I am learning about structure of atom, in which i saw right from J.J. Thomsom atomic model to modern nuclear atomic model all are spherical in shape.

I have seen how different discoveries help to develop the atomic model but i have't came through a single discovery which proves that atoms are spherical in shape.

So, Why atoms are spherical in shape?

  • $\begingroup$ Related: chemistry.stackexchange.com/q/9561 $\endgroup$ – Nicolau Saker Neto Jun 20 '14 at 11:14
  • 1
    $\begingroup$ atoms are not perfectly spherical. $\endgroup$ – CognisMantis Jan 5 '15 at 7:40
  • $\begingroup$ Only isolated atoms can be considered spherical, and even then as approximation. E.g. Hydrogen atoms in $\mathrm{H_2O}$ are very far from being spherical in shape. $\endgroup$ – juanrga Mar 1 at 13:37

Isolated atoms are spherically symmetric.

The s-orbitals are spherically symmetric.

Probabilty contour plots of the three p-orbitals are dumbbell shape individually; however, adding the probability functions of all three results in a spherically symmetrical distribution.

Likewise, the sum of the probabililty functions of each subshell of orbitals (such as all three 2p orbitals, all five 3d orbitals, etc.) is also spherically symmetric.

For the above reason it is apparent that atoms with filled or half filled valence shells will have spherical symmetry, but what about the remaining atoms?

In 1965, Journal of Chemical Eduction published an article "Shapes of Atoms" volume 42 page 145, stating that atoms with filled or half filled valence shells are spherical, while the remainder are other shapes. A few months later a correction was published at volume 42 page 397, explaining that the original article failed to consider the fact that a valence electron in a isolated atom does not occupy a particular orbital; instead the wave function of the electon is a "hybrid state" of all the orbitals of the valence subshell because they are degenerate.

See also Ruslan's comments below regarding the second J. Chem. Ed. article, that the proper description would be: the atom is in a "mixed state".

  • $\begingroup$ In fact that correction is an unfair one. The authors claim that due to degeneracy the atom must be in a particular state such that the result is spherically symmetric. But in fact all the other eigenstates with the same energy are equally possible. Moreover, why would one even say that the correct orbital is a sum of eigenstates? It could be e.g. $\left|m=1\right\rangle+i\left|m=0\right\rangle-\left|m=-1\right\rangle$, which is also a superposition of all valence subshell states, but isn't spherically symmetric. $\endgroup$ – Ruslan Jan 13 '15 at 10:45
  • $\begingroup$ In isotropic environment the actual state doesn't depend that much on what the environment the atom is in now - what's important is how the atom was prepared in some energy eigenstate. If e.g. the atom was prepared in a magnetic field, then the field was released, the atom would remain in that state with definite $m$ until some interaction changes it. Thus the atom with incompletely filled valence shell can be of very different shapes, depending on the history of its state preparation. $\endgroup$ – Ruslan Jan 13 '15 at 10:48
  • $\begingroup$ Actually, the correction is true in spirit, although appears to use wrong terms: the electron being in a superposition of all the degenerate states can't make the atom spherical. It's the mixture of states, where the probability densities are added, which can. The atom can be in a mixed state, because we don't know any details of how it was prepared in the given ground state. And assuming isotropic environment and a gas of such atoms, we can expect the state to be an even mixture of all possible degenerate states, which is symmetric. $\endgroup$ – Ruslan Jan 14 '15 at 16:05
  • $\begingroup$ @Ruslan it would be my fault for using the word "superposition", the reference doesn't use that word, it uses the terms "degeneracy", "hybridization", "hybrid state" and "resonance hybrid". $\endgroup$ – DavePhD Jan 14 '15 at 16:20
  • $\begingroup$ Maybe you're right, but the paper uses a suspicious normalization factor of $\frac1{\sqrt3}$, which would only make sense for a superposition of states. So I guess it's their mistake (or misunderstanding). Could you improve your answer to address this? $\endgroup$ – Ruslan Jan 14 '15 at 16:24

I think you should think this question a little through:

  • The easy answer is that atoms are not spherical in shape. Sphere is a solid object, atoms to not have well defined boundaries with "solid cutoffs". Atoms are more like clouds consisting distributions of electrons, neutrons and protons.

  • Atoms, per definition, are electrons + a single nucleus, ie. the huge positive charge of a point-like object (nucleus) pulls together the whole thing. That result is that electrons approximately move in a central field. Such central field generates spherical symmetry of the electronic states, and that is why you have e.g. spherical harmonics when you describe the atomic orbitals. Check the solution of the Schrodinger equation how it leads to states, and how wave-functions with only radial dependence (i.e. sphere-like) appears.

+++++ Incorrect part of the answer was deleted based on the comments of DavePhD.

  • $\begingroup$ well I don't agree as High resoultion electron microscopy shows surfaces of elements like Pt to be spherically bumpy, and with STED it is even more conforming $\endgroup$ – yawar Dec 9 '14 at 17:14
  • $\begingroup$ Why "only the atoms with field or half-field electron systems"? and what do you mean by "S states"? What is the connection between being half-field (filled?) and being an S state? $\endgroup$ – DavePhD Jan 5 '15 at 2:25
  • $\begingroup$ Thanks, corrected the typo. Electronic states have symmetry symbols, and some of them has higher, some of them has lower symmetries (en.wikipedia.org/wiki/Term_symbol). In case of isolated atoms, typically only filled and half-filed orbitals will lead to S states due to the Hund-rule (en.wikipedia.org/wiki/Hund%27s_rules). $\endgroup$ – Greg Jan 5 '15 at 7:28
  • 1
    $\begingroup$ @Greg, ok but you still haven't explained the only part of "only the atoms with filled or half-filled...", why the others are not spherical by the electron wavefunctions being in a superposition of all the degenerate orbitals of the partially filled subshell, why you disagree with the statement in J. Chem Ed. vol. 47, page 397 that "all isolated atoms are spherical". pubs.acs.org/doi/abs/10.1021/ed042p397 $\endgroup$ – DavePhD Jan 5 '15 at 12:46
  • $\begingroup$ @DavePhD OK, I see you point now about the multireference nature of L>0 states. Thank you for the correction. Sorry, I haven't read through your answer previously, as your answer was added months later. You are right. $\endgroup$ – Greg Jan 5 '15 at 16:58

This is a delightfully interesting question, and @DavePhD has put up an excellent answer. I will try and contribute my own perspective to it, although more or less in line with what has already been said; I have also included some illustrations to make concepts a bit more tangible, and perhaps, palatable


How do we determine if something is spherical? Well, we do so by "looking at", measuring its dimensions etc. For large macroscopic objects, that usually means photons of light scatter off the object, our ruler etc. and are captured by our eyes. The signals generated are converted into visual information by our brains. Essentially, all processes involved are quantum mechanical, however, for macroscopic objects their unintuitive, quantum weirdness does not manifest itself.

Similar, without going into details, to determine the "spherical-ness" of an atom we need to probe some physical observable, with which we associate an operator $\hat{\mathcal{O}}$. For a given quantum state $|\psi\rangle$, the expectation value for this observable, i.e. something we expect to get would be given by $\langle \psi|\hat{\mathcal{O}} |\psi\rangle$ (depending on your level of education, you may need to look this up. You can read some details here). Anyway, we can also talk about the probability density of finding an electron in some region of space, and that would be given by $\langle \psi|\psi\rangle \equiv |\psi|^2 $

So essentially, one never really sees the "wavefunction".

Also, one needs to point out is that space is isotropic. Basically, this means there is nothing special about any particular direction in three-dimensional space. Also, even from classical electrostatics we know that Coulombic interactions behave the same way: if you have a negative charged place at some point in space, it will attract positive charges all around it in the same fashion regardless of whether they are placed to its right or two its left.

Now, we know electrons in atoms occupy "orbitals" which are obtained by solving Schrödinger's equation. It is complicated to solve for many-electron systems, but we can do it for hydrogen. Writing down the equation, we see,

$\left( -\frac{\hbar^2}{2\mu} \nabla^2 - \frac{e^2}{4\pi \epsilon_o}\right)\psi (r,\theta, \phi) = E \psi (r,\theta, \phi) $

Here, $\mu$ is the reduced mass of the system, this allows us to treat the problem in centre of mass coordinates (remember Coulomb forces are central). $\nabla^2$ is the laplacian operator, which basically contains the second-derivatives along x,y,z in cartesian space. It has a more complicated form in spherical coordinates which is what we are using for our wavefunction $\psi$, but basically, the important fact is that this first term is kinetic energy, and then the second term is the coulombic potential due to the interaction of electron and nucleus.

This is a separable, partial differential equation which can be solved in terms of special functions. The normalised position wave functions, given in spherical coordinates are

$\psi_{nlm} (r,\theta, \phi) = \sqrt{\left(\frac{2}{na_0^*}\right)^3 \left(\frac{(n-l-1)!}{(2n+l)!}\right)} e^{-\rho/2}\rho^l L^{2l+1}_{n-l-1} (\rho) Y^m_l( \theta, \phi) $

Where, $\rho = \frac{2r}{na_{0}^{*}}$,

$a_{0}^{*} = \frac{4 \pi \epsilon _{0}\hbar ^{2}}{\mu e^{2}}$ (reduced Bohr radius)

$L$ and $Y$ denote the generalised Laguerre Polynomial and Spherical Harmonic respectively, and n,l,m are the quantum numbers you must be familiar with.

Anyway, we are interested only in the functional forms, so we can ignore all the constants out in front. We also note that the wave function is a product of two-functions, radial and angular.

$\psi_{nlm} (r,\theta, \phi) = R(r)Y^m_l(\theta, \phi)$

Now, essentially, spherical symmetry demands that there will be no angular dependance, and any properties measured or probability of finding electrons etc. only depends on distance from the centre r. This means whenever you measure $\langle \psi|\hat{\mathcal{O}} |\psi\rangle$, it will again only depend on $r$ and not on the angular variables.

We only need to focus our attention on the Spherical Harmonics, as the contain the angular information.

For the purpose of this discussion, we can crudely assume that even in many-electron atoms, the spatial wave functions are similar to hydrogenic wave functions (which is only a one electron system). This is because it is only possible to obtain analytic, functional solutions for the hydrogen atom.

Main Argument

For say an $s$-orbital like l = 0, m = 0. It is trivial to see that the wave function, and consequently $|psi|^2$ is spherically symmetric. i.e. depends only on $r$ because $Y^0_0 = \frac{1}{2\sqrt{ \pi}}$.

Now, let us look at the $p$-orbitals where $l = 1$ and $m$ takes values $-1,0,1$

$$ Y^0_1 = \frac{1}{2} \sqrt{\frac{3}{\pi}} cos \left( \theta \right)$$ $$ Y^{\pm 1}_1 = \mp \frac{1}{2} \sqrt{\frac{3}{\pi}} sin \left( \theta \right)e^{\pm i \phi}$$

Now, clearly $\sum\limits_{j} |Y^j_1|^2 = \frac{3}{4\pi}$ is independent of $\theta$ and $\phi$

The same analysis can be repeated for $d,f$ -orbitals as well. A table of spherical harmonics can be found here

So essentially atoms with electrons in valence $s$-orbitals and/or completely filled or half-filled $p,d,f$ shells would be spherically symmetric.

Now, what if we have something like $2p^1$ type situation. A single $p$-orbital is not spherical. But given that space is isotropic, we have no external fields etc., all the $p$-orbitals corresponding to different $m$ values are essentially the same (degenerate). Nature can't tell the difference, and neither can you when you perform an experiment to measure the "Spherical-ness" of an electron.

This does not mean we are naively taking a superposition of the $p$-orbitals. If you see proof by pictures section, this is clearly not true

Proof by Pictures

Note: The spherical harmonics are complex valued functions, so they are hard to plot, the individual $p$ orbitals conventionally plotted are generated by taking linear combinations of them which are still solutions to the Schrödinger's equation. Given below is $p_z$ orbital, $p_x$ and $p_y$ look the same just are oriented along the $x$ and $y$ axis. enter image description here

A naive sum of $p_x$, $p_y$ and $p_z$ orbitals.

enter image description here

Plot of $\sum\limits_{j} |Y^j_1|^2 $. Remember that observable quantities are related to $|\psi|^2$.

enter image description here


Essentially because the Coulomb interaction is isotropic. Physics would have been pretty interesting if a spherically symmetric potential results in a less symmetric wavefunction -- what is there to skew the distribution of electrons?

  • $\begingroup$ Spherical symmetry applies when one ignores the fine details of real-world. Those symmetric wavefunctions are derived by assuming the nucleus is a point-like particle, and Coulomb interaction taken classically. Remember the physicists' joke about spherical cows. Real world is far from spherical symmetries. $\endgroup$ – juanrga Mar 1 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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