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. $\endgroup$ – Ivan Neretin Feb 21 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 at 14:20
  • $\begingroup$ @orthocresol My impression was exactly opposite. $\endgroup$ – Ivan Neretin Feb 21 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 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 at 14:51

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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