Abstract
AbstractThe C–H bond of the methylidyne radical, CH•, is abnormally long and weak, even longer and weaker than that of methane, CH4. This is remarkable given the fact that the C–H bond has been shown to contract as the number of substituents around the pertinent carbon atom decreases (e.g., from ethane to ethene to ethyne) because of the accompanying reduction in steric congestion. To elucidate the origin of this anomaly, we have analyzed the C–H bonding mechanism of quartet CH•••and doublet CH•and compared this with the sterically more encumbered triplet CH2••, doublet CH3•, and singlet CH4, using quantitative (Kohn‐Sham) molecular orbital theory. Our analyses reveal that, depending on the effective electronic configuration of the methylidyne radical, its relatively long and weak C–H bond originates from: (i) the position at which the maximum electron‐pair bonding overlap is achieved (quartet CH•••); and (ii) the destabilizing steric Pauli repulsion between the valence orbitals on the interacting fragments (doublet CH•).Key pointsThe C–H bond of the methylidyne radical is remarkably longer and weaker than that of the sterically more encumbered methane molecule.The position of maximum electron‐pair bonding overlap determines the long C–H bond of the methylidyne quartet CH•••.The C–H bond of the methylidyne doublet CH•expands even further, due to its effective 2s22p2electronic configuration, yielding a significant destabilizing Pauli repulsion between the valence orbitals on the interacting fragments.
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