Dear everyone,
I have some confusion about setting the quantum number to calculate the HOMO and LUMO of two-probe system. for example, for the molecule line C4BPH12, there are 40 outer-shell electrons, usually based on that two electrons occupy one orbit, so the 40 outer-shell electrons will occupy 20 orbits, and the quantum number of HOMO should be 19 (the quantum number of the first one lowest energy orbit is 0 ), for LUMO is that 20. however, it seems not like that through calculating the energy spectrum, as following,
0 -4.64
1 -4.36
2 -4.14
3 -3.8
4 -3.77
5 -3.76
6 -3.65
7 -3.06
8 -2.94
9 -2.7
10 -2.45
11 -2.19
12 -2.09
13 -2.07
14 -1.86
15 -1.67
16 -1.38
17 -1.31
18 -1.24
19 -0.71
20 -0.27
21 0.77
22 1.54
23 2.04
24 2.12
25 2.26
26 2.76
27 2.84
28 3.01
is it possible that two special electrons occupy two orbits, so, the quantum number of HOMO should be setted to 20(-0.27eV) .
or, the energy -0.27 could be considered as 0, so, the quantum number of HOMO and LUMO should be setted to 19 (-0.71eV) and 21(0.77eV).
how to explain it resonablely?
Another, for the molecule owning odd outer-shell electrons, ie. Si5Li cluster. there are total 23 electron, do they should occupy 12 orbits, and which orbit is occupied with only electron, HOMO ?
Thank you!