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Hello,
I have carried out the second order perturbation analysis and a small section of the output is (showing the involved NBOs and Stabilization Energies):
21. BD(1) C 25-H 28 /121. LP*(7)Fe 30 9.06
21. BD(1) C 25-H 28 /122. LP*(8)Fe 30 11.23
Now NBO No. 121 and 122 are almost pure P orbitals as can be seen from below
121. (0.09383) LP*(7)Fe 30 s(1.96%)p49.61(97.4%)d 0.30(0.60%)f 0.00(0.00%)
122. (0.08905) LP*(8)Fe 30 s(1.16%)p84.81(98.31%)d 0.46(0.53%)f 0.00(0.00%)
My question is which of the two NBO(121 or 122 or both) should I consider? Do I need to add both values of stabilization energies (11.23+9.06) to make comparison with other systems?
Regards,
Anjali
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Dear Anjali,
Both of them should be considered. However, the two values cannot be added together, there will be an overcounting problem.
Best,
Tian
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Then how am I supposed to make comparison with other system since the other system also has a stabilization energy of 18.87 and 0.80 respectively for the same donor-acceptor pairs.
Added to this there are contributions involving s-orbital.
Here am giving an example of how the systems differ in their stabilization energy values.
System 1
donor-->p-orbital = 11.23
donor-->p-orbital = 9.06
donor-->s-orbital= 4.83
System2
donor-->p-orbital = 18.87
donor-->p-orbital = 0.80
donor-->s-orbital= 15.06
The donor orbital is the same in each case. It is a C-H sigma bond.
Last edited by Anjali Ganai (2023-06-15 11:43:02)
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Thanks a lot.
Regards,
Anjali
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