How to compare size of subshells?

Question

For the same principal quantum number, on increasing the value of the azimuthal quantum number does the average radius of the subshell increase or decrease?

In other words, which out of, say, 3s, 3p, 3d would have the largest average radius and why?

Answer 1

For a hydrogen atom, according to Hertel and Schulz [1]:

$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$

and

$$\langle r^2\rangle_{n,\,l}=\frac{a_0^2 n^4 \left(\frac{3}{2} \left(1-\frac{l (l+1)-\frac{1}{3}}{n^2}\right)+1\right)}{Z^2},$$

where $\langle r\rangle_{n,\,l}$ is the average distance of the electron from the nucleus as a function of n and l, and $\langle r^2\rangle_{n,\,l}$ is the average distance squared of the electron from the nucleus as a function of n and l, $a_0$ is the Bohr radius, and Z is the nuclear charge (Z = 1 for hydrogen). These are known as expectation values, because these are the expected values based on the probabilistic distribution of electron density.

Using these, we can compare $\langle r\rangle_{n,\,l}$ and $\langle r^2\rangle_{n,\,l}$ for the $3s \,(n = 3, l = 0), 3p \,(n = 3, l = 1) \text{ and } 3d\, (n = 3, l = 2)$ subshells:

$$\langle r\rangle_{3,\,0}= \frac{27 a_0}{2 Z}$$ $$\langle r\rangle_{3,\,1}= \frac{25 a_0}{2 Z}$$ $$\langle r\rangle_{3,\,2}= \frac{21 a_0}{2 Z}$$

$$\langle r^2\rangle_{3,\,0}= \frac{207 a_0^2}{Z^2}$$ $$\langle r^2\rangle_{3,\,1}= \frac{180 a_0^2}{Z^2}$$ $$\langle r^2\rangle_{3,\,2}= \frac{126 a_0^2}{Z^2}$$

You can see that, as the azimuthal quantum number (l) increases, the electron's average distance from the nucleus decreases. Note, however, that in hydrogen (or any single-electron atom, such as $\ce{He^+}$ or $\ce{Li^{2+}}$, etc.), all orbitals in the same shell have the same energy. Thus the energies of the 3s, 3p, and 3d orbitals are identical. As a consequence, a hydrogen electron excited to the third energy level is actually in a weighted linear combination of all of these orbitals, and its actual average distance would likewise be a weighted average of these.

But still, why does the average distance decrease with increasing azimuthal quantum number? Well, quantum is not my field, but I believe this is what's going on (perhaps a quantum expert can let us know if this is correct):

As I noted, in any single-electron atom, the 3s, 3p, and 3d orbitals have the same energy. The electron's energy is a combination of potential energy due to its distance from the nucleus, and the energy due to its angular momentum. As the azimuthal quantum number increases, the electron's angular momentum increases (it has no angular momentum in 3s, some in 3p, and more in 3d), and thus the energy associated with this angular momentum increases. But since the total energy is constant, as the energy due to angular momentum increases, the potential energy associated with distance from the nucleus must decrease, and hence the average distance from the nucleus gets lower.

Finally, you can see from the formula that $\langle r\rangle$ is much more sensitive to changes in n than l. For instance, if we compare the sizes of the s, p, and d subshells in the 3rd shell (n = 1, l = 0 to 2), $\langle r\rangle$ ranges from $\frac{27 a_0}{2 Z}$ to $\frac{25 a_0}{2 Z}$ to $\frac{21 a_0}{2 Z}$. By contrast, if we compare the sizes of the s-subshells in the first three shells (l = 0, n = 1 to 3), $\langle r\rangle$ ranges from $\frac{3 a_0}{2 Z}$ to $\frac{12 a_0}{2 Z}$ to $\frac{27 a_0}{2 Z}$.

1. Hertel, Ingolf V., and Schulz, Claus-Peter. Atoms, molecules and optical physics. Berlin: Springer, 2015

Answer 2

Before we start answering the question, we should first understand what the radius of electron is. According to quantum mechanics, the electron density is smeared across the space, and strictly speaking, the density becomes radially only if you have radial nodes, and at $r=0$ (at the nucleus) and at $r=\infty$. This means that you cannot specifically say that this is the radius of my shell. In quantum mechanics, the average radial distance of the wave function is defined as an "average radius", and the formula is given as $$\langle r\rangle=\frac{\langle\psi|r|\psi\rangle}{\langle\psi|\psi\rangle}$$ For hydrogen atom, the average radius of the $1s$ orbital is known as Bohr radius ($0.529 Å$).

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Considering this idea, you can take the wave function of the system you like and calculate the average radius. However, you have to keep the following in mind:

1. The notion of orbitals ($1s,2s$ etc.) exists for hydrogen or hydrogenic atoms (one electron systems) only. Beyond that, the analytical form of the wave function (for multi-electron systems) is impossible to obtain. In fact, it is one of the major challenges in the field of quantum mechanics. In many cases, people use approximate wave function to estimate the orbitals and their energies, and in such cases, it does not make sense to speak about the hydrogenic orbitals at all.
2. Does it really make sense to talk about average radius in an non-radially symmetric system? This is because, if you change the value of $\theta$ and $\phi$ for a non-radially symmetric system, you might get different values for average radius.

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I know I have not answered your question directly, but in case you are interested in orbitals of hydrogen (or hydrogenic systems) only, you can get the expression on internet. Combine that with a bit of knowledge of mathematics and/or coding, you will get the average radius of any (or any property) of any orbital you wish to know.