关于AdNDP分析键级的问题

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文献当中用AdNDP得到D5h-Cr2Li5的成键轨道，并通过图中数据得到Cr-Cr之间是5.5重键，请问老师：1：这个结论是否是根据（成键数目*占据数的加和-反键数目乘以占据数的加和）/2得到？也就是（2*1.99+1.99+2*1.86+1*1.93-1.01）/2=5.30。2：文献左列右下角提到反键轨道占据数为1.01，请问老师该如何寻找反键轨道啊？While for HOMO−2 to HOMO−5,they unambiguously represent 2 δ, 2 σ and 2 π bondingbetween Cr–Cr atoms, respectively. Adaptive natural density partitioning (AdNDP) analysis was also performed and the results were (Fig. S5†) similar: there are 5 distinct two-center two-electron (2c-2e) Cr–Cr bonds including 2 π, 1 σ and 2 δ with occupation numbers (ONs) of 1.99, 1.99 and 1.79, respectively, a 7-center 2-electron (7c-2e) Cr–Cr σ bond with ON = 1.99 and a 7-center 1-electron (7c-1e) σ anti-bond with ON = 1.01.In contrast, the additional two 5-center 2-electron (5c-2e) patterns with ON = 1.94 only contain five Li centers rather than Cr centers, though they look like two Cr–Cr π bonds.Consequently, there is a 5.5-fold bond in D5h-Cr2Li5。图片有点模糊，为了方便观看我把那段文字粘上了。

sobereva

1 应该是，自己手动算一下和文章对照便知

2 在Multiwfn的AdNDP功能里你可以指定对哪些原子搜索，指定成那两个Cr，并且把占据数阈值设得明显小于1，就能搜出占据数是1左右的轨道