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[求助]
为什么旋轨耦合后单重态与三重态的禁阻会被打破
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However, 3d84s2 → 3d94s1 promotion permits sd hybridization of Ni, and this hybrid allows to form the new bonds in the NiC2H4 complex [33,34]. While the interaction of C2H4 and the 3F ground state of Ni results in a diabatic repulsive potential energy surface (Fig. 3), the 3d94s1(1D) excited state is well suited to bind to C2H4 and it provides diabatic surfaces that are attractive. Crossing of the two diabatic surfaces of different spin will be weakly avoided and spin–orbit coupling will generate two newadiabatic surfaces, shown as the bold lines in Fig. 3. On the lower adiabatic surfaces, the electron configuration and electron spin changes smoothly in the course of the reaction. In a first approximation, the barrier of the reaction has been related to the s → d promotion energy, and the latter can be approximated by the atomic’s excitation term Esd [35]. Ni reacts efficiently because Esd amounts only to 6 kcal/mol, compared with 67 for Mn, 34 for Fe, and 20 kcal/mol for Co, metals which are inert at 300 K. Observation of the NiC2H4 formation at this temperature implies an adiabatic barrier height no larger than 5 kcal/mol, which correlates well with a calculated 10 kcal/mol barrier for the diabatic curve crossing above ground-state reactants.The reason why atomic Sc, Ti and V – despite their large Esd which are comparable to the excitation energies of the inert metals Mn, Fe, and Co – react slowly with C2H4 is due to the fact that these ‘early’ transition metals need not change their electron spin to bind to C2H4 (Fig. 3a for Sc). As crossing of diabatic surfaces of the same spin will be more strongly avoided than crossing between surfaces of different spin, for a given Esd value one can reasonably expect smaller adiabatic barrier heights and, consequently, enhanced reactivity.有大侠帮解释下文献中这段话怎么回事么 发自小木虫Android客户端 |
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