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What is the first ionization energy for hydride?

Since there are two electrons in the s orbital, I suppose it should be more stable than hydrogen, despite the extra electric charge, so the ionization energy should be greater than the 13 eV needed to ionize hydrogen. However, I could not find any table with the energies for hydride.

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  • $\begingroup$ The negative ion of hydrogen used in biology has no relation whatsoever to the one used in chemistry. $\endgroup$ Feb 21, 2019 at 13:57
  • $\begingroup$ From what I can tell, OP is interested in the chemical species, not the biological equivalents of it. The mention of hydride as a biological reductant was only there to justify their posting on Biology.SE. Since the question has been migrated to Chemistry, I removed that reference. $\endgroup$
    – orthocresol
    Feb 21, 2019 at 14:20
  • $\begingroup$ @orthocresol My impression was exactly opposite. $\endgroup$ Feb 21, 2019 at 14:35
  • $\begingroup$ @orthocresol is right, I am not interested in the biological properties of hydride. I would like to understand whether an ion with 2 electrons in the orbital s is more stable than an atom with 1 electron in the same orbital. From what I have seen, hydride is not very stable since a photon of light can displace the extra electron. This does not happen with hydrogen... $\endgroup$
    – Gigiux
    Feb 22, 2019 at 14:29
  • $\begingroup$ @Gigiux, careful there. A photon can strip a hydrogen atom of an electron. You just need a photon that matches the first ionization energy. $\endgroup$
    – Zhe
    Feb 22, 2019 at 14:51

2 Answers 2

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You probably don't look at the "ionization energy" for hydride. Instead, you have the reverse reaction:

$\ce{H + e- -> H-}$

You'd measure the enthalpy change here as the electron affinity (EA) of the hydrogen atom. The Wikipedia data page says $72.8\ \mathrm{kJ}/\mathrm{mol}$.

You're basically asking for the opposite of that, so the enthalpy change has the same magnitude but opposite sign. Note that this is not an ionization energy because you're not creating an ion.

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The first ionization energy of the hydride ion is $\pu{0.7542 eV}$ as calculated by Mills (Ref.1) and measured experimentally by Lykke, et al. (Ref.2).

This is much less than the well-known $\pu{13.6 eV}$ ionization energy of Hydrogen and follows the general rule that ionization energies increase as you remove more electrons.

In this case, the positive charge of the nucleus is entirely screened by the first electron, so there's no additional electrical attraction - only a relatively small magnetic attraction between the spin-paired electrons. The weaker binding energy is reflected in a larger radius: $1+\sqrt{s(s+1)}$ times the Bohr radius of the neutral Hydrogen atom (where $s$ is the electron spin of $\frac{1}{2}$), according to Mills' theory (eq. 7.66 and 7.69).

References:

  1. Dr. Randell L. Mills, In The Grand Unified Theory of Classical Physics, 3 Volumes; 2018 edition; Copyrights: Dr. Randell L. Mills, 2018 (ISBN: 978-0-9635171-5-9)
  2. K. R. Lykke, K. K. Murray, W. C. Lineberger, "Threshold photodetachment of $\ce{H−}$," Phys. Rev. A 1991, 43(11), 6104 (https://doi.org/10.1103/PhysRevA.43.6104).
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