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Can we define a size for the H atom orbitals which are not spherically symmetric, e.g. $p, d$ etc? For example, is it meaningful to say that the $(n+1)p$ orbital is larger in its extent than $np$ orbital ($n$ is the principal quantum number)?

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  • $\begingroup$ The orbitals are spherically symmetric: en.wikipedia.org/wiki/Spherical_harmonics#/media/… $\endgroup$ – safesphere May 26 at 7:44
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    $\begingroup$ Only s orbitals are spherically symmetric. $\endgroup$ – Poutnik May 26 at 8:18
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    $\begingroup$ Why should make less sense to speak about the size of p,d,,f orbitals than about the size of s orbitals? $\endgroup$ – Poutnik May 26 at 8:24
  • $\begingroup$ @Poutnik as you said, only s orbitals are spherically symmetric. so what do you mean by size? $\endgroup$ – mithusengupta123 May 26 at 8:33
  • $\begingroup$ @mithusenguota123 It depends what we mean by orbital 1/particular solution of the wave equation 2/ quantum state of electron 3/ geometrical shape described by the surface of the same arbitrary value of $\Psi \cdot \bar \Psi$. But the size of the 3/ can be taken either as the average radius as in one of the answers, either as the maximal radius with the given orbital surface probability limit. $\endgroup$ – Poutnik May 26 at 9:47
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It makes perfect sense to define the average radius of any hydrogen orbital $\psi_{nlm}(\vec{r})$, regardless of whether it is spherically symmetric ($l=0$) or not:

$$ \langle r\rangle = \int r|\psi_{nlm}(\vec{r})|^2 d^3r $$

Evaluating this integral for the hydrogen orbitals $\psi_{nlm}(\vec{r})$ you get (see FAMU-FSU Col­lege of En­gi­neer­ing - Expectation powers of $r$ for hydrogen) $$ \langle r\rangle = a_0 \frac{3n^2 - l(l+1)}{2}$$ where $a_0$ is the Bohr ra­dius, about 0.53 Å.

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  • $\begingroup$ Another criteria can be the maximum extend in a single dimension, a kind of a length, for given arbitrarily chosen probability value defining the surface of the orbital 3D geometrical shape. This has direct applicability for creation molecular orbitals with sp^n hybridization. $\endgroup$ – Poutnik May 26 at 9:34
  • $\begingroup$ It's interesting that it decreases with $l$ for fixed $n$. The opposite is true for the wavefunctions in nuclear physics. I guess the difference is because the $1/r$ potential is "soft," so by putting some of the KE into radial motion, you can move the classical turning point farther out. In nuclear physics, the potential is more like a hard wall. $\endgroup$ – Ben Crowell May 26 at 13:46
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You can define $\langle r\rangle$ or $\langle r^2 \rangle$ for different orbitals and compare them. You can check that the average distance of electron form the nucleus is larger for higher orbitals.

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