Recently I came across a textbook of physical chemistry which stated the following:

In the formation of a stable hydrogen molecule, two hydrogen atoms share a pair of electrons between them and two $s$-orbitals overlap along the same axis resulting in the formation of sigma $(\sigma)$ bond. As the shared electrons are attracted by both the nuclei, the size of hydrogen molecule is smaller than that of hydrogen atom.

However, the actual radius of hydrogen molecule is $120~\text{pm}$ which is greater than that of hydrogen atom i.e. $53~\text{pm}$. So, what does the author actually mean when he says that the size of hydrogen molecule is smaller than that of hydrogen atom? Moreover, does it imply that the electron density is maximum at the COM of the hydrogen molecule? Any help would be appreciated.

  • 3
    $\begingroup$ The hydrogen molecule is not a sphere, so it does not have a radius. $\endgroup$
    – Karsten
    Commented May 15, 2019 at 15:22
  • 1
    $\begingroup$ The hydrogen atom is not a hard sphere, so you have to be careful to define what you mean by size and radius. $\endgroup$
    – Karsten
    Commented May 15, 2019 at 15:30
  • $\begingroup$ The figures in this article might be helpful to debunk the textbook statement: resonance.is/direct-imaging-wave-function-hydrogen-molecule $\endgroup$
    – Karsten
    Commented May 15, 2019 at 15:35
  • 5
    $\begingroup$ Maybe they meant: As the shared electrons are attracted by both the nuclei, the size of hydrogen molecule is smaller than the sum of the sizes of two hydrogen atoms. $\endgroup$
    – Jan
    Commented May 15, 2019 at 16:25
  • $\begingroup$ Since an isolated hydrogen atom is spherically symmetric, perhaps the original claim refers to the difference in the effective radius looking along the bond in dihydrogen rather than the radius looking perpendicular to the bond. Is the claim true in that context? $\endgroup$
    – matt_black
    Commented Nov 20, 2023 at 10:19

1 Answer 1


You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen:

Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids.

From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is eerily similar to $4 \times \pu{0.53 Å}= \pu{2.12 Å}$.

However this is a distraction from the main point of the statement in the book, which is simply that concentration of electron density between the two $\ce{H}$ nuclei results in an "end-to-end" distance (the distance, using some measure to determine the edges of the molecule, along the internuclear axis) that is less$^\dagger$ than $\pu{2.12 Å}$, and that the internuclear distance is less than $\pu{2\times\pu{0.53 Å}= 1.06 Å}$ (which it is: the H-H bond length is $\pu{0.74 Å}$). An internuclear distance shorter than twice the lone H radius is consistent with bonding, a process by which electrons and nuclei are allowed to occupy lower-energy regions of the potential than they would when the atoms are far appart.

But the answer to the question is no, a hydrogen molecule is still bigger than a hydrogen atom.

$\dagger$ But just barely. Define the edge of an atom or molecule as the point where the electron density equals that at a distance of $\pu{1 Bohr}$ ($\pu{0.529 Å}$) from the nucleus in a hydrogen atom. Next find the edges in a simple MO model of $\ce{H2}$ ([3]). The edges are $\pu{1.056 Å}$ from the center of the molecule, that is, the molecule along the longest axis has a diameter of $\pu{2.11 Å}$, barely less than $4\times$ the Bohr radius.

enter image description here

[3]: Pearson, Ralph G., and Palke, William E.. Proc.Natl.Acad.Sci.USA, Vol. 77, No.4, pp.1725-1727, April 1980.


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