I was reading JD Lee's Concise Inorganic Chemistry and it has given some values of mean radial distance which is increasing from s to p to d orbital. I had earlier read that the most probable radial distance of an orbital has the order s>p>d>f. This makes me wonder what is the difference between the mean radial distance and most probable radial distance. Also, why is the order of mean distance reverse from that of most probable radial distance? Also can you please explain why is the energy of an orbital proportional to mean radial distance as given in the book? Source: Concise Inorganic Chemistry by JD Lee; sorry for the bad photography

Image Source: Concise Inorganic Chemistry by JD Lee

  • $\begingroup$ If $\psi$ is the radial part of the wavefunction the mean or average radial distance is $\langle r \rangle=\int r\psi^*\psi dr/\int\psi^*\psi dr$ where the denominator is the normalisation. The most probable is (presumably) the value of $r$ at the maximum of $\psi^*\psi$. ( The $\psi$ is an equation involving $r$ and the * indicates the complex conjugate. If $\psi$ is real is just the same as $\psi$ and if not replace $i$ with $-i$ where $i=\sqrt{-1}$). These values are not the same as $\psi$ is not 'symmetrical' as $r$ increases. $\endgroup$
    – porphyrin
    Commented Apr 11, 2018 at 18:51
  • $\begingroup$ Do you have a source for the most probable radial distance ordering? slideplayer.com/6234381/20/images/36/… $\endgroup$
    – Tyberius
    Commented Apr 11, 2018 at 19:31
  • $\begingroup$ @Tyberius As it can be seen in the image you provided, the most probable radius of 3s>3p>3d. $\endgroup$ Commented Apr 11, 2018 at 19:43

2 Answers 2


In multi-electron atoms the energy of the s, p and d orbitals for a given $n$ is determined primarily by shielding by inner electrons and penetration of the orbital towards the nucleus. As s orbitals have some probability of being at the nucleus their energy is lowest, next is p then d orbitals in increasing energy. Also there is spin-orbit coupling which splits sub-levels of p and d orbitals or any orbital which has angular momentum so this does not effect s orbitals. This effect is much smaller in d than p orbitals and generally smaller than the separation of s to p to d.

The statements in the book extract you give seem contradictory. The mean radial value of an H type orbital is calculated using $\displaystyle \int_0^\infty r^3R_{n,l}^*R_{n,l} dr $ where $R_{n,l}$ is the (normalised) radial part of the wavefunction for an atom with quantum numbers $n,\;l$. The $r^3$ arises because the volume element in the integration is $r^2\sin(\theta)d\phi d\theta$ in spherical coordinates. The other $r$ is how the average is defined, i.e. $\int r\psi^*\psi d\tau$ for some normalised wavefunction $\psi$.

The radial parts of the wavefunction for the 3s,3p and 3d are define with $q=Zr/a_0$ ($a_0$ is the Bohr radius) as $$\begin{align} R_{3,0}&=N(27-18q+2q^2)exp^{-q/3}\\ R_{3,1}&=N(6q-q^2)e^{-q/3}\\ R_{3,2}&=Nq^2e^{-q/3} \end{align}$$

and $N$ is the normalisation which is different for each wavefunction. The integrals are not hard to evaluate but very tedious as there are many terms. However, there is a general formula for the average of H atom type orbitals and is

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

The average values are

$$3s \;\langle r \rangle = 0.476 \text{ nm,}\quad 3p \;\langle r \rangle = 0.441\text{ nm,}\quad 3d\; \langle r \rangle = 0.370\text{ nm}$$

The s orbital value is the same as in your question but the others differ both in value and in order of size. Perhaps there are typos in the text you give or some other unstated factors producing these values.

The maximum values have to be calculated by making the derivative (of $q^2R_{n,l}^2$) zero and solving the equation. Alternatively the value can be taken with sufficient accuracy from a plot of the function which is far easier. The values are

$$3s \;r_{max} = 0.46 \text{ nm,}\quad 3p \;r_{max} = 0.43\text{ nm,}\quad 3d\; r_{max} = 0.32\text{ nm}$$

the plot below shows the radial functions $q^2R_{n,l}^2$ each normalised to the same maximum height. The mean is greater than the maximum in each case as expected as the main part of the radial probability is 'lopsided' towards larger distances.

radial distribution

(In evaluating the integrals $\displaystyle \int_0^\infty r^ne^{-\beta r}= \frac{n!}{\beta^{n+1}}$ is useful.)

  • $\begingroup$ Thanks for the answer! I think the average radius value that you calculated differs from the book because the book has given mean radius for a phosphorus atom and not hydrogen. Now I don't know why the s orbital value matched because intuitively, it feels like they should differ. $\endgroup$ Commented Apr 13, 2018 at 9:34
  • $\begingroup$ Also, according to you since the mean radius value for H atom is in the order 3d<3p<3s, the statement that energy of an orbital is proportional to the mean radius is false, right? $\endgroup$ Commented Apr 13, 2018 at 9:37
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    $\begingroup$ I used Z=15 for phosphorus so all values should fit. It seems that that statement is not correct or at least needs to be explained. The picture above seems to contradict it. But the figure is for H type atoms not one with multiple electrons but for an atom I don't see why this would have a big effect on the radius such a changing the ordering of the average sizes. $\endgroup$
    – porphyrin
    Commented Apr 13, 2018 at 10:16

First of all,in the radial plot of probablity distribution of $3s$ and $3p$ subshells have more than one peaks. So, if we take the highest peak as the most probable radius, then we have to solve ,
$1$. the maxima for $\ce{P(r_{3s}) }$ which is proportional to $\ce{|$\Psi$_{3s}|^2r^2}$. So, actually, we have to solve $$\ce{\frac{d}{d$\rho$}($\rho$^2(27 - 18$\rho$ + 2$\rho$^2)^2e^{-\frac{2\rho}{3}}) = 0, where $\rho$ =\frac{Zr}{a_0}}$$Solving this and the global maxima of $\rho$ will come at approximately $13$.
$2$. Similarly, for $\ce{P(r_{3p})}$ which is proportional to $\ce{|$\Psi$_{3p}|^2r^2}$ (note, here we are concerned aboutr radial part which is same for all $p_x,p_y, p_z$ orbitals). So, here we have to solve,$$\ce{\frac{d}{d$\rho$}($\rho$^2(6 - $\rho$)^2e^{-\frac{2\rho}{3}}) = 0, where $\rho$ =\frac{Zr}{a_0}}$$ Solving this and the global maxima of $\rho$ will come at exactly $12$
$3$. and finally, for $3d$ orbitals, the maxima will be a single peak and that will come from the equation, $$\ce{\frac{d}{d$\rho$}($\rho$^4e^{-\frac{2\rho}{3}}) = 0, where, $\rho$ = \frac{Zr}{a_0}} $$ This will give the maxima at $\rho$=$6$.
So, it is clear that the most probable radius follows the order, $r_{mp}(3d) < r_{mp}(3p) < r_{mp}(3s) $
But when you come to the mean radius, we just cannot bring same analogy as the most probable radius. Because the mean radius is actually an overall effect of radial probability distribution throughout the entire space. So, unless we calculate the integral,$$<\Psi|r|\Psi> =\int_{0}^{\infty} r|\Psi|^2 r^2drsin\theta d\theta d\phi$$ it is difficult to say, which will have the larger mean radius, and which one will have smaller. It totally depends on the integral. So, we actually have to evaluate the integral to tell the answer (which can be done but very much calculative. You can use the idea of gamma functions to evaluate this kinds of integral $\int_{0}^{\infty}r^n e^{-\alpha r }dr$ easily which will appear lot in this mean radius calculation).
Coming to the energy, the statement is said in a loose way i.e, As an orbital becomes farther from the nucleus, its energy becomes lesser negative and increases towards zero.

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    $\begingroup$ Mean radius is actually no more calculative than what you just did. Factor away the angular part, and you'll be left with a relatively manageable integral. $\endgroup$ Commented Apr 11, 2018 at 20:20
  • $\begingroup$ Also, you might want to use \langle...\rangle for $\langle\Psi|r|\Psi\rangle$. $\endgroup$ Commented Apr 11, 2018 at 20:25
  • $\begingroup$ Thanks! So it is not necessary that the mean radius of a d orbital will be greater than that of p and that of s, am I correct in saying this? $\endgroup$ Commented Apr 11, 2018 at 21:05
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    $\begingroup$ So when we look at the energy of an orbital, do we look at the mean radius? If we were to at the most probable distance from the nucleus, then the reverse trend of energy would be predicted. $\endgroup$ Commented Apr 11, 2018 at 23:03

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