With the aid of DFT calculations, we now propose one possible mechanism of degradation of ZnSe LHs. As previously reported scanning electron diffraction measurements suggest that degradation occurs from the basal plane edge, we simulated a single layer fragment of ZnSe(L) (Figure , State 1) and calculated the enthalpy of water adsorption at this edge (Figure , State 2). The energy difference between the pristine fragment and that with water adsorbed is -99.2 kJ/mol, suggesting such adsorption is quite favourable. To degrade the material, both H2O and O2 must dissociate (eventually yielding the degradation product ZnO); we therefore hypothesize that H2O dissociates at the basal plane edge to form OH -and H + , which become directly bound to the LH (Figure , State 3). The formation of OH -in particular is quite plausible given the close proximity of the capping amines. These amines may serve as proton acceptors, mediating the formation of OH -that can then bind to the Zn ions of the ZnSe fragment. Furthermore, we observed experimentally that exposure of ZnSe(ba) to pH 6.8 water reduced its degradation rate; water of pH 7.0 or 7.2 led to more rapid degradation, suggesting that OH -and not H + is critical for LH degradation. Dissociation of H2O is exothermic with an energy difference of -8.2 kJ/mol before and after dissociation (Figure ). Adsorption of oxygen at the fragment is then necessary to drive degradation; this too is exothermic by -6.1 kJ/mol (Figure , State 4). We propose the next step involves the oxidation of Se 2+ by O2, producing elemental Se 0 and Zn(OH)2 (incorporating O from both H2O and O2). Directly simulating this process would be challenging due to the unphysical, transient nature of the lone Se 0 atoms and Zn(OH)2 molecules that would be produced in the minimal supercell used for DFT calculations. Nonetheless, the chemical plausibility of this step is supported by the difference in redox potentials of ZnSe oxidation (eq. 1) and O2 reduction (eq. 2):