself splits the p-levels in [tex]p_{1/2}[/tex] and [tex]p_{3/2}[/tex].
so-crates
#5
Oct16-04, 04:33 PM
P: 309
The Pauli principle states that no two electrons can have the same set of principle quantum numbers. These numbers are n, l, m_l, and m_s.

n is the principle quantum number that would be identified as the energy level, starting at 1. This is the first number in your electron configuration. For example hydrogen with a configuration of 1s2. The 1 is the principle quantum number.

l is your angular quantum number. Its range is 0...n - 1. This represents your “s,p,d,f” orbitals. For an s orbital, l = 0, for a p orbital, l = 1, and so forth. This number represents the angular momentum of the orbit. At a given energy, a circular path has minium angular momentum and that is why S orbitals with appear spherical. The more eccentric the orbit (more elliptical) the more angular momentum it posseses at a given energy. Thus as l increases, orbitals get more eccentric and spike-like.

m_l is the magnetic quantum number, whose range is -l(“l” not one)... -1, 0, 1 ... l. We see here that for l = 1 (the p orbital), l can be either -1, 0 or 1. Thus at the we have three different p orbitals which are conveniently called x, y and z.

m_s is the spin magnetic quantum number, also called "electron spin", which can be +/- 1/2.

The key here is that unless the atom is in a magnetic field, you cannot distinguish between the different m_l and m_s values.

Also keep in mind we are talking about individual atoms and not molecules. For molecules, things get more complicated with hybridized orbitals and bonding/anti-boding orbitals.
Pieter Kuiper
#6
Oct17-04, 06:06 PM
P: 143
The preceding post is not correct.

For one thing, s-orbitals have no angular momentum, and in a classical picture they are highly elliptical. The orbits would look like those of periodic comets.
p-orbitals have angular momentum of [tex]\hbar[/tex], d-orbitals of [tex]2\hbar[/tex], etcetera. Sometimes orbitals where [tex]l=n-1[/tex] (3d, 4f, 5g) are called circular orbitals.
reilly
#7
Oct22-04, 01:54 PM
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so-crates is absolutely correct. Any state in an s-wave state is, by definition an L=0 state, is spherically symmetric -- the angular wave function is a constant, a zero-order Legendre Polynomial. A classical orbit with zero angular momentum is impossible -- there are two ways to get a L=0 state: a particle at rest, a particle in linear motion incident on the proton/CM. So the classical picure and the QM picture are very different for small angular momentum, but, necessarily quite similar for large L (See, for example, Edmunds book on angular momentum, in which he looks at the asymptotic forms of high L wave functions.)

Any non-speherical pertubing force will split the p states. The reasoning is very similar to that underlying multipole expansions in electrodynamics.
Regards,
Reilly Atkinson
marlon
#8
Oct22-04, 04:36 PM
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Indeed so-crates is completely right...

Great post, reilly

marlon
Pieter Kuiper
#9
Oct23-04, 05:39 AM
P: 143
Quote Quote by so-crates
At a given energy, a circular path has minium angular momentum and that is why S orbitals with appear spherical. The more eccentric the orbit (more elliptical) the more angular momentum it posseses at a given energy. Thus as l increases, orbitals get more eccentric and spike-like.
These statements are false in classical mechanics, as demonstrated on
http://www.physics.csbsju.edu/orbit/orbit.2d.html

There seems to be confusion in the literature. For example:
"for a given energy, the larger the angular momentum the more elongated the orbit."
http://www.telescope.org/nuffield/solar/solar9d.html
"For a given energy, the orbit with the minimum angular momentum is a circular one."
http://astro.pas.rochester.edu/~aqui.../Lecture2b.pdf
But these statements are wrong. For a circle the eccentricity [tex]e=0[/tex]. The angular momentum is proportional to [tex]\sqrt{1-e^2}[/tex]. So it is maximal for a circle.

Quote Quote by reilly
A classical orbit with zero angular momentum is impossible -- there are two ways to get a L=0 state: a particle at rest, a particle in linear motion incident on the proton/CM.
I agree, but the classical analogy is useful when one wants to explain why s-orbitals at the same time have a high electron density at the nucleus and extend further from the center of the atom.
what_are_electrons
#10
Oct24-04, 02:37 AM
P: n/a
Perhaps we should wait until someone has managed to measure the shape of an s-orbital before we declare what the shape is?

Please note that the electrons must interact with the nucleons which suffer some slippage and have a somewhat different shell structure that lead to asymmetry in the electron orbitals.
ZapperZ
#11
Oct24-04, 08:27 AM
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Quote Quote by Pieter Kuiper
I agree, but the classical analogy is useful when one wants to explain why s-orbitals at the same time have a high electron density at the nucleus and extend further from the center of the atom.
Again, we need to be a bit careful here. While it is true that the radial wavefunction R_nl for l=0 has a probability density that peaks at r=0, if we want to say anything about the electron position, then what is relevant is the product of rR_nl. This is because you need to actually find the probability of the average position via |<R_nl|r|R_nl>|^2.

Already, you can tell that r=0, something else happens. In fact, it is ZERO at r=0. So the electron really does not have a substantial probability to be right at nucleus. I think that people who have gone through a QM class at some point had to either solve for, or even plot the rR_nl function (I know I did). This is exactly the reason why.

Zz.
Pieter Kuiper
#12
Oct24-04, 10:16 AM
P: 143
Quote Quote by ZapperZ
Again, we need to be a bit careful here. While it is true that the radial wavefunction R_nl for l=0 has a probability density that peaks at r=0, if we want to say anything about the electron position, then what is relevant is the product of rR_nl. This is because you need to actually find the probability of the average position via |<R_nl|r|R_nl>|^2.
Of course, that is the way to calculate the expectation value of the electron distance to the nucleus. The outcome is n^2 times the Bohr radius.

Already, you can tell that r=0, something else happens. In fact, it is ZERO at r=0. So the electron really does not have a substantial probability to be right at nucleus.
The product rR_nl is a way to calculate the radial density P(r) for finding the the electron at a distance r from the nucleus. It is related to the probability density by [tex]r^2 |\Psi(x,y,z)|^2[/tex]. The radial probability is zero at the origin because a point has zero surface and volume.

In contrast to orbitals with [tex]l \neq 0[/tex], s-orbitals do have a contact density at the nucleus. This is measurable. It is responsible for the Knight shift in nuclear-magnetic-resonance experiments and for the isomer shift in Mössbauer spectroscopy.

The comparison with classical orbits is illuminative. Comets in their highly elliptical orbits spend very little time close to the sun. But planets in their circular orbits get never close to the sun.
ZapperZ
#13
Oct24-04, 10:41 AM
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Quote Quote by Pieter Kuiper
In contrast to orbitals with [tex]l \neq 0[/tex], s-orbitals do have a contact density at the nucleus. This is measurable. It is responsible for the Knight shift in nuclear-magnetic-resonance experiments and for the isomer shift in Mössbauer spectroscopy.
I'm not sure how you can tell, based on the Knight shift measurement, that there is a "contact" density of the s-orbital electron with the nucleus. The shift in the resonance frequency, especially in metals, have NOTHING to do with any "contact" between any electrons and the nuclei in that metals. This has everything to do with the nuclear interaction with the conduction electrons, which certainly is WAY out of the nucleus since these are delocalized electrons.

The issue here is still that the expectation value of finding an s-orbital electron at the nucleus DROPS towards zero as one approaches the nucleus for a radius less than some mean value. At the very least, it is inaccurate to say that "....s-orbitals at the same time have a high electron density at the nucleus."

Zz.
Pieter Kuiper
#14
Oct24-04, 11:21 AM
P: 143
Quote Quote by ZapperZ
I'm not sure how you can tell, based on the Knight shift measurement, that there is a "contact" density of the s-orbital electron with the nucleus. The shift in the resonance frequency, especially in metals, have NOTHING to do with any "contact" between any electrons and the nuclei in that metals. This has everything to do with the nuclear interaction with the conduction electrons, which certainly is WAY out of the nucleus since these are delocalized electrons.
In a very simple picture, delocalized electrons are like a free electron gas, so they are everywhere, also at the nuclei. If you put in the attractive Coulomb potential, the density at the nucleus increases, and these electrons have local s-character.

The Knight shift measures contact density of the wave function at the Fermi level, so one can only do this for metals. The isomer shift in Mössbauer is a bit easier to interpret. It "arises due to the non-zero volume of the nucleus and the electron charge density due to s-electrons within it." http://www.rsc.org/lap/rsccom/dab/mo...spec/part2.htm

The issue here is still that the expectation value of finding an s-orbital electron at the nucleus DROPS towards zero as one approaches the nucleus for a radius less than some mean value. At the very least, it is inaccurate to say that "....s-orbitals at the same time have a high electron density at the nucleus."
You are wrong. The radial density (the chance of finding an electron in a shell with thickness dr and radius r) decreases as r decreases, simply because the volume (the surface) of the shell decreases. But the electron density of s-electrons increases towards the nucleus.
http://www.chemistry.mcmaster.ca/esa...section_2.html
ZapperZ
#15
Oct24-04, 01:25 PM
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Quote Quote by Pieter Kuiper
In a very simple picture, delocalized electrons are like a free electron gas, so they are everywhere, also at the nuclei. If you put in the attractive Coulomb potential, the density at the nucleus increases, and these electrons have local s-character.
Er... no. These delocalized electrons are everywhere in the material, but this is a "cartoon" picture as an approximation. There are no indication that they actually penetrates right to the nucleus, simply because the core-level electrons are still there! The fact that they are the outermost electrons is the reason why their atomic orbitals have been hybridized with neighboring atomic orbitals and thus, they do not retain any isolated energy levels.

The Knight shift measures contact density of the wave function at the Fermi level, so one can only do this for metals. The isomer shift in Mössbauer is a bit easier to interpret. It "arises due to the non-zero volume of the nucleus and the electron charge density due to s-electrons within it." http://www.rsc.org/lap/rsccom/dab/mo...spec/part2.htm
Again, the fact that there IS a Fermi level implies that these electrons are no where near the nucleus. Just by mentioning that these are "free" electrons already imply that they experience NO potentials from the lattice ions, much less, the nucleus. So saying they still interact with the nucleus is contradictory. The Knight shift does not give any information about the proximity of these conduction electrons to the nucleus.

You are wrong. The radial density (the chance of finding an electron in a shell with thickness dr and radius r) decreases as r decreases, simply because the volume (the surface) of the shell decreases. But the electron density of s-electrons increases towards the nucleus.
http://www.chemistry.mcmaster.ca/esa...section_2.html
Again, there appears to be a misconception regarding R_nl and rR_nl. I have indicated that the probability density of the radial part of the wave function for all l=0 DOES peak at r = 0. However, you cannot use this to say that this then indicates that the s-electron spends MOST of its time at the nucleus, or has the highest probability of being found at the nucleus. The mathematics just does not agree with that.

Zz.
Pieter Kuiper
#16
Oct24-04, 02:12 PM
P: 143
Quote Quote by ZapperZ
Er... no. These delocalized electrons are everywhere in the material, but this is a "cartoon" picture as an approximation. There are no indication that they actually penetrates right to the nucleus, simply because the core-level electrons are still there! The fact that they are the outermost electrons is the reason why their atomic orbitals have been hybridized with neighboring atomic orbitals and thus, they do not retain any isolated energy levels.
Of course their wave functions have to be orthogonal to the wave functions of the core electrons. Close to the nucleus the wave functions of the conduction electrons in say sodium are very similar to atomic Na 3s wave functions. Qualitatively (number of radial nodes) these wave functions look like hydrogen 3s wave functions.

The clearest indication that s-symmetry wave functions penetrate core-level electron clouds comes from the periodic table: the energy of the 4s orbital in potassium and calcium is lower than the energy of their 3d orbitals. This energy difference is due to the stronger coulomb interaction of the nucleus with the 4s electrons. The 3d electrons are in circular orbits (classically speaking) and do not get that close to the atom's core.

Again, there appears to be a misconception regarding R_nl and rR_nl. I have indicated that the probability density of the radial part of the wave function for all l=0 DOES peak at r = 0. However, you cannot use this to say that this then indicates that the s-electron spends MOST of its time at the nucleus, or has the highest probability of being found at the nucleus. The mathematics just does not agree with that.
Your retreat is rather ungracious.

I never said "most of its time". I made the comparison with comets. Most of the time they are frozen at distances beyond Pluto. But once in a while they come for a short time close to the sun.

I just said that the probability density of s-electrons is high at the nucleus.
And earlier you said that was inaccurate.
ZapperZ
#17
Oct24-04, 06:10 PM
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Quote Quote by Pieter Kuiper
Of course their wave functions have to be orthogonal to the wave functions of the core electrons. Close to the nucleus the wave functions of the conduction electrons in say sodium are very similar to atomic Na 3s wave functions. Qualitatively (number of radial nodes) these wave functions look like hydrogen 3s wave functions.

The clearest indication that s-symmetry wave functions penetrate core-level electron clouds comes from the periodic table: the energy of the 4s orbital in potassium and calcium is lower than the energy of their 3d orbitals. This energy difference is due to the stronger coulomb interaction of the nucleus with the 4s electrons. The 3d electrons are in circular orbits (classically speaking) and do not get that close to the atom's core.
But this is highly misleading. The 4s orbital has a smaller radius ON AVERAGE, and thus, is less screened than the 3d. This say nothing about their penetration to the nucleus. This is what I have been arguing about. If you are using this argument, then you cannot ignore the expectation value calculations, which clearly stated that at r=0, |<r>|^2 is zero!

Secondly, you are also contradicting yourself. If what you said about the 4s and 3d orbitals is true, then transitions metals which have 3d valence shell should, in fact, be farther away and ARE the conduction electrons that are responsible for the Knight shift. Yet, we all know the d orbital does not get anywhere near the nucleus. The fact that you can still get a Knight shift from them CLEARLY indicates that the Knight shift has nothing to do with electrons coming in contact with the nucleus.

Your retreat is rather ungracious.

I never said "most of its time". I made the comparison with comets. Most of the time they are frozen at distances beyond Pluto. But once in a while they come for a short time close to the sun.

I just said that the probability density of s-electrons is high at the nucleus.
And earlier you said that was inaccurate.
Ungracious?

To say that there is a "high density" of anything means that there is a high probability of finding it there. |<r>|^2 approaches zero for the s-orbital at r=0. Maybe you have a different definition for |<r>|^2, but it is the expectation value that determine how likely the electron is found at a given location.

If you mean "high density" as in the probability density, well I have stated that in my first posting in this thread. However, the probability density of the radial wavefunction R_nl does NOT tell you the probability of finding anything anymore than the prob. density of the wavefunction |<psi>|^2. Without an observable operator involved, this tells you nothing.

Zz.
Pieter Kuiper
#18
Oct25-04, 12:45 AM
P: 143
Quote Quote by ZapperZ
If you are using this argument, then you cannot ignore the expectation value calculations, which clearly stated that at r=0, |<r>|^2 is zero!
Let me concede this trivial truism, and end my discussion with you.


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