摘要
选用6-311++G(3df,2p)基组,在二级微扰的理论下,对四角锥形结构化合物BeB4X4(X=H,F,Cl),HBB4H4与BB4H4+的分子振动频率,及原子间的相互作用进行了计算,作用能的计算使用了CCSD(T)方法。结果显示HBB4H4与BB4H4+是违反韦德规则的另两个特例,它们表现稳定的原因与芳香性有关。
Using the 6-311++G(3df,2p) basis set,the clusters BeB4X4(X=H,F,Cl),HBB4H4and BB4H4+ which all have a square pyramidal shape were got at the second-order Mller-Plesset perturbation(MP2) level with all frequencies real.The vertexes of these square pyramidal clusters are Be,BH and B separately.They all act as a two electron donor for B4X4,and the B4X4 act as a bowl to accommodate the pair of electrons.The interaction energies between the vertexes and the B4X4 bowls were calculated at the CCSD(T) level.The interactions between bowls with different terminal atoms and transferred electrons were discussed.The character of the B4 ring is affected deeply with the terminal atoms.For HBB4H4 and BB4H4+,they are four and six electrons short of the Wade magic electron count,but they all have their strong bonding here.
出处
《无机化学学报》
SCIE
CAS
CSCD
北大核心
2012年第9期1950-1958,共9页
Chinese Journal of Inorganic Chemistry
基金
吉林化工学院博士启动基金资助项目