I have seen that $\ce{H2}$ can dissociate on a metal-oxide surface by two methods:

  1. Heterolytically, forming a proton-$\ce{O}$ bond ($\ce{OH}$ group) and a hydride-metal bond ($\ce{M-H}$) on a metal-oxide surface.

  2. Homolytically, where two $\ce{OH}$ groups are formed (adsorbed protons) and the two electrons of the $\ce{H2}$ molecule are transferred to low-lying $\mathrm{d}$ or $\mathrm{f}$ states of the cations. This leads to a change in oxidation state and a reduction of the oxide.

However, the reasons for the different behaviours are not obvious to me. Assuming two comparative MO compounds have the same crystal symmetry, how does the electronic structure of the metal impact which process dominates?


I am particularly interested in the example given by: Differences in the Existence States of Hydrogen in $\ce{UO2}$ and $\ce{PuO2}$ from DFT + U Calculations

Here, the preferential formation of a hydride ion $\ce{[(UO_{2})_{n}]^{+}H^{-}}$ or hydroxide ion $\ce{[Pu_{n}O_{2n-1}]^{+}[OH]^{-}}$ has been shown. The authors give the following explanation:

The differences in the existence states of atomic hydrogen in the two dioxides are proposed to be dependent on the nature of $\mathrm{5f}$ electrons of uranium and plutonium; that is, uranium $\mathrm{5f}$ electrons are more delocalized and more favorable to participate in chemical bonding than plutonium $\mathrm{5f}$ electrons.

Why should the higher degree of electron itinerary (which enables electrons to more easily form chemical bonds) favour hydride formation?

This seems to conflict with: Interaction of hydrogen with actinide dioxide (111) surfaces.

In this paper and others, $\ce{PuO2}$ has the more reactive surface. Here, the formation of a hydroxide is more favourable on $\ce{PuO2}$, and less favourable on $\ce{UO2}$. Therefore, I am confused as to how this availability of electrons changes.


This reference$\ce{^{[1]}}$ shows what can happen even in a single oxide, in this case magnesia, with different metal environments.

Dissociation of hydrogen on magnesia, from Ref. 1

Magnesia is, of course, strongly ionic, so the dissociation of hydrogen to form $\ce{O^{2-}-H^+}$ and $\ce{Mg^{2+}-H^-}$ on the surface might well be expected. Putting the magnesia as a thin film on silver does not change this. But on gold the hydrogen dissociates in a redox fashion, forming only $\ce{O^{2-}-H^+}$ as the electrons are injected into the gold. The lowered Fermi energy of the gold causes this transfer as the electrons "fall" into the lower energy state. It's an example of gold acting as an electronegative element, as it does in compounds such as $\ce{CsAu}$, ultimately caused by the effects of relativistic mechanics on the electron structure of gold.

In the case of bulk uranium and plutonium dioxides, electron donation or withdrawal must come from the oxide itself. With uranium (IV), the 5f orbitals are at relatively high energy, and where they have electrons those electrons are readily available for bonding and for donation -- but conversely, empty 5f orbitals are relatively poor electron acceptors. So we get metal hydride formation like the $\ce{Mg^{2+}-H^-}$ bonds in plain magnesia or magnesia/silver.

With plutonium in the same +4 oxidation state, enough increased effective charge and orbital contraction have occurred to make the plutonium 5f subshell a better electron acceptor, making the $\ce{O^{2-}-H^{+}}$ bonding mode more favorable.


  1. Hsin-Yi Tiffany Chen, Livia Giordano, and Gianfranco Pacchioni, "From Heterolytic to Homolytic H2 Dissociation on Nanostructured MgO(001) Films As a Function of the Metal Support", J. Phys. Chem. C, 2013, 117, 20, 10623–10629.

2. Chemical Forms of Uranium". Argonne National Laboratory. Archived from the original on 22 September 2006. Retrieved 18 February 2007. Link

  • 1
    $\begingroup$ Thank you @Oscar Lanzi for your insightful answer. I have now made my question more specific as you suggested; however, your input is still a valued answer. $\endgroup$ – Wychh Jun 6 '20 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.