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So I was studying some graphics of the representation of orbitals in terms of radial dependence. My graphic is similar to this one: https://www.google.pt/search?q=radial+dependence+of+orbitals&espv=2&biw=1280&bih=728&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi8kvOTuZvNAhWBshQKHYuNBw8Q_AUIBigB#imgrc=XTvxODPVjMkSHM%3A

Anyway, my book says that "the distance from the nucleus of the maximum value more distant from the nucleus is the radius of the orbital".

Well according to the graphic the maximum value of the orbital 3s corresponds to a greater value of the radius of the orbital than the orbital 3d. But that doesn't make sense, right? The orbital 3s should be closer to the nucleus, right?

This is making me confused I might be misreading the sentence? Can someone help me?

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    $\begingroup$ Maximum value means less than nothing. It is the average value that matters. $\endgroup$ – Ivan Neretin Jun 9 '16 at 18:26
  • $\begingroup$ Sorry I didn't understand what you mean can you please elaborate more? $\endgroup$ – Granger Obliviate Jun 9 '16 at 19:12
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The answer depends on your definition of orbital radius or "closer".

Orbitals correspond to the probability distribution of finding an electron in a certain place around the nucleus. So you may define the radius as

  • the distance where the probability to find the electron has its maximum, or
  • the average distance where the electron can be found, or
  • the radius of the sphere wherein the electron can be found with 50%, or 90%, or 99% probability, or
  • some other fancy measure.

Which definition is best suited may depend on the type of comparison you make.

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  • $\begingroup$ Ok I think I messed up the definitions of orbit and orbital. When I said closer I was thinking abou circular orbits. So based on your answer I would say that: if radius is the distance where the probability to find the electron has its maximum than the definition is correct, right? If radius is the average distance where the electron can be found, then the definition of the book, isn't right. Am I correct? $\endgroup$ – Granger Obliviate Jun 9 '16 at 19:48

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