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Laplacian bond order this method is an extension of the qtaim (quantum theory of atoms in molecules) concept of using the laplacian of the electron density $\nabla^2\rho$ to characterize bonding

Standard qtaim just uses the value of $\nabla^2\rho$ at bond critical points (minimum electron density along the bond, max along orthogonal directions), but these values are unintuitive and often don. You could also run quantum calculations for all possible pairs (or higher order clusters) of atoms or hit the literature to look up experimental bond strengths for greater accuracy. Hi @magic_number, after running with more recorded timestep, i think the main reason is because the molecule pass through zlo and have atom deleted, as result, the bond connecting that atom is lost But i know having a fixed z boundary is possible to run because that's what i found from several lammps papers. I need some cutoff radii to count bonds between different atoms in my system When a.cif file is opened in vesta, there are some default values of min and max bond lengths between two atomic speci.

When you are scanning two bond lengths in gaussian, you step once through the first bond scan, and complete stepping through the second bond scan If you are using gaussian to run your calculations, just output the calculated wave functions and use the multiwfn to run the topology analysis. In the image you posted in the question, the bond between 5 and 6 is long, so during the scan, the bond length would have to decrease which is why the step size is negative. A bond is present when the distance between two atoms is lower than the sum of the van der waals radii of the two atoms Van der waals radii are typically looked up from a preexisting table, not calculated We can find such tables in chemical reference books or online

For instance, here is a link that supplies the following van der waals radii.

From the qtaim analysis we got the the lead make a bond with one nitrogen atom at the nanostructure surface This is due to the interbasin path together with the bond critical point between both atoms The electron localization function is shown in fig According to its definition 1, elf took values among 0 and 1. How to choose sigma and epsilon value for carbon to calculate the bond coeff (lj units) in lammps Ask question asked 3 years, 5 months ago modified 3 years, 5 months ago