Heats of formation of krypton fluorides and stability predictions for KrF4 and KrF6 from high level electronic structure calculations

Atomization energies at 0 K and heats of formation at 0 and 298 K are predicted for KrF+, KrF-, KrF2, KrF3+, KrF4, KrF5+, and KrF6 from coupled-cluster theory (CCSD(T)) calculations with effective core potential correlation-consistent basis sets for krypton. To achieve near chemical accuracy (+/-1 kcal/mol), three corrections were added to the complete basis set binding energies based on frozen core coupled-cluster theory energies: a correction for core-valence effects, a correction for scalar relativistic effects, and a correction for first-order atomic spin-orbit effects. Vibrational zero point energies were computed at the coupled-cluster level of theory. The calculated value for the heat of formation of KrF2 is in excellent agreement with the experimental value. Contrary to the analogous xenon fluorides, KrF2, KrF4, and KrF6 are predicted to be thermodynamically unstable with respect to loss of F2. An analysis of the energetics of KrF4 and KrF6 with respect to fluorine atom loss together with calculations of the transition states for the intramolecular loss of F2 show that fluorine atom loss is the limiting factor determining the kinetic stabilities of these molecules. Whereas KrF4 possesses a marginal energy barrier of 10 kcal/mol toward fluorine atom loss and might be stable at moderately low temperatures, the corresponding barrier in KrF6 is only 0.9 kcal/mol, suggesting that it could exist only at very low temperatures. Although the simultaneous reactions of either two or four fluorine atoms with KrF2 to give KrF4 or KrF6, respectively, are exothermic, they do not represent feasible synthetic approaches because the attack of the fluorine ligands of KrF2 by the fluorine atoms, resulting in F2 abstraction, is thermodynamically favored over oxidative fluorination of the krypton central atom. Therefore, KrF6 could exist only at very low temperatures, and even the preparation of KrF4 will be extremely difficult.