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使用相对论赝势从头计算方法和成键能判据研究了模型化合物ＭＣＯ和ＭＣＯＮａ＋（Ｍ＝Ｒｕ，Ｐｄ）的电子结构，讨论了其中的化学键及Ｎａ＋的助催化作用．得出在单独Ｒｕ、Ｐｄ情况下ＣＯ不被活化，其原因在于金属与ＣＯ的主要作用是ＣＯ的弱反键占据轨道５σ电子到金属空ｄ轨道的配位，ＣＯ的弱反键轨道上减少电子的占据不仅不能使ＣＯ的键削弱反而有少许增强；当有Ｎａ＋参与时增强了金属ｄ电子到ＣＯ反键轨道２π的反馈能力，从而使ＣＯ键被削弱而得到活化．计算还表明Ｎａ＋的作用相当于空间电荷
The electronic structures of the model compounds MCO and MCONa + (M = Ru, Pd) have been studied using the ab initio method and the bond energy criterion of the relativistic pseudopotential. The chemical bonds and the promoters of Na + have been discussed. The results show that CO is not activated in the presence of Ru and Pd alone. The reason is that the main role of metal and CO is that the weak antibonding of CO occupies the coordination of the orbital 5σ electrons to the metal d orbital, and the weak antibonding orbital of CO decreases The occupation of electrons not only can not weaken the CO bond, but enhances it slightly. When Na + participates, the feedback ability of the metal d electron to CO antibonding orbit 2π is enhanced, so that the CO bond is weakened and activated. Calculations also show that the role of Na + is equivalent to the space charge