Theoretical calculations with DFT, MP2 to MP4(SDQ), and CCSD(T) methods clearly display that the Cp 2 Zr-mediated coupling reaction of two acetylene molecules easily takes place through a nonsymmetrical transition state with nearly no barrier and significantly large exothermicity but the M(PH 3 )-mediated reaction (M ) Ni, Pt) takes place through a symmetrical transition state with either a considerably large activation barrier for M ) Pt or a moderately large activation barrier for M ) Ni. The charge-transfer (CT) interaction between the d orbital of the transition-metal center and the π*-π* bonding couple of two acetylene molecules plays an important role for the C-C bond formation in the M(PH 3 )mediated coupling reaction, which needs a symmetrical transition state structure. On the other hand, a CT interaction between the d π -π* back-donation orbital of Cp 2 Zr(C 2 H 2 ) and the π* orbital of the second acetylene molecule is strongly formed in the Cp 2 Zr-mediated coupling reaction, which leads to a nonsymmetrical transition state structure. These differences are reasonably interpreted by the d π -π* back-donation of Cp 2 Zr(C 2 H 2 ) being much stronger than that of M(PH 3 )(C 2 H 2 ). This is because the early-transition-metal element has d orbitals at higher energy than does the late-transition-metal element. The Ni(PH 3 )mediated coupling reaction takes place with a smaller activation barrier than does the Pt-(PH 3 )-mediated reaction. This result is interpreted by noting that the 3d transition-metal element has d orbitals at higher energy than does the 5d transition-metal element. Also, (MeO) 2 Zr-mediated coupling reaction proceeds with nearly no barrier, in spite of the fact that the d orbital is at lower energy in (MeO) 2 Zr(C 2 H 2 ) than that in Cp 2 Zr(C 2 H 2 ). This is because the MeO group is flexible and gives rise to little steric repulsion between ligands and substrates.

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