What is the radius of the hydride $(\ce{H-})$ ion? There are several sources with conflicting reports on the same.

Inorganic Chemistry: Principles of structure and Reactivity cites Pauling’s 1960 study of the hydride ion, stating that

it has a radius of 208 pm compared to 216 pm for the iodide ion

This would imply that it is larger than the fluoride, chloride and even the bromide ion! This article in nature disproves the above and states:

the ionic radius of an $\ce{H-}$ ion is 134±2 pm, which is slightly larger than the radius of $\ce{F^-(II)}$ (128.5 pm) and almost the same as the radius of $\ce{O^2-(II)}$ (135 pm). This evaluation agrees with the generally accepted trend that the radius of an $\ce{H-}$ ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion.

Which one of these is correct? Also, could additional citations be given to tell the exact size of the hydride ion?


3 Answers 3


Unfortunately, there is no caliper-like tool to measure ionic radii directly. They are determined based on experimental data such as crystal structure or crystal lattice energy using empirical relationships. The exact value would inevitably deviate from one method to another.

Pauling [1, pp. 150–152] used a hard-sphere model which doesn't account for overlap in crystal lattice and was predicting higher interatomic distances compared to the crystallographic data, hence an overestimated value of the anionic radii.

Several limitations of the hard-sphere model can be resolved by applying a soft-sphere model. According to the recent work by Lang and Smith [2], who also compared applicability of both hard- and soft-sphere models, $r_\mathrm{i}(\ce{H-}) = \pu{139.9 pm}.$

CRC Handbook of Chemistry and Physics [3, p. 12-22] lists the value of thermochemical radius determined using Kapustinskii equation from a set of known lattice energies: $r_\mathrm{i}(\ce{H-}) = \pu{(148 ± 19) pm}.$


  1. Pauling, L. The Chemical Bond; Cornell University: Ithaca NY, 1960.
  2. Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.
  3. Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

Since the question is about the exact size — and additional citations are requested — it may be worth mentioning Randell Mills' model which calculates the hydride radius analytically as $1+\sqrt{s(s+1)}$ times the Bohr radius, where $s$ is the electron spin of $\frac{1}{2}$. This works out to 99pm and represents the radius of a free hydride ion.

This is different from the ionic radius in a crystal lattice. Depending on your needs, one value or the other may be useful. For example, Mills' radius allows him to calculate the ionization energy of $\ce{H-}$ to within experimental accuracy with a simple, analytical formula (see also here).


Randell L. Mills, The Grand Unified Theory of Classical Physics; 2018 (ISBN: 978-0-9635171-5-9)


Well since the ion looks like the helium atom but without 1 proton and since the helium core is approximately 4 times heavier than the hydrogen core and if we neglect the magnetic interactions between core and electrons i would expect it to be 4 times the radius of helium = roughly 125 pm

  • $\begingroup$ I wanted to say ion instead of atom in the first row. $\endgroup$
    – Tom
    Commented Dec 30, 2019 at 9:44
  • 5
    $\begingroup$ I'm afraid that's not how it works. $\endgroup$
    – andselisk
    Commented Dec 30, 2019 at 10:00
  • 3
    $\begingroup$ You can edit your answer. You should certainly expand your answer and tell which formula and what theory you used. $\endgroup$
    – Zenix
    Commented Dec 30, 2019 at 10:01
  • $\begingroup$ @Andselisk how did he reach to that conclusion? I mean what (which theory or formula) did he use? $\endgroup$
    – Zenix
    Commented Dec 30, 2019 at 10:02
  • 6
    $\begingroup$ @Zenix How am I supposed to know? Maybe a linear proportion, from the look of it. $\endgroup$
    – andselisk
    Commented Dec 30, 2019 at 10:04

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