mult value calculation for the DFT-NEB calculation


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Hi Niri,

Many thanks for your response. I didn’t understand how the number of unpaired electron becomes one in this case. Electronic structure of C and N are 1s22s22px12py1 and 1s22s22px12py12pz1. So in the C atom there are 2 unpaired electrons in the 2px and 2py orbitals whereas in the N atoms there are 3 unpaired electrons in the 2px, 2py and 2pz orbital. Therefore in my sense, total unpaired electron is 2+3 = 5.

Is it like that there are total 1 (in 2px orbital of the C) + 1 (in 2px orbital of the N) = 2 electron in the 2px orbital of the C and N system and so they are paired?

Thank you.
Kind regards,
Sanjib