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Highly Symmetrical Palladium Cluster Complexes with Either Anticuboctahedral or Cuboctahedral Pd13 Core: Theoretical
Yu Tian1, Bo Zhu2, Tetsuro Murahashi3
1Institute of Theoretical Chemistry, College of Chemistry, Jilin University, Changchun, 130023, P. R. China.
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-μ; C16H16 = [2.2]paracyclophane) has an anticuboctahedral Pd13 core unlike [Pd13(μ4-C7H7)6]2+ with cuboctahedral Pd13 core. DFT calculations show that Anti-μ 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-μ,μ and Cubo-μ,μ, respectively) and [Pd13(μ-Cl)3(μ3-C16H16)6]+ with distorted icosahedral Pd13 core (dis-Ih-μ). Not the stabilities of [Pd13(μ-Cl)3]+ core and (C16H16)6 ligand-shell but rather the interaction energy (Eint) 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 Eint value in Anti-μ 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-μ) 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-μ 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+.
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