1. The correct radial probability distribution curve for the hydrogen atomic orbital with principal quantum number,$ n = 3$ and azimuthal quantum number, $l = 1$ is: ($4πr^2ψ^2 $= radial probability density function and r = radial distance from the nucleus) enter image description here Ans (3), source

Clearly, we are dealing with $3p$ orbital, now we know that there is one angular node and one radial node. This leads me to think there would be two minima for the graph, however, in the correct answer 3., there is only one minima. So, I suppose either the angular node or radial node was considered.

But, even then, supposing there is only one minima, how do we further distinguish between 3 and 4?

I would appreciate answers/ comments which would point me to the ideas I need to understand to solve this.

  • 4
    $\begingroup$ The form of the distribution always has the form of fig 1, 2 or 3 not 4. This follows because the $r^2$ in the probability increases rapidly with distance $r$ but is eventually curtailed by the exponential part in the wavefunction. $\endgroup$
    – porphyrin
    Commented Aug 5, 2021 at 8:28
  • $\begingroup$ Do all orbital have such an exponential factor attached in it's mathematical representation? @porphyrin $\endgroup$ Commented Aug 5, 2021 at 9:29
  • $\begingroup$ @Buraian Kindly Do not use math jax etc. In the title as it is not possible for the search engine to find the question. Thanks! $\endgroup$
    – Rishi
    Commented Aug 5, 2021 at 14:03
  • $\begingroup$ yes, a decreasing exponential such as $e^{-ar}$ in the radial part that is. You can look up the equations in your phys chem textbook. $\endgroup$
    – porphyrin
    Commented Aug 5, 2021 at 14:12
  • $\begingroup$ Here is a 3D diagram of the 3p orbital. You get the corresponding radial probability by considering probability at a certain distance in any direction. $\endgroup$
    – Karsten
    Commented Aug 5, 2021 at 20:00

2 Answers 2


The wave function of the H atom have a number of nodal surfaces equal to the first quantum number. But these nodal surfaces are not always spheres, as you think. They can be planes or still different surfaces.

In general the quantum number $n$ gives the total number of nodal surfaces. The second quantum number $l$ gives the number of nodal non-spherical surfaces of the particular atomic orbital. As a consequence, $n -l$ gives the number of nodal spheres, that you are calling "minima" in your drawing.

Let's talk about the atomic orbitals with $n = 3$, as it is your desire. The $2$nd quantum number can be $0$, $1$, or $2$.

If $n = 3$ and $l = 0$, the orbital is $3s$ : there are one nodal sphere at infinite distance, and two nodal spheres near and around the nucleus. And there is only one such orbital, as the third number $m$ is equal to zero.

If $n = 3$ and $l = 1$, the orbital is called $3p$, and it is your problem. As the third number $m$ can be $-1$, $0$, or $+1$, there are three such orbitals $3p$, called $3p_x$, $3p_y$ $3p_z$. For each orbital, there is $l = 1$ nodal plane per orbital, and $n - l = 2$ spherical nodes in each orbital, one being at infinite distance, and one near the nucleus. There are three such orbitals, with the same number of spherical and planar nodes : one has a nodal plane perpendicular to the $Ox$ axis, the second perpendicular to the $Oy$ axis, and the third perpendicular to the $Oz$ axis.

If $n = 3$ and $l = 2$, the orbitals are called $3d$. As the third quantum number $m$ can have 5 values, being $-2$. $-1$. $0$, $+1$ or $+2$, it makes five $3d$ atomic orbitals. These $5$ orbitals have all $n - l = 1$ spherical node, which is at infinite distance from the nucleus. Remains two nodal surfaces per each one of these 5 orbitals. For the first three of these five orbitals, the $2$ non-spherical nodal surfaces are two mutually perpendicular planes, namely $Oxy - Oyz$, then $Oxy - Oxz$, and $Oyz - Oxz$. The two other $3d$ orbitals must have two nodal surfaces which are neither spheres, nor planes. They may be conical, but they are difficult to explain without drawing.

This long development explains why the distribution curve of $3p$ orbitals have two minima (one at ∞), so that your figures $3$ and $4$ are equally valid. To explain why Fig. 3 is to be preferred, you should use the radial wave function for $3p$ orbital, which is $$\ce{R(r) = constant · (4 - {r/a}) ({r/a}) e^{-{r/2a}}}$$ When calculating $\pu{R(r)}$ for different values of $\pu{r/a}$ going from $1$, to $2$, $3$, $4$, etc. up to $10$, you will find that $\pu{R(r)}$ takes the following values : $$ 1.82, 1.47, 0.67, 0.00, 0.41, 0.59, 0.63, 0.586, 0.50, 0.404$$ This shows clearly that the maximum value of $\pu{R(r)}$ must be near the nucleus, and before the minima observed at $\pu{r/a = 4}$. Now if you multiply this value by $\pu{r^2}$, to get your wanted distribution curve, you find that $\pu{R(r)·r^2}$ takes the following values : $$\pu{1.82, 5.89, 6.03, 0.00, 10.25, 21.24, 30.87, 37.5, 40.5, 40.4}$$ This shows clearly that your distribution curve has a maximum far away from the nucleus, and far away form the minima (observed at $\pu{r = 4a}$).

As a consequence, your curve $3$ has to be preferred.

  • $\begingroup$ Hmm not exactly, I felt tyberius has written an answer closer to what I wanted $\endgroup$ Commented Aug 5, 2021 at 3:10
  • 2
    $\begingroup$ This is a nice and well written answer but I suggest writing comments ("is this what you are looking for?") before posting such a lengthy answer that does not address the question. Because of that last sentence this might be removed as a comment. $\endgroup$
    – Buck Thorn
    Commented Aug 5, 2021 at 8:38
  • 1
    $\begingroup$ I think it should be $n-1$ and $n-1-l$ for the node values $\endgroup$ Commented Aug 5, 2021 at 9:33
  • $\begingroup$ @Buraian. The number of nodal surfaces is $n$ and $n-l$ if the infinite is counted. It is $n-1$ and $n-1-l$ if you count the finite surfaces. $\endgroup$
    – Maurice
    Commented Aug 5, 2021 at 16:33
  • $\begingroup$ Why doesn't angular node not appear? $\endgroup$ Commented Aug 6, 2021 at 10:44

These plots are with respect to $r$, so it will only show radial nodes. An angular node would appear in a plot with respect to $\theta$. So for 3p, you will only have one node.

To distinguish between options 3 and 4, you just need to realize that larger $n$ will have it's maximum further from the nucleus.

  • $\begingroup$ The second point, does that property have a name/ can you suggest where I could read more about it? (The larger n rule) $\endgroup$ Commented Aug 5, 2021 at 3:09
  • $\begingroup$ is R only the radial part of the probabiltiy i.e: is there constribution from azimuthal and polar part as well? $\endgroup$ Commented Aug 5, 2021 at 13:42
  • $\begingroup$ @Buraian I don't know if this property has a name, but you can recognize this pattern by visualizing the orbitals. You can see that as $n$ increases, you get a sequence of $n-1$ shells with gradually increasing size as you move away from the nucleus. The radial distribution function already integrates over the azimuthal/polar parts, so you are just seeing how the distance affects the probability. Its basically averaged over all the azimuthal/polar angles. $\endgroup$
    – Tyberius
    Commented Sep 3, 2021 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.