摘要
在分子束—气实验装置上,用直流放电法产生亚稳态Ba(~3D),分别与气体o,p,m—C_6H_4Cl_2和o,m,p—C_6H_4ClCH_3反应,产生电子激发态BaCl(A,B_C)产物,通过化学发光测量得到化学发光光谱、各反应的碰撞总截面和相对化学发光反应截面。实验结果表明:Ba(~3D)+o,m,p-C_6H_4Cl_2三个反应生成BaCl各电子态的相对发光反应截面互不相同,与两个Cl原子在苯环上的位置有关,可以解释为反应经过两个通道:一是与基态Ba(~1S)情况类似,Ba(~3D)的价电子首先进入苯环的π反键轨道,再经由振动耦合进入C-Cl键σ反键轨道形成BaCl;另一通道是Ba的价电子直接进入C—Cl键的σ反键轨道。
Chemiluminescent reactions of Ba(~3D) state prepared by DC discharge with o, m, p-C_6H_4Cl_2, o,m,p-C_6H_4ClCH_3 which resulted in electronically excited states A^2∏ , B^2∑ and C^2 ∏ of BaCl were investigated via a molecular beam-gas apparatus. Chemiluminescence spectra, total collision cross sections and relative chemiluminescence cross sections of the reactions were presented in this paper.The results showed that in the reactions of Ba(~3D)+o,m,p-C_6H_4C1_2 the relative chemiluminescence cross sections depend on the position of atom Cl on benzene ring. It seems reasonable to assume that the reactions proceeded via two different channels simultaneously: (1) atom Ba attacks directly the C-Cl bond of C_6H_4C1_2 leading to form the BaCl; (2) a valence electron of Ba transfers into a π orbit of benzene ring at first, and then the electron of π orbit of benzene ring transfers into a σ orbit of C-Cl bond via vibrational coupling leading to form the BaCl.
基金
国家自然科学基金
