Why metal hydrides have low lattice enthalpy

Question

Lattice enthalpy is inversely proportional to the distance between the ions. And the trend of lattice enthalpies is: $\ce{MF>MCl>MBr>MI}$. But why do metal hydrides have lesser lattice enthalpy than flourides?

Data

Answer 1

As suggested in the comments, hydride ion size is founded on shifting sands, and even the size of the cation is not really fixed. Everything in ordinary chemistry is at least somewhat squishy.

We should be looking at distances between ion centers and crystal structures. Where the crystal structure is the same we can assess the distance from Lattice constants, which are themselves a little squishy but at least accessible by direct (X-ray diffraction) experiments.

Let's look at sodium compounds. Lattice constants in Angstroms (Å) are from the Wikipedia articles on the respective compounds, lattice energies in kJ/mol are quoted from https://www.wiredchemist.com/chemistry/data/lattice-energies.

$\ce{NaH}: \pu{4.98 Å}, \pu{811 kJ/mol}$

$\ce{NaF}: \pu{4.62 Å}, \pu{904 kJ/mol}$

$\ce{NaCl}: \pu{5.64 Å}, \pu{769 kJ/mol}$

We would expect that for the same ion charges and crystal structure (the latter true of all the compounds cited above), we would expect smaller lattice constant to correspond to more lattice energy, and this we see. Essentially, the hydride ion holds its valence electrons so loosely that they wander over slightly more volume than the greater number of, but more tightly held, electrons in fluoride ions.

The similar lattice constants for alkali and alkaline earth fluorides versus hydrides facilitates the formation of solid solutions, which have been considered for hydrogen storage applications.