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Below is a picture of the radial component of the wave function distribution for the 1 through 3s orbitals. It makes sense to me that there are points where the wave function is 0, since by definition, a radial node is where the wave function = 0.

What confuses me is the following graph of the 2p orbital wave function distribution. Since the 2p orbital is shaped like a tear drop, if you went a certain distance r from the nucleus, there would be areas that the tear drop doesn't cover (since a tear drop isn't spherical). Wouldn't the probability of finding an electron in those empty space areas be 0? How can you draw a radial wave function when the 2p orbital depends on angles as well?

enter image description here

Source of images: http://faculty.uml.edu/ndeluca/84.334/topics/topic1.htm

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  • $\begingroup$ You can certainly plot the radial part of the function, but do need to keep the angular portions in mind. $\endgroup$ – Jon Custer Oct 20 '20 at 19:41
  • $\begingroup$ @JonCuster How can you plot the radial part though? The 2p orbital isn't like a sphere where you can arbitrarily go out a distance r and be within the orbital. With the 2p, if you go out at the wrong angle, won't you be in empty space, outside of the orbital? $\endgroup$ – Jay Oct 20 '20 at 22:39
  • $\begingroup$ Look at the form of the spherical harmonic solutions, and see how $r$ enters into them. Yes, they vary with $\theta$ and $\phi$ as well. $\endgroup$ – Jon Custer Oct 20 '20 at 23:35
  • $\begingroup$ @JonCuster So if we look at only the radial part, doesn't that imply we are fixing some value for theta and phi? $\endgroup$ – Jay Oct 22 '20 at 2:07

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