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Palladium hydride is an example of a hydride which contains H atoms in palladium's crystal lattice (which is fcc).

I would like to know if it's possible to predict if another hydride, $\ce{NbH_{x}}$ could exhibit the same behavior.

I know that in an ionic solid, the fraction $$\frac{r_{+}}{r_{-}}$$ Can be used to predict the geometry around a cation. Can I use this expression for niobium hydride? In the case of niobium, the crystal structure is bcc, so it contains tetrahedral and octahedral interstial sites.

J. LS
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Isaiah G.
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1 Answers1

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Your intuition is correct; $\ce{NbH_{x}}$ retains the bcc Nb lattice, with the hydrides in the tetrahedral interstital sites, up to x = 2, when the compound switches to a fcc Nb lattice (the Fluorite structure). This is pretty typical behaviour for a transition metal hydride.

I don't think radius ratio rules are going to be useful in predicting the behaviour of these systems; there is a lot of covalency going on (hydride-hydride interactions and Nb-Nb interactions), and the metal lattices will end up being distorted by the correlated interstital occupancies of the hydrides. Hydride has a radius of around 130 pm while Niobium has a radius of around 150 pm so you wouldn't predict the observed behaviour based on radius ratios.

More information on this particular system can be found in Hauck (1976), Ordering of Hydrogen in Niobium Hydride phases.

J. LS
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