Highly Symmetrical Palladium Cluster Complexes with Either Anticuboctahedral or Cuboctahedral Pd13 Core: Theoretical Insight into Factors Determining Symmetrical Structure

Abstract

One of the important open questions is what factor(s) determines the symmetry of the structure of the metal nanocluster complex. [Pd₁₃(μ-Cl)₃(μ₄-C₁₆H₁₆)₆]⁺ (Anti-μ₄; C₁₆H₁₆ = [2.2]paracyclophane) has an anticuboctahedral Pd₁₃ core unlike [Pd₁₃(μ₄-C₇H₇)₆]²⁺ with cuboctahedral Pd₁₃ core. DFT calculations show that Anti-μ₄ is more stable than isomers, [Pd₁₃(μ-Cl)₃(μ₃-C₁₆H₁₆)₃(μ₄-C₁₆H₁₆)₃]⁺ and [Pd₁₃(μ-Cl)₃(μ₂-C₁₆H₁₆)₃(μ₄-C₁₆H₁₆)₃]⁺ with cuboctahedral Pd₁₃ core (Cubo-μ₃,μ₄ and Cubo-μ₂,μ₄, respectively) and [Pd₁₃(μ-Cl)₃(μ₃-C₁₆H₁₆)₆]⁺ with distorted icosahedral Pd₁₃ core (dis-Ih-μ₃). Not the stabilities of [Pd₁₃(μ-Cl)₃]⁺ core and (C₁₆H₁₆)₆ ligand-shell but rather the interaction energy (E_int) between [Pd₁₃(μ-Cl)₃]⁺ and (C₁₆H₁₆)₆ ligand-shell determines stabilities of these complexes. μ₄-C₁₆H₁₆ coordination bond is stronger than μ₂- and μ₃-coordination bonds, leading to a larger E_int value in Anti-μ₄ than in isomers bearing μ₂- or μ₃-coordination bond. An icosahedral Pd₁₃ core is not favorable for these Pd₁₃ complexes because of the absence of a Pd₄ plane. [Pd₁₃(μ-Cl)₃(μ₄-C₁₆H₁₆)₆]⁺ with cuboctahedral Pd₁₃ (Cubo-μ₄) is not stable despite the presence of six Pd₄ planes, because its three Pd₄ planes with μ-Cl ligand cannot form μ₄-C₁₆H₁₆ coordination bond due to steric repulsion of C₁₆H₁₆ with the μ-Cl ligand. In contrast, Anti-μ₄ is stable because it has six Pd₄ planes with no Cl ligand to form strong μ₄-C₁₆H₁₆ coordination bonds without steric repulsion. Also, discussion is presented on the difference in symmetry between [Pd₁₃(μ-Cl)₃(μ₄-C₁₆H₁₆)₆]⁺ and [Pd₁₃(μ₄-C₇H₇)₆]²⁺.