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

Question

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?

Answer 1

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

Answer 2

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