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Topic: orientation of d orbitals  (Read 3478 times)

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Offline vwcanter

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orientation of d orbitals
« on: December 25, 2013, 05:03:01 PM »
"d" electron orbitals have two nodal lines. It is easy to see that the first three orbitals, the ones shaped like four leafed clovers, could be oriented three ways, one for each plane in three dimensional space. However, there is supposedly a fourth orientation, just like the first one, rotated 45 degrees, in the same plane as the first. However, if you could fit in a solution there, why couldn't you do the same to the other two orientations? That would give six orientations for the four leafed clover shape. But everyone says there are only four. I have not attempted to solve the wave equation. But it should be possible to picture it, knowing only the number of nodal lines. Any insight is appreciated.

Offline vwcanter

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Re: orientation of d orbitals
« Reply #1 on: December 25, 2013, 06:18:24 PM »
OK, so the clover shaped d orbitals must not be at 90 degrees to each other, like the p orbitals. They seem to be saying they can be spaced equally at 45 degrees from each other, but I'm still trying to work that out. And the fifth orbital, the dumbbell with the doughnut around the middle- that one has only one orientation, because the middle doughnut part would overlap if it were rotated?

Offline Enthalpy

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Re: orientation of d orbitals
« Reply #2 on: December 28, 2013, 03:25:29 PM »
http://winter.group.shef.ac.uk/orbitron/
http://winter.group.shef.ac.uk/orbitron/AOs/3d/index.html
On the picture there, you see that what they call x2-y2 is orthogonal to xy, xz and yz, but if you rotate also xz or yz by 45° the new one is not orthogonal to x2-y2.

With "orthogonal" I mean the integral over xyz of the product of the wavefunctions is zero.

One can pick many different sets of wavefunctions as a base. Maybe other bases are more symmetric than this one.

Offline Irlanur

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Re: orientation of d orbitals
« Reply #3 on: January 01, 2014, 12:30:40 PM »
The spherical harmonics are solutions to the eigenvalue-problem of the angular momentum operators lz and l2. It can be shown, that for degenerate eigenfunctions (degenerate: same eigenvalues), every linear combination of these eigenfunctions are eigenfunctions to the same eigenvalue. They don't need to be orthogonal, but it's always possible to choose orthogonal eigenfunctions. and that's convenient.

In simple words: there are infinite possibilities to choose the d-orbitals, but it's common to use the xy,xz,yz,z2 and x2 -y2

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