# How do you calculate the correct charge on F after F2 dissociates?

When you ionize $$\ce{F2}$$ to $$\ce{F2^5+}$$, it quickly dissociates. Experiments show that the products of dissociation are $$\ce{F^2+}$$ and $$\ce{F^3+}$$.

I have tried to model the dissociation of $$\ce{F2^5+}$$ using Gaussian$$09$$ (DFT), specifying a 10 Å separation between products.

No matter which functional I use ($$\omega$$B97X, PBE, M06,etc.), I get a $$+2.6$$ charge on one $$\ce{F}$$ ion and $$+2.4$$ on the other. Why don't these functionals give $$\ce{F^2+}$$ and $$\ce{F^3+}$$ ions the corresponding formal charges as shown by experiment? And how do you get the correct charges?

• Comments are not for extended discussion; this conversation has been moved to chat. – jonsca May 8 '19 at 19:59
• @Feodoran It is based on the wavefunctions described in the program, not your wavefunctions. In the calculation packages, the wavefunctions are localised to some extent - they can't be distributed all over the universe. That's why the results showing non whole numbers are wrong. – XYZ May 8 '19 at 20:00
• One should be cautious about attempting to compare partial atomic charges with formal oxidation states (see nature.com/articles/s41563-018-0165-7). As a trivial example, I just ran an M06/def2-TZVP calculation of the HO2 molecule, which formally should assign charges of H+ and O2-. Based on Mulliken charges, you get a summed partial charge of -0.35 for the O atoms rather than -2 (of course, we then get +0.35 on H). Clearly, this is against the formal oxidation states one would assign. It's not specific to a Mulliken population analysis either. – Argon May 12 '19 at 22:55
• @Argon But how far apart are the atoms in your calculation? – XYZ May 12 '19 at 22:59
• Fully optimized! I turned this into an "answer" to provide more detail. – Argon May 12 '19 at 23:03

If you have two fragments that are very far apart, certain quantum chemistry methods can cause spurious effects from including both fragments in the same calculation. A well known example of this is dissociation curves for $$\ce{H2}$$ with Restricted Hartree Fock.

By using a density based method, we are less likely to localise the electrons on any one atoms in integer quantities. A wavefunction method such as Hartree Fock, which places integer electrons in well defined molecular orbitals is more likely to create this localisation. If you need higher accuracy, MP2 or coupled cluster may be of more use.

In fact I have achieved Mulliken charges of +2 and +3 on the two different atoms at 10 angstrom separation using Hartree Fock, with a cc-PVTZ basis set in Gaussian 09.

If you insist on using DFT you should use some chemical/physical intuition to partition this system. From experiment, i.e. time of flight spectrometry, we know that the atoms have integer charge when they strike a detector, and dispersion interactions over that distance are going to very small compared to the coulomb attract of the other atom as a point charge. So instead you can calculate the energies of the energies of the individual $$\ce{F^n+}$$ for $$n=0,1,2,3,4,5$$. The combination of atoms with total charge $$+5$$ and the lowest energy, considering the classical coulomb attraction between them, is probably the product of the dissociation.

Gaussian input used:

%chk=FF_HF.chk
%nproc=1
%mem=4GB
# UHF/CC-PVTZ NoSymm

F2 5 plus Ion

5 2
F      0.00000000    0.00000000   -5.00000000
F      0.00000000    0.00000000    5.00000000


• Thanks mate. UHF does show the correct charges. But in terms of energetics, UHF is horrendous, so I opt not to use it. – XYZ May 12 '19 at 22:06
• By the way, I think cc-pvtz is a bit much for a UHF calculation. – XYZ May 12 '19 at 22:07

One should be cautious about attempting to directly compare partial atomic charges with formal oxidation states (see this Nature Materials article for a nice discussion on the topic). As a trivial example, I just ran an M06/def2-TZVP calculation of the HO2 molecule (see Gaussian 16 input file below), which formally should assign charges of H+ and O2-. Based on Mulliken charges, you get a summed partial charge of -0.35 for the O atoms rather than -2 (of course, we then get +0.35 on H). Clearly, this is against the formal oxidation states one would assign. It's not specific to a Mulliken population analysis either. Hirshfeld charges provide nearly the same result.

In my experience, partial charges really should not be thought of as quantitative analogues of formal oxidation states. Based on your example, it is clear that the redox states of the two cations are different. If you wanted to gauge whether they could truly be assigned a 2+ and 3+ oxidation state, your best bet is to identify a molecule where conventional wisdom states the fluorine atoms are in the 2+ or 3+ oxidation state, run a population analysis on that molecule, and compare the partial charges with what you get.

#P M06/def2TZVP opt freq Integral=UltraFine

test

0 2
O  9.9430695560218645  9.9578012177096120 11.3238448457162892
O 10.0393101063805297 10.0291419761442047  9.9836293222874293
H 10.6898654975975802 10.5116551761461796 11.6656176919962338

• Thanks for your effort mate but I think you got my question wrong. What I'm talking about in my question is when the cations are far far apart from each other - in that case the cations should have the formal charges in the calculations. – XYZ May 12 '19 at 23:05
• Ah! Apologies about that. I now see where the prior answer was going with this. – Argon May 12 '19 at 23:07
• LOL, maybe I need to rephrase my question...But anyway, if you know the reason why DFT shows me the wrong charges, you are welcome to say something about that. – XYZ May 12 '19 at 23:11
• By the way, I saw you wrote scf=YQC before you edited the answer. Why did you use that? I've never seen anyone use that. – XYZ May 12 '19 at 23:13