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P orbitalsby Dual Op Amp
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#1
Aug2904, 10:26 AM

P: 151

How would three P orbitals, X,Y,Z, in the same subshell, give the electrons different quantum energies?



#2
Aug2904, 11:15 AM

P: 4,006

Because of the spinorbitcoupling. It is this coupling that gives rise to the 3 porbitals. When you would apply an extern electric field, the three orbitals will position themselves in three different ways, each corresponding to a different energylevel. regards marlon 


#3
Oct1104, 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.



#4
Oct1104, 01:21 PM

P: 143

P orbitals
But this is independent from spinorbit splitting, which in itself splits the plevels in [tex]p_{1/2}[/tex] and [tex]p_{3/2}[/tex]. 


#5
Oct1604, 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 spikelike. 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/antiboding orbitals. 


#6
Oct1704, 06:06 PM

P: 143

The preceding post is not correct.
For one thing, sorbitals have no angular momentum, and in a classical picture they are highly elliptical. The orbits would look like those of periodic comets. porbitals have angular momentum of [tex]\hbar[/tex], dorbitals of [tex]2\hbar[/tex], etcetera. Sometimes orbitals where [tex]l=n1[/tex] (3d, 4f, 5g) are called circular orbitals. 


#7
Oct2204, 01:54 PM

Sci Advisor
P: 1,082

socrates is absolutely correct. Any state in an swave state is, by definition an L=0 state, is spherically symmetric  the angular wave function is a constant, a zeroorder 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 nonspeherical pertubing force will split the p states. The reasoning is very similar to that underlying multipole expansions in electrodynamics. Regards, Reilly Atkinson 


#9
Oct2304, 05:39 AM

P: 143

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{1e^2}[/tex]. So it is maximal for a circle. 


#10
Oct2404, 02:37 AM

P: n/a

Perhaps we should wait until someone has managed to measure the shape of an sorbital 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. 


#11
Oct2404, 08:27 AM

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Sci Advisor
PF Gold
P: 29,239

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. 


#12
Oct2404, 10:16 AM

P: 143

In contrast to orbitals with [tex]l \neq 0[/tex], sorbitals do have a contact density at the nucleus. This is measurable. It is responsible for the Knight shift in nuclearmagneticresonance 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. 


#13
Oct2404, 10:41 AM

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PF Gold
P: 29,239

The issue here is still that the expectation value of finding an sorbital 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 "....sorbitals at the same time have a high electron density at the nucleus." Zz. 


#14
Oct2404, 11:21 AM

P: 143

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 nonzero volume of the nucleus and the electron charge density due to selectrons within it." http://www.rsc.org/lap/rsccom/dab/mo...spec/part2.htm http://www.chemistry.mcmaster.ca/esa...section_2.html 


#15
Oct2404, 01:25 PM

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P: 29,239

Zz. 


#16
Oct2404, 02:12 PM

P: 143

The clearest indication that ssymmetry wave functions penetrate corelevel 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. 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 selectrons is high at the nucleus. And earlier you said that was inaccurate. 


#17
Oct2404, 06:10 PM

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P: 29,239

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. 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 sorbital 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. 


#18
Oct2504, 12:45 AM

P: 143




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