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Shapes of orbitals

  1. Dec 19, 2004 #1
    Just how was it determined from electron wavefunctions calculated by schrodingers equation?

    For example, if i take the 1S and 2P wavefunctions ( ground state wavefunction against distance from nucleus)

    http://www.shef.ac.uk/chemistry/orbitron/AOs/1s/wave-fn.html

    http://www.shef.ac.uk/chemistry/orbitron/AOs/2p/wave-fn.html

    Just how does this correspond to the actual shape of the orbital?



    i apologize of the question is a bit dumb...
     
  2. jcsd
  3. Dec 19, 2004 #2

    Gokul43201

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    You can not deduce the shape of the orbital from just the radial part of the wave function. You also need the angular parts. Then for a fixed value of [itex]|\psi| ^2 [/itex], you can plot [itex] (r, \theta, \phi) [/itex] that give you that value, and this 3D plot will give you the shape of the orbitals.

    PS : The question is certainly not "dumb" and this is often misunderstood by a lot of people.

    If I find the time, I'll see if I can show how a 2p orbital looks like a dumbell...or if you'd like me to do this, say so.
     
    Last edited: Dec 19, 2004
  4. Dec 20, 2004 #3
    You can do that without much math.Wow.Do it please.
     
  5. Dec 20, 2004 #4
    actually, i think it DOES require a lot of maths. But he is (i think) clever than most of us and can do it..

    It would really appreciative if you could show us, although if you don't want to, you dont have to.
     
  6. Dec 20, 2004 #5

    Gokul43201

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    The trick is in being able to draw polar plots roughly, from the spherical harmonics.

    Consider the state, n=2, l=1, m=0
    The polar part of the wavefunction is given by :
    [tex] |Y_1^0|^2 = \frac{3}{4 \pi} cos^2 \theta [/tex]

    Ignoring the constant term, the functional behavior is [itex] cos ^2 \theta [/itex], which we want to plot against [itex]\theta [/itex]

    Let's draw the 2D version of this plot (or you can simply graph this using Mathematica or a calculator that can do polar plots). Draw the X (theta = 90, -90) and Z (theta = 0, 180) axes, and draw a large number of lines, all passing through the origin - like the spokes on a bicycle wheel. These lines represent the different values of theta. Now on each line, place a point at a distance (from the origin) given by [itex]cos^2 \theta [/itex]. Finally, join all these points, neighbor to neighbor. Remember, the X-axis represents [itex] \theta = 90 [/itex] and the Z axis represents [itex] \theta = 0 [/itex]

    At [itex]\theta =90,~ cos^2 \theta = 0[/itex]. So, the points along the horizontal spokes are at the origin. At [itex] \theta [/itex] increases, or decreases, [itex] cos ^2\theta[/itex] increases till it reaches maxima at [itex] \theta = 0, 180 [/itex]. So, for spokes above and below the X-axis, the points move farther and farther out, reaching a maximum at the +/-Z-axis. Join these points and you'll find it looks like a vertical dumbell oriented along the Z-axis. So, clearly this is the [itex]2p_z[/itex] orbital. While this is not all of it (you must now combine the radial part with the polar part), and the best way involves using some 3D plot software, I can't really do that here.

    For the s-orbitals, the total wavefunction is something like [itex] |\psi _n(r, \theta \phi)|^2 ~=~A r^{2n}e^{-2r/na_0} [/itex]. Since these orbitals have no polar ([itex]\theta[/itex]) or azimuthal ([itex]\phi[/itex]) dependence, they are spherically symmetric, and all the "equipotential" surfaces are spheres.

    If this was too confusing, I'll try and attach a picture, when I find a little more time.

    PS : Not more clever...maybe a little more experienced, that's all ! :biggrin:
     
    Last edited: Dec 20, 2004
  7. Dec 20, 2004 #6
    Can you explain how you got that .
    Thanks.
     
  8. Dec 20, 2004 #7

    Gokul43201

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    Looking for a link to spherical harmonics, I found this excellent page, which not only gives you the general formula, it also provides a list of values of [itex]Y_l^m[/itex] for different (l,m), but especially nice, is that it also has beautiful 3D plots of the orbitals.

    But anyway, you'll find that [itex]Y_1^0 = \frac{1}{2} \sqrt {\frac{3}{\pi}} ~cos \theta [/itex].

    If you want to know why the polar part of the 2p orbital is given by this equation, you'll have to go through the entire process of learning how to solve the Schrodinger equation in different geometries.
     
    Last edited: Dec 20, 2004
  9. Dec 21, 2004 #8
    Thanks
    Something to do this summer .I hope! :biggrin:
     
  10. Dec 21, 2004 #9
    Doesn't solving for the wave function involve legendre polynomials or something like that? And from that you can get a function who's plot will give the shape of the oribitals?
     
  11. Dec 21, 2004 #10

    Gokul43201

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    Yes, that's roughly right.

    The polar part of the solution to the Time Independent SE in spherical co-ordinates is indeed something very much like the Legendre Polynomials. To be precise, they are the Associated Legendre Polynomials.

    The function I used above for the p_z orbital is just the Associated Legendre Polynomial, [itex]P_1^0(cos \theta) [/itex].
     
  12. Dec 21, 2004 #11
    Thanks for the detailed explanation.

    I was always curious how scientists got the answer for that..And i always thought our chem teachers were being too dogmatic about this kinda stuff.
    But now i understabd why (pretty complicated stuff!)

    At roughly what age did you learn this stuff?
     
  13. Dec 21, 2004 #12

    Gokul43201

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    Your chem teachers were dogmatic about this stuff because they had no idea about it themselves. You learn this in Physics, not chemistry.

    Only theoretical/atomic chemists will know anything about Quantum Mechanics and solutions of the commonly occuring partial differential equations.

    I learnt this stuff pretty late - my first year of Grad School - because my major in College was Engineering. But I think most physics majors will learn this stuff towards the end of college.
     
  14. Dec 21, 2004 #13
    I see.

    I personally don't do Physics, but am very, very curious about quantum mechanics. Do you reckon its possible to learn QM without doing physics (in high school, this is)

    My presumption would be that we don't learn about quantum mechanics in high school anyway. But in uni..
     
  15. Dec 21, 2004 #14

    Gokul43201

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    You might be able to learn the basics, but without a knowledge of physics, it would have no ....aroma.

    It would be like learning to play music simply by memorizing a sequence of keys - that would be meaningless.
     
  16. Dec 21, 2004 #15

    dextercioby

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    Partly right,my friend.I won't go and quote Feynman saying noone will ever understand QM,but i just wanna say i've never met (and honeslty never will) a chemist,a "theorist" kind (:tongue2:) with whom to discuss mathematical and axiomatic structure of Copenhagen/textbook version of QM.They probably know a bit about the works of Schroedinger,Heisenberg,Born,Dirac,Bethe & Salpeter (the famous book),Ritz,Legendre,Heitler,London,Hueckel and Pauling and names like von Neumann,Weyl,Stone,Borel are totally unfamiliar or probably unknown.

    Generalizations are always dangerous... :wink:

    It's true for most of the physics teachers as well.


    Daniel.

    PS.PROBABLY (it's QM,right?? :wink:) none of chemistry "theorists" thinks about Whittaker functions and Gauss confluent hypergeometric functions when seeing the PDE of the (the most simple model) of Hydrogen atom.
     
  17. Dec 21, 2004 #16

    dextercioby

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    You don't do physics in HS??Too bad.The simplicity of the exposure and a talented proffessor would have made you love this science of nature.
    Decent understanding of QM would require strong knowledge of both mathematics and physics.As a theoretical physicist,i would emphasize the role of mathematics.Without proper knowledge of advanced mathematics,it's impossible to understand the axiomatical structure,i.e.the 6 (my count) principles.

    In school,HS that is,the results of quantum mechanins regarding atomic structure are presented and students are required to them for granted.There are some universities in which the course is being taught sloppy and u'll have to search through (maybe old and dusty) books for the answers to your questions.

    Daniel.

    PS.Like many chapters of physics,quantum mechanics is just about combining advanced mathematical notions,an axiomatic structure and a lot of imagination and patience...
     
  18. Dec 22, 2004 #17
    Well the deal goes like this.
    I currently do Biology , chemistry, Maths and DT(AS level). I do them all for a year, then i can drop 1 subject.(For the other 3 subjects to carry on to A level)

    But i CAN drop 1 and then just do AS physics. This probably sounds gobblygoop for you people who don't know about the system here in UK, but i think it just might make me do physics..

    I am thinking of doing physical chemistry, as it does both of the science branches. But as i said, no physics = :(
     
  19. Dec 22, 2004 #18

    dextercioby

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    I'm only 200Km away from Britain and i find what u're telling me to be very curious.I mean the "dropping" science disciplines,it's really weird and i should say illogical at the same time... :rolleyes: Anyway,i assume it's only up to to take the WISE decision.U have to chose among preferences,maybe,and,if u've chosen "physical chemistry",then that should be it...I wish u only luck...

    Take as it a good thing:physical chemistry involves an "itsy-bitsy" amount of Quantum Mechanics,at least in the latters's results in atomic and molecular structure.

    Daniel.
     
  20. Dec 22, 2004 #19
    Well, i was thinking of dropping Design technology.. :blushing:
     
  21. Dec 22, 2004 #20
    Honestly, I have never taken a physics course so far in my university career and have taken a hardcore class on quantum chemistry (Physical Chemistry II) and did extremely well (although I am a math major too, which helped a lot). P-Chem is extremely instense if you don't have a very decent math background. P-chem is the bridge between chemistry and physics. I wouldn't write P-chem off as having an itsy bitsy amount of QM, I think it really depends on how it is taught.
     
    Last edited: Dec 22, 2004
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