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What are the shapes of atomic orbitals |
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| Jul16-12, 03:57 AM | #1 |
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What are the shapes of atomic orbitals
I know that the shape of s orbital is sphere, p orbital is dumbbell shaped and d orbital is like a doughnut but why do these orbitals have these shapes why not some other shape.
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| Jul16-12, 08:51 AM | #2 |
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hi shivamdabas! welcome to pf!
![]() i think the easiest way to understand the shapes is to consider the nodes the nodes are where the probability is zero … so the individual shapes have to fit between them! ![]() the nodes are planes, spheres, and/or a cone for s, (and for n ≥ 2) all the nodes are (concentric) sphere(s), so the shapes that fit between them must be spheres or spherical shells for p, the nodes are one plane and (for n ≥ 3) sphere(s), so the shapes that fit are blobs on either side of the plane, and smaller blobs to fit inside the sphere(s) for d, the nodes are two planes or a cone and (for n ≥ 4) sphere(s), so the shapes that fit are blobs in the four regions between the planes, or two blobs "inside" each half-cone and one donut fitting "round" the cone, and smaller blobs to fit inside the sphere(s) for f, i'm not sure where the nodes are |
| Jul16-12, 09:01 AM | #3 |
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what about 1s ?
it has no nodes, sooo …i say, i say, i say, my 1s orbital has no nodes |
| Jul16-12, 12:00 PM | #4 |
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What are the shapes of atomic orbitals
Thanks for you reply
Now, I know that the shapes of the orbitals are governed by nodes which are the places where probability of finding an electron is zero but how can we know about the shape of these orbitals. |
| Jul16-12, 01:45 PM | #5 |
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i think it's …
the number of spherical nodes is n-l-1, and the number of plane nodes is l (or l-2 plus a cone node) |
| Jul16-12, 02:19 PM | #6 |
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For s orbitals, there is no angular momentum, and therefore the orbitals will have spherical symmetry. This means at r=0 the wavefunction can (and generally will) be nonzero. For other orbital types, the angular momentum is nonzero, and the angular wavefunction will therefore have alternating regions of positive or negative value, separated by nodes. Of course, for the wavefunction (and its first few derivatives) to be continuous at r=0 the radial wavefunction must approach zero at r=0. Each additional increment in angular momentum about the z axis adds a nodal plane containing the z axis. Each increment in angular momentum perpendicular to the z axis will add a cone-shaped node with an axis parallel to the z axis. If there are an odd number of "cones", then one of them will actually be a plane containing the equator of the atom. |
| Jul16-12, 02:38 PM | #7 |
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Recognitions:
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The node count mentioned only works for real orbitals. As the OP mentioned doughnut shaped d-orbitals, I think he talks about complex orbitals with definite z-component of angular momentum.
In these toroidal shaped orbitals the phase of the wavefunction changes, at least in the simplest cases, like exp(imf) where f is the angle in the x-y axis. The real orbitals are combinations of complex orbitals with +m and -m, hence they change like cos(mf) or sin(mf). |
| Jul16-12, 02:41 PM | #8 |
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1. You solve the Schrodinger equation for, say, the hydrogen atom. You get the wavefunction. 2. In doing #1, you end up with two parts: the radial solution R(n,l), and the orbital solution P(l,theta). 3. If you plot the wavefunction, you get those shapes! So to know where those shapes came from, you have to solve for the wavefunction. Check out this link: http://panda.unm.edu/Courses/Finley/.../WaveFcns.html I have no idea what you already know, or if this is what you are asking for. But practically everything here is derived from some First Principles starting point. Zz. |
| Jul16-12, 02:47 PM | #9 |
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| Jul16-12, 02:56 PM | #10 |
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But the standard way to "plot" an orbital is to pick a cutoff value for the probability density such that the combined areas above the cutoff value have a volume that's 90% (as an example) likely to contain the electron. By analogy, using a similar method to plot the wavefunction for a particle in an infinite square well would look like a dashed line rather than a sine wave. |
| Jul16-12, 03:07 PM | #11 |
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I was reading the questions posted by the OP carefully, trying to decipher what was being asked! Here, look at them again! The bold emphasis are mine. I'm answering the question the way I think it is being asked. Zz. |
| Jul16-12, 03:16 PM | #12 |
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I'm not just being pedantic or trying to get the last word in...from my experience teaching, it's a non-trivial step for some people to follow. |
| Jul16-12, 04:20 PM | #13 |
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I was describing where and how those things are derived. You don't have to plot such things if you can visualize it in your head. Most of us can't, and that's why we plot it out to see what it looks like. But that isn't the essence to the answer. Zz. |
| Jul16-12, 06:16 PM | #14 |
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n, m, l are Principal, Magnetic and Azimuthal Quantum Numbers a_0, I am assuming is the Bohr Radius. e, in these equations, I am also assuming is Euler's number. (As opposed, for instance, to the elementary charge) R_ml, Y_lm and r are what exactly? TIA, AC |
| Jul17-12, 12:01 AM | #15 |
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Thanks everyone especially ZapperZ and PhaseShifter, if you can then please explain me what is radial wave function and angular wave function.
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| Jul17-12, 05:47 AM | #16 |
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Separability is a mathematical "operation". You write a function F as factors of its components, i.e. F = X(x)Y(y)Z(z). For the solution to the Schrodinger equation, one writes this as F = R(r)P(theta)L(phi), meaning the radial and orbital components are separated out. See this: http://hyperphysics.phy-astr.gsu.edu...um/hydsch.html Zz. |
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