According to the Schrödinger wave equation, Bohr's model of atomic orbitals is disproved and a new formalism that describes atomic orbitals is introduced.

It follows that the definition of atomic radius - "the distance between the center of the nucleus to the outermost orbit" - doesn't make sense, given that the new theory does not describe atomic orbitals as discrete entities in space.

If an orbital has no exact boundary, then what exactly is the definition of atomic radius and how can I calculate it?

  • 6
    $\begingroup$ The most probable distance of the distribution is used. $\endgroup$
    – MaxW
    Sep 15 '16 at 16:30
  • 1
    $\begingroup$ I'm not sure of the answer that's why I'm answering in the comments. The atomic radius is measured in following forms:1) Covalent radius 2)Metallic radius 3)Vander waals radius. Measurement of any of these radius does not involve the concept of orbitals. It involves the distance between the atoms in bonded state. The half of the bond length gives the approx radius. That's why the radius of halogen is much larger than expected(as Vander Waals radius is measured in their case). $\endgroup$ Sep 15 '16 at 16:45
  • $\begingroup$ How do you measure radius of elements in their elemental stste??? $\endgroup$
    – Bharat
    Sep 15 '16 at 17:20
  • 1
    $\begingroup$ I'm going to keep this simple and not give a book length explanation. A Cl atom would have many kinds of radii. First what would be the radius of a Cl atom out in free space. Second what would the nuclear distance be between Cl atoms in the $\ce{Cl2}$ molecule? Half that distance would be another radius. Third what would be the distance be between a $\ce{Cl^-}$ ion and various cations such as $\ce{Na^+}$, $\ce{K^+}$, $\ce{Rb^+}$, $\ce{Ca^{2+}}$ and so forth. // For diatomic gases you could measure the distance by the rotational moments of the molecules. For solids you'd use x-ray diffraction. $\endgroup$
    – MaxW
    Sep 15 '16 at 18:59

As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system. In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus ($r$) and two angles ($\phi$ and $\theta$). [If you pick a particular $r$, the angles $\phi$ and $\theta$ work in a way similar to how we locate points on the surface of the earth (latitude and longitude).]

This being said, the image below may clarify things and answer your question (source):

enter image description here

The graph deals only with s atomic orbitals of hydrogen, the 1s highlighted in red.

See the x-axis? It is measured in multiples of $a_0$ a.k.a. Bohr radius,

$$a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} = \frac{\hbar}{m_e c \alpha}$$

This is the radius obtained from Bohr's theory for the 1s orbital of hydrogen. How can it the useful as a unit in this graph?

Well, the square of a wave function is a probability distribution, $\rho(r, \phi, \theta) = \Psi^2 (r, \phi, \theta)$. If you had a dice with probability distribution $\bar{\rho}(n) = 1/6$ (a fair dice) for each $n = 1,..., 6$, you would get, throwing that dice, on average,

$$\sum_{n=1}^6 \bar{\rho}(n) \times n = \sum_{n=1}^6 \frac{n}{6} \\= \frac{1}{6} + \frac{2}{6} + \cdots + \frac{6}{6} = 3.5$$

So $3.5$ is the expected value of throwing a dice. See how the last equality is just the usual average we're familiar with? But the first one is most useful in our context. Since our variables $r$, $\phi$ and $\theta$ are all continuous, in order to get an average radius, we integrate instead of sum:

$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [\rho(r, \phi, \theta) \times r] r^2 \sin(\phi) d r d \phi d \theta $$

$r^2 \sin(\phi) d r d \phi d \theta$ is just the volume element in spherical coordinates (in our dice example we had an implicit $1$ there as "volume"), what matters is just what is in squared brackets.

Now 1s ($n = 1$, $\ell = 0$) orbitals do not depend on angles, they are perfectly spherical. That means that $\Psi_{n \ell m_\ell}(r, \phi, \theta) = R_{n \ell} (r)$ with $n = 1$, $\ell = 0$ and $R_{n \ell} (r)$ is some function describing the radial electron wave function. Our integration simplifies to:

$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 \sin(\phi) d r d \phi d \theta \\ = \int_0^{2 \pi} \int_0^{\pi} \sin(\phi) d \phi d \theta \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\ = 4 \pi \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\ = \int_0^\infty [ (4 \pi r^2 R^2_{n \ell} (r)) \times r ] d r $$

This surely looks like the expected value of throwing a dice! The variable being averaged is $r$ and the new probability distribution is $\rho_\text{radial} (r) = 4 \pi r^2 R^2_{n \ell} (r)$, called radial probability distribution. That's what's being plotted in the figure above.

If you go further and carry out the last integral ($R_{1 0} (r)$ can be found here), you'll see the answer is just $a_0$, the Bohr radius. That's expected, after all, since Bohr's theory was quite successful for the hydrogen atom. Nothing has been disproved.

Thus, the atomic radius of an atom can be obtained from quantum mechanics by calculating the expected value of the distance between the outermost electron and the nucleus, as MaxW has pointed out.


I am unaware of ANY definition of atomic radius that states it is the Bohr radius, please provide the citation for your source for that claim. The Bohr model works fairly well with the H atom in/near the ground state, why do you claim it is "disproved"? Again, please provide the citation. It is true that the Bohr model fails (badly) for more complex atoms (is this what you mean by "disproved"??) As already noted, "atomic radius" has a number of different definitions. Interatomic distance depends on temperature and pressure, as you probably know. You ask what the definition is, and we can't answer without you also describing the context - what use would you make of it? (There is no one definition that fits all uses.) The crudest measure would be to determine the volume of a single atom (or molecule) based on its density. g/cc→moles/cc→cc/mole→cc/particle You could then assume each unit volume is a cube and its radius is ½ of the length of one edge, or you could assume that the units are spheres and have maximum packing ( = π/(3√2)) of that volume. X-Ray or neutron (etc.) diffraction determines the structure and spacing of crystals, which can easily give interatomic nearest neighbor distance in solids. In some liquids and amorphous (and semi-crystalline) solids too (using synchrotrons). The concept of radius breaks down in considering ions in solution, since ions are generally enveloped in a solvation sphere so that anion to cation distance is a statistical value determined by both of the ion-solvent interactions. There is also a definition of cross-section (πr²) for atoms in the vapor phase (gasses and plasmas). And once covalent interactions become significant, hard sphere models of structure aren't going to work very well at all. Comparing elements crystal structures at STP (see Wikipedia Periodic_Table_ (crystal_structure)) indicates that most of the elements you will encounter have near optimal packing, unfortunately the elements of the "most" chemical interest to us (Groups 13-17, aka "the p-block" elements) rarely have optimal packing, so the above rough calculation isn't very good for them.


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.