摘要
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．
The relativistic Pseudopotential ab initio and the bonding energy criterion were used on MCO and MCONa + Complexes (M=Ru,Pd).The bonding energy Eb (i)of M.O.and Eb (AB)between atoms had been obtained,It was found that CO was not activated easily by pure Ru or Pd,but CO was activated greatly when adding Na +to the system RuCO or PdCO.The calculation results also showed that the main effect of Na +is equivalent with space charge.
出处
《辽宁师范大学学报（自然科学版）》
CAS
1996年第3期236-239,共4页
Journal of Liaoning Normal University:Natural Science Edition
关键词
成键能
助催化作用
模型化合物
一氧化碳
活化
the relativistic effective core potential,ab initio,promoted catalysis,the bonding energy criterion
