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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.
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.
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:
