Explain more about your system. Does your system contains cation...cation or anion...anion charged species, then you can expect electrostatic energy in positive value.

Sir, I am working on non covalent interactions, especially halogen bonding of radicals. Typically,

A-X+R=A-X...R

A is any electron withdrawing group, X is halogen (Iodine here) and R is free radical (OH radical here).

Tüm enerji çeşitleri, matematiksel ifade ile hesaplansa da ve bulunan bu çözümlerin sonuçlarının negatif veya pozitif olması, sistemin ortama vermesi (negatif) ortamdan alması pozitif olması, enerjinin yönünü işaret eder. Enerji değişimleri evrensel olarak mutlak değerdir.

I found this in google translate, trying to understand what it says Sir, Ertan Çiftci .

Although all types of energy are calculated with mathematical expression and the results of these solutions are negative or positive, the system gives the environment from the (negative) environment is positive, indicates the direction of energy. Energy changes are universally absolute values.

In the LMO-EDA approach the total interaction energy is decomposed into different contributions (See Su and Li, J. Chem. Phys. 131, 014102 (2009); doi: 10.1063/1.3159673). The electrostatic energy Eele is only one contribution to the total energy and it may be positive or negative depending on the arrangement of the interacting particles. There are further contributions like the exchange energy, Eex, the relaxation energy, Erel, (which is frequently called polarization or induction energy) and the dispersion energy, Edisp, which are always negative. Furthermore there is the repulsion energy which - as its name says - is always repulsive. So there is no reason that the sign of the total energy and the electrostatic energy have to be related to each other.

However, positive electrostatic energies can only be found if e.g. the dipole moments of two systems point in the same direction. Such orientations are energetically unfavourable and thus only rarely investigated.

Ok sir, I will check the dipole moment and get back to you. Thank you.

Reinhold F Fink

Sir, I can understand that total interaction energy depends upon all the factors that you have mentioned. I have only found two instances where this Eels is found +ve, where in all the other cases it is around -0.44 to 14.0 kcal/mol. So I take that the dipole moment might be pointing towards the same direction. So it can be said that the dipole moments of the individual monomers.

But, I have one monomer as Cl radical. How can it have any dipole moment?

Did you see the e.g. in my answer?

Please take a look at some references on electrostatic interactions (e.g. the book A. Stone, The Theory of Intermolecular Forces, Oxford University Press, 2012) for more details about electrostatic forces. Please check for the keyword "charge penetration" as well.

I will not answer in depth as I am not particularly familiar with the LMO-EDA method, but from what I know of chemistry consider that radicals do not react or interact driven by electrostatics, but rather by the exchange-pairing of the single electron. Approaching a radical to an anion might be electrostatically repulsive and still favored overall.