Abstract
AbstractThe triel bond between the π‐hole on the triel atom of TrR3 (Tr = B, Al, Ga; R = H, F, Cl, Br) and a lone pair on the Te atom of H2Te is examined using ab initio methods. For Tr = B, the triel bond is weakened as the R substituent becomes more electronegative, while the opposite pattern is noted for Al and Ga. The weakest triel bond of all occurs in H2Te‐BF3 (2.9 kcal/mol) but is much stronger for all the other complexes (>11 kcal/mol). The placement of electron‐releasing OH, NH2, and CH3 R′ substituents on the R′2Te base strengthens the triel bond, whereas the opposite occurs for the electron‐withdrawing CN. Despite its high electronegativity, the F substituent causes a strengthening of the interaction, which is due in large part to the formation of secondary chalcogen bonds that involve the σ‐holes on Te in F2Te. However, the F atom on TrF3 is unable to act as an electron donor in these chalcogen bonds, leading to weakened F2Te··TrF3 interactions.
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