Deuteration favours the polymorph with shorter O-H bonds: a stretch-bend rule tested across 220 inorganic polymorph pairs

Deuteration can change which polymorph of a hydrogen-bonded crystal is stable. We derive a rule for the direction and size of this effect and test it across compounds. The H/D free-energy difference between two polymorphs is, to first order, $\Delta\Delta F_{\mathrm{iso}} \simeq S_{\mathrm{str}}(1-C)\Delta\bar{r}$, where $\Delta\bar{r}$ is the difference in mean O-H length, $S_{\mathrm{str}} \simeq \frac{1}{2}hc(1-1/\rho)s$ at room temperature comes from the softening $s$ of the O-H stretch with bond length, and $C$ is the fraction of the stretch term cancelled by bend and libration. While $C<1$, deuteration favours the polymorph with the shorter O-H bonds, that is, with the weaker hydrogen bonds; where $C(T)$ crosses unity the preference reverses. In harmonic calculations with three universal machine-learned potentials on 220 inorganic polymorph pairs, $\Delta\Delta F_{\mathrm{iso}}$ rises by 0.11 to 0.13 meV per H per mÅ at 300 K, with $C$ about one half. For moderate to strong hydrogen bonds (a boundary chosen after inspecting the data) the rule is right in 74 of 84 pairs and 27 of 28 formulas; the exceptions share one Al(OH)$_3$ structure in which bend and libration outweigh the stretch at room temperature ($C>1$ in two of the three models). The rule fails for weak hydrogen bonds and did not transfer in a prospective test on new pairs. Exploratory quasi-harmonic calculations put the diaspore to $\delta$-AlOOH transition about 1 GPa higher for AlOOD than for AlOOH, a shift that high-pressure experiments can test.