Competition Between the Two σ‐Holes in the Formation of a Chalcogen Bond

Abstract A chalcogen atom Y contains two separate σ‐holes when in a R 1 YR 2 molecular bonding pattern. Quantum chemical calculations consider competition between these two σ‐holes to engage in a chalcogen bond (ChB) with a NH 3 base. R groups considered include F, Br, I, and tert‐butyl (tBu). Also examined is the situation where the Y lies within a chalcogenazole ring, where its neighbors are C and N. Both electron‐withdrawing substituents R 1 and R 2 act cooperatively to deepen the two σ‐holes, but the deeper of the two holes consistently lies opposite to the more electron‐withdrawing group, and is also favored to form a stronger ChB. The formation of two simultaneous ChBs in a triad requires the Y atom to act as double electron acceptor, and so anti‐cooperativity weakens each bond relative to the simple dyad. This effect is such that some of the shallower σ‐holes are unable to form a ChB at all when a base occupies the other site.