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Radius of hydride (H⁻) ion

Alan Koenigsberg  Follow

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: $r_\mathrm{i}(\ce{H-}) = \pu{(148 ± 19) pm}.$

References

  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.

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David Daminic  Follow

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

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FAMILYDELi  Follow
your answer. You should certainly expand your answer and tell which formula and what theory you used.More
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John Walker  Follow
@Andselisk how did he reach to that conclusion? I mean what (which theory or formula) did he use?More
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Joseph Halaby  Follow
You can More
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Eyob Muluneh  Follow
Im afraid thats not how it works.More
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Glenn Janot  Follow
I wanted to say ion instead of atom in the first row.More
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Bert Hickman  Follow

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).

References:

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

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