I am trying to determine Gibbs free energy of solvation for a sodium salt of an alkylcarboxylate. B3LYP/6-31g. The minimized structure makes covalent bonds between sodium and carboxylate oxygens. How do I prevent this? Thanks
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2$\begingroup$ How did you come to the conclusion that the bonds are covalent? Even an inaccurate level of theory like B3LYP/6-31G doesn't produce these kinds of errors. $\endgroup$– Martin - マーチン ♦Apr 22 at 14:54
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2$\begingroup$ I think it's important to remember that DFT calculations (e.g., Gaussian) produce a set of atoms and XYZ coordinates. We like to see "bonds" in visualization tools, but they are inferred based on interatomic distances. $\endgroup$– Geoff HutchisonApr 22 at 17:00
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$\begingroup$ see my answer below about covalent $\endgroup$– Nonclassical CarbocationApr 22 at 18:52
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How do I prevent this?
You don't.
Ionic bonding is an idealization we do not expect to be realized in pure form between individual atoms, even in cases where the electronegativity differential would seem to be quite large. As described here, purely ionic bonding is more closely approached between multiatomic entities for which the molecular-orbital structure favors such bonding.
Moreover, the boundary between "ionic" and "covalent" bonding is really indistinct; what appears to be ionic bonding may in many cases have a covalent interpretation. The lattice structure of calcium carbide offers one such case.
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$\begingroup$ Reminds me about chemistry.stackexchange.com/q/40729/4945 However, I think OP is not interpreting their calculation correctly, and to me the question isn't really clear. $\endgroup$– Martin - マーチン ♦Apr 22 at 15:13
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$\begingroup$ how do I know it is covalent? when I look at fchk file, visualization draws two bonds between oxygens and sodium and the atoms no longer bear any charge. i have no problem modeling ammonium bromides, just getting stuck with sodium carboxylates. $\endgroup$ Apr 22 at 18:51
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1$\begingroup$ @NonclassicalCarbocation a visualisation tool will create an image based on interatomic distances. At least with the information usually included in a fsck file. You need to analyse the calculation further to judge the bonding situation. You must not equate a line in a program between two balls as a covalent bond. Study your system, one way of doing that is described in the question I've linked, but there are others. However, be aware that there will always be room for interpretation. As with your original goal, you should follow the literature to find the best approach. $\endgroup$– Martin - マーチン ♦Apr 22 at 19:26