影响屏蔽常数大小的因素有二:(1)同产生屏蔽作用的电子的数目及它所处原子轨道的大小,形状有关.(2)同被屏蔽的电子离核的远近和运动状态有关.我们可根据斯莱特提出的规则来近似求算屏蔽常数6.《无机化学》教材中斯莱特估算屏蔽常数6的方法为:(1)先将电子按内外次序分组:1S;2S;2P;3S,3P,3d;4S,4P,4d,4f,5S,5P,5d,5f 等.(2)外层电子对内层电子没有屏蔽作用,各组的σ=0.(3)同一组,σ=0.35(但1S,σ=0.30)(4)(n-1)组对 nS,nP There are two factors that affect the size of the shielding constant: (1) It is related to the number of electrons that generate the shielding effect and the size and shape of the atomic orbit on which it is located. (2) It is related to the distance and movement of the shielded electrons from the nucleus. The shielding constant can be approximated by the rule proposed by Slater. 6. The method of estimating the shielding constant 6 of Slater in “Inorganic Chemistry” textbook is as follows: (1) The electrons are grouped according to the order of 1S, 2S, 2P, 3S, 3P, 3d; 4S, 4P, 4d, 4f, 5S, 5P, 5d, 5f, etc. (2) The outer electrons have no shielding effect on the inner electrons, σ = 0 in each group. 0.35 (but 1S, σ = 0.30) (4) (n-1) For pair nS, nP