Solvation of uranyl formate/acetate

Question

Uranyl formate has a structure something like:

$\ce{UO2-(OCHO)2}$

So the structural formulae I've seen have two double-bonded oxygens directly attached to the U and each formate group is attached by single bonds to the oxygens. Like this. The empirical formula for the dry powder is written as a monohydrate.

Can someone explain how this thing gets solvated in water and what the solvated ions are, exactly? I just realized I was assuming it dissociates into a 'uranyl' ion (whatever that looks like) and two formate ions$\ldots$but then I'm confused about where the formate's oxygen goes.

Answer 1

This is a uranyl ion, one of the most common forms of dissolved uranium:

Because the two U=O bonds can form a single delocalised region, there's a strong tendency towards linearity (c.f. carbon dioxide). It tends to dissolve in water by gaining equatorial water ligands (4 or 5), some of which it may then hydrolyse, depending on the pH:

The axial oxygen atoms are also valid targets for hydrogen bonding, typically gaining one hydrogen bond each.

The difficulty in dissolving uranyl formate in particular suggests that the formate ion is quite well-bound, and it seems that in this case you get a combination of water and formate ligands in the complex around the uranyl. I was expecting the formate to be a bi-dentate ligand, but according to the results of a combined computational and experimental study 1, it's uni-dentate (which incidentally allows it to form polymeric chains in solution).

Here's a picture from that study showing a computationally obtained structure of a mixed water/formate-ligated uranyl structure.

1. 2013, Dalton Trans., Christian Lucks, André Rossberg, Satoru Tsushima, Harald Foerstendorf, Karim Fahmya, Gert Bernharda, DOI:10.1039/C3DT51711J