如何在NBO计算中限制分子中某个原子的电子态

zidu113

各位大神好：     近来看了一篇文献进行NBO计算，分析结果如下面的表4所示。 大意是总共9个结构为Ph3P−C−L 分子（如下图1所示），通过常规的（非限制的）NBO计算得到图2所示的路易斯结构。可以看出分子1和分子4中的中心碳原子上有两个孤对电子。其余结构的中心碳原子上只有一对孤对电子。然后，作者应该是对其余的这些碳原子上只有一对孤对电子的分子进行了强制NBO计算，也就是强制让这些碳原子上带有两对孤对电子（计算结果应该是下表1中的带有括号中的数值）。不太明白作者是怎么强制让这些碳原子上带有两对孤对电子进行NBO计算的。  


Figure 1：Investigated compounds (moieties L with a net negative charge lead to a negatively charged molecule; the donor lone pair is given for all ligands).


     The bonding situation in compounds 1–9 was next investigated by NBO analysis.  Both the regular NBO and the enforced NBO analysis (with two lone pairs fixed at the central carbon atom in accordance with the molecular orbitals) were performed.

Table 1.NBO results (BP86/TZVP) for compounds 1–9. E refers to the atom of the ligand L that is adjacent to the central carbon atom. Partial charges q and orbital populations are given in
electrons. Values in parenthesis correspond to Lewis structures with two lone pairs at the central carbon atom, which have been enforced in the NBO calculation.                [td]

Compound	L	Charges			WBI[a]		LP(C)σ		LP(C)π	Residual density
		q(C)	q(P)	q(E)	C−P	C−E	Occ[c]	[%s㎡rbital]	Occ	[%]
1	SO2PyH	−1.13	1.50	2.07	1.23	1.08	1.72	46.5	1.22	3.3
2	SO2Ph−	−1.29	1.48	2.07	1.42	1.13	1.64	42.8	(1.45)	3.2 (3.2)
3	IMe	−0.96	1.51	0.33	1.37	1.54	1.55	16.5	(1.31)	2.9 (3.2)
4	IMe−	−0.97	1.47	0.26	1.40	1.34	1.68	35.8	1.26	3.3
5	PPh2NH2	−1.43	1.55	1.68	1.32	1.34	1.62	36.7	(1.49)	3.3. (3.3)
6	PPh2NH−	−1.42	1.51	1.70	1.45	1.13	1.62	36.2	(1.46)	3.2 (3.1)
7	SiMe3−	−1.50	1.45	1.60	1.58	1.01	1.59	41.0	(1.43)	2.8 (2.8)
8	CN−	−0.99	1.46	0.18	1.43	1.41	(1.57)	(34.9)	1.33	3.9 (3.6)
9	PPh3	−1.39	1.55	1.55	1.32	1.32	1.61	38.6	(1.49)	3.4 (3.4)

[a] WBI=Wiberg bond indices.  LP=lone pair. [c] Occ=occupancy.





Figure 2. Lewis structures according to the NBO analysis (note that single bonds may also be dative bonds and double bonds can also be considered as dative bonds with strong π

sobereva

用$CHOOSE指定Lewis结构，详见手册

zidu113

嗯，sobereva 老师，谢谢您。