Diagram of P orbital vs. S orbital distance from nucleus

My chemistry book explains that even though electrons in the $\mathrm{2p}$ orbital are closer to the nucleus on average, electrons from the $\mathrm{2s}$ orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?

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    $\begingroup$ All, this is for CrazyEtc to say for sure, but I think his main question was a bit simpler: Why do orbitals that have less energy sometimes extend farther out into space than ones with higher energy? That's an interesting question, and one that's worth trying to answer as directly as possible. $\endgroup$ May 12, 2012 at 16:05
  • $\begingroup$ Due to shielding. $\endgroup$
    – BigSack
    Aug 25, 2012 at 6:02

7 Answers 7


I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g., $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{3d}$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms? Although AcidFlask already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information. I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states (which are in turn lower in energy than $\mathrm{3d}$ states).

$\ell$-degeneracy of the hydrogen atoms energy eigenvalues

The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the Kepler problem known from classical mechanics. This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant. Now, it is known from physics that symmetries of a system lead to conserved quantities (Noether Theorem). For example from the rotational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the magnetic quantum number $m$, which are all energetically equivalent as long as the system maintains its rotational symmetry. So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom. Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry. The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential. (It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.) This dynamical symmetry conserves the Laplace-Runge-Lenz vector $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found here.)

Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms?

As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g., by applying a magnetic field, the $\ell$ degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form. This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus. (Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)

Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?

Two effects are important here:

  • The centrifugal force puts an "energy penalty" onto states with higher angular momentum.${}^{1}$ So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.

    1. The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$ \begin{equation} \bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0 \end{equation} \begin{equation} {}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad \end{equation} The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.
  • Core repulsion (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.

    1. Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.

The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance between centrifugal force and core repulsion, that both act against the nuclear Coulomb attraction. In the real atom the balance between centrifugal force and core repulsion is broken, The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons. Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states which are in turn lower in energy than $\mathrm{3d}$ states.

Of course, one has to be careful when using results of the hydrogen atom to describe effects in multielectron atoms as AcidFlask mentioned. But since only a qualitative description is needed this might be justifiable.

I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.

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    $\begingroup$ Why do you say that the "l-degeneracy" of the hydrogen atom is accidental? $\endgroup$ Apr 20, 2019 at 8:09
  • $\begingroup$ @TanYongBoon I mean "accidental" in the sense that it just so happens to mark a balance between centrifugal force and core repulsion. Of course, this is the result of an underlying symmetry and in that sense it is not accidental. $\endgroup$
    – Philipp
    Apr 20, 2019 at 11:52

General chemistry textbooks tend to explain atomic structure exceedingly poorly using a hodgepodge of obsolete concepts. Your chemistry book provides such a typical example - the notion of penetration only makes sense in the ancient Bohr-Sommerfeld model that has been obsolete since the discovery of quantum mechanics! The idea was that orbits of electrons in $\mathrm{p}$ states were more elliptical than those in $\mathrm{s}$ states because they had more angular momentum, and so on average there will be some fraction of the time where $\mathrm{p}$ electrons would be closer to the nucleus than those in the $\mathrm{s}$ orbits, i.e. they would penetrate the $\mathrm{s}$ orbit.

We now know that electrons do not actually have classical orbits and that the Bohr-Sommerfeld model is wrong. However there remains one correct statement of fact, which is that electrons in $\mathrm{p}$ orbitals by definition have more angular momentum than those in $\mathrm{s}$ orbitals (The orbital angular momentum is $\ell=1$ in $\mathrm{p}$ and $\ell=0$ in $\mathrm{s}$). Nonetheless it is completely misleading, since $\mathrm{2s}$ and $\mathrm{2p}$ orbitals are only well-defined in hydrogen and other ions with only one electron, and furthermore that all orbitals of the same principal quantum number are degenerate in energy in hydrogenic systems. Thus angular momentum alone cannot be the answer.

It is only due to electron-electron interactions that the orbitals that correspond to $\mathrm{2s}$ and $\mathrm{2p}$ in multi-electron atoms split and become nondegenerate. The effect of having more angular momentum therefore turns out to be only indirectly responsible via the different many-electron interactions occurring in the analogs of $\mathrm{2p}$ and $\mathrm{2s}$ orbitals in many-electron atoms. The simplest explanation that is even remotely correct is the screening effect, where the $\mathrm{2s}$ electrons by their very presence make the $\mathrm{2p}$ electrons feel a lower effective nuclear charge than if they were absent. A more precise quantum mechanical statement is that Slater's variational treatment of the nuclear charge in hydrogenic orbitals results in a lower effective nuclear charge for $\mathrm{2p}$ electrons than $\mathrm{2s}$ orbitals. There is unfortunately no good a priori physical principle for this; it is merely a statement of a result from a quantum mechanical calculation.

Summary: $\mathrm{2p}$ electrons are higher in energy than $\mathrm{2s}$ ones due to screening effects that result from electron-electron interactions. Explanations involving penetration of classical orbits or purely angular momentum considerations are wrong, even though they tend to show up in general chemistry textbooks.

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    $\begingroup$ General chemistry textbooks tend to explain EVERYTHING using over simplified concepts. It always reminds me of Jack Nicholson's line in "A few Good Men" - You can't handle the truth!! The truth in this case being too complicated for a "simple" explanation. $\endgroup$
    – MaxW
    Oct 22, 2015 at 2:10

It seems like it should be the average distance that matters

No. It is the average energy that matters. Note that this stuff about "spends so much time here and so much there..." is really just a (not particularly good) way of describing a quantum-mechanical wave function's absolute square. The electron is actually never at any particular place in the orbital, rather it's everywhere at the same time – until you decide to measure its precise position (thereby destroying the orbital, that has to do with the Heisenberg uncertainty principle), in which case there would be a certain probability of finding it here or there. That probability can be calculated from the wave function $\psi(\mathbf{x})$ by taking its absolute square, which gives the radial distribution function in your plots.

The energy of any quantum state is determined by the expectation value of the Hamiltonian operator, which very roughly means "measure the energy at all places in space, multiply with the probability the electron is found at that place, and add all those values up". That should sound quite intuitive. In actuality it's more complicated: the electron also has some kinetic energy at every place in space, which is itself encoded in the wave function – but you need to consider its complex phase, which is a whole extra dimension that you simply don't see in your plots and certainly not in the average distance.

Calculating the exact energies for atomic orbitals is a lot of maths that probably doesn't belong here, but it's possible to carry it out*, and it should be obvious now that the result doesn't necessarily have much to do with simple "energy at the average distance". The quantities such as angular momentum mentioned in CHM's answer can be used to make it a bit simpler again (the Hamiltonian separates into a radial and spherical part), but that alone doesn't really explain much I suppose.

*As a matter of fact, it's not possible to calculate it exactly for anything more complicated than a hydrogen atom! You have to do it with certain approximations.

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    $\begingroup$ For small systems, it actually is possible to compute it exactly. You just have to do it numerically. $\endgroup$
    – Nick
    Oct 30, 2013 at 16:28

I think the simplest way to explain what is going on is by way of a classical analogy.

Picture an elastically bound ball that whizzes back-and-forth along the $x$ axis. Next, whack the moving ball hard with a $y$-aligned impulse. The result is an elliptical or figure-8 path, one for which the maximum radius of motion along the $x$ axis will, in a perfect world, remain unchanged.

The single-axis ball is the classical analog to an $s$ orbital, and the elliptical or figure-8 ball corresponds to a $p$ orbital. Notice that for a given maximum radius, the $p$ orbital will include two orthogonal oscillation components, $x$ and $y$, and so will always have more energy than an $s$ (or $x$-only) oscillation with the same maximum radius. That also means that you can construct an $s$ orbital that extends beyond the maximum range of the $p$, yet will still have less energy than the $p$.

Or even more succinctly, think of a rocket with a fixed energy budget. The $s$ orbital is using all of its energy to achieve the maximum possible "launch altitude" from the nucleus. It can get away with that because in quantum land a light object can head straight back towards the nucleus without ever hitting it. In contrast, the $p$ orbital diverts some of its available energy into achieving a more conventional concept of an orbit, which makes its maximum radius lower than that of the do-or-die $s$ electron. I suspect that the textbook was trying to convey this "short cut" aspect of the $s$ orbital, rather than a special attraction per se near the nucleus.

P.S. - "Conventional orbit" is a loaded phrase in this context. The peculiarities of the quantum world require orbital solutions that look like figure eights and other loopy forms, rather than conventional elliptical orbits. These loopy forms are all spherically distorted standing-wave solutions, which can in turn be thought of as akin to distorted multiple loops on a skip rope. These skip ropes rotate in the complex plane, though, not in ordinary space.

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    $\begingroup$ Answers involving purely angular momentum considerations cannot be correct, as they cannot explain why the $2s$ and $2p$ orbitals are degenerate in energy in the hydrogen atom. $\endgroup$ May 12, 2012 at 5:26

Because of the penetration, that is, because the $\mathrm{2s}$ electron can spend time near the nucleus, it is less effectively shielded by the core electrons than a $\mathrm{2p}$ electron. Because it is less effectively shielded, a $\mathrm{2s}$ electron experiences a higher effective nuclear charge and is held closer to the nucleus than a $\mathrm{2p}$ electron which gives the $\mathrm{2s}$ orbital a lower energy.


This is not a thorough answer. It holds when comparing orbital energies of the same shell number $n$ (in your case 2) in atoms containing a low number of electrons. This question contains more information on this.

The letter of the orbital ($\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, etc.) is a nameplace for the azimuthal quantum number (orbital angular momentum, $\ell$). The $\mathrm{s}$ orbital has $\ell=0$, $\mathrm{p}$ has $\ell=1$. The angular momentum being higher, the energy also is.


Could not a simple way of rationalising this result be to use the formula for a classical Coulombic attraction between point charges?

Given the radial distribution as shown, and the scaling of electrostatic force with r² rather than r, any time spent closer to the nucleus is vastly more important than time spent further away in terms of energy minimisation. This means that as radius increases (going left to right on the graph) what the lines do gets increasingly irrelevant unless they have exhibited very similar behaviour previously.


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