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Fluoroantimonic acid is the strongest acid because its conjugate base $\ce{SbF6-}$ is exceptionally stable. $\ce{SbF6-}$ is in perfect octahedral symmetry wherein the negative charge is equally shared between the 6 fluorine atoms. Fluorine is the most electronegative element which makes it difficult to pull out its lone pairs (protonation), and the high oxidation state of antimony further reduces the polarizability of fluorine atoms.

What’s I don’t understand is that elements in the same column can form similar structures ($\ce{PF6-}$, $\ce{AsF6-}$, and $\ce{BiF6-}$) except for nitrogen, but what makes $\ce{SbF6-}$ so unique? $\ce{SbF6-}$ is neither the lightest nor the heaviest. The d block contraction and inert pair effect makes the high oxidation states of arsenic and bismuth unusually electronegative, but what is special about antimony? The only weaker conjugate base is $\ce{AuF6-}$ because $\ce{Au^{V}}$ is extremely electronegative, but $\ce{AuF6-}$ is unstable under acidic environment and quickly releases fluorine gas.

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    $\begingroup$ +1 //Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Jun 2, 2023 at 10:09
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    $\begingroup$ Not that there is $\ce{[H2F . nHF][SbF6. m SbF5]}$, with n,m >= 0 $\endgroup$
    – Poutnik
    Commented Jun 2, 2023 at 10:14
  • $\begingroup$ onlinelibrary.wiley.com/doi/10.1002/anie.201308023 $\endgroup$
    – Mithoron
    Commented Jun 2, 2023 at 11:53
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    $\begingroup$ Arnold Schwarzenegger is pretty strong, but is he the strongest man alive? Is he unique among all who ever lived or will live? No and no. Same thing with HSbF6. It is not unique. $\endgroup$
    – Ivan Neretin
    Commented Jun 2, 2023 at 13:48
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    $\begingroup$ Is "unique" the correct word here? You may call it "special" if you want. Just because it is a superacid doesn't make it "unique". $\endgroup$
    – Nilay Ghosh
    Commented Jun 2, 2023 at 17:27

1 Answer 1

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The answer is surprisingly simple, or perhaps elegant.

If we consider the radius ratios of nitrogen(V), phosphorous(V), arsenic(V), antimony(V), and bismuth(V) in VI-coordination with that of fluoride ion in I-coordination (for comparing gas phase acidic behaviour), we obtain the following:

E Radius Ratio Value Octahedral (with ionic bonds)
$\ce{N}$ $\dfrac{r_\ce{N^{+5}}}{r_\ce{F-}}$ 0.10 No
$\ce{P}$ $\dfrac{r_\ce{P^{+5}}}{r_\ce{F-}}$ 0.29 No
$\ce{As}$ $\dfrac{r_\ce{As^{+5}}}{r_\ce{F-}}$ 0.35 No
$\ce{Sb}$ $\dfrac{r_\ce{Sb^{+5}}}{r_\ce{F-}}$ 0.46 Yes
$\ce{Bi}$ $\dfrac{r_\ce{Bi^{+5}}}{r_\ce{F-}}$ 0.59 Yes

These values are in close agreement with Pauling's values as well and, accordingly, $\ce{SbF6-}$ is the most natural octahedral hexafluoride complex.$^\text{1}$ Bismuth also shows a natural tendency to form octahedral hexafluoride complex, but with a slightly loosely packed structure. Nonetheless, antimony and bismuth are expected to form more stable octahedral hexafluoride complexes $\ce{EF6-}$ and stronger corresponding acids $\ce{HEF6}$ than other group-5 elements. This is supported by:

  1. Fluoride Ion Affinities
  2. Conductivity Measurements

Fluoride Ion Affinities$^\text{2}$

The fluoride ion affinity of $\ce{SbF5(g)}$, $\pu{-506 \pm 63 kJ mol^{-1}}$, is higher than that of $\ce{AsF5(g)}$, $\pu{-421 \pm 22 kJ mol^{-1}}$, suggesting $\ce{HSbF6(g)}$ is more strongly acidic than $\ce{HAsF6(g)}$.

Conductivity Measurements$^\text{3}$

The order of acidity of group-5 pentafluorides, based on conductivity measurements is:

$$ \ce{SbF5>BiF5>AsF5>PF5} $$

Suggesting the following order of acidity.

$$ \ce{HSbF6>HBiF6>HAsF6>HPF6} $$

References

  1. Clifford, A. F., Beachell, H.C., and Jack W.M. (1967). The hydrogen fluoride solvent system—I A qualitative survey of acids. J. Inorg. Nucl. Chem., 5, 57. 10.1016/0022-1902(57)80081-5.
  2. Jenkins, H. D. B., Roobottom, H. K., and Passmore, J. (2003). Estimation of Enthalpy Data for Reactions Involving Gas Phase Ions Utilizing Lattice Potential Energies:  Fluoride Ion Affinities (FIA) and pF- Values of mSbF5(l) and mSbF5(g) (m = 1, 2, 3), AsF5(g), AsF5·SO2(c). Standard Enthalpies of Formation:  ΔfH°(SbmF5m+1-,g) (m = 1, 2, 3), ΔfH°(AsF6-,g), and ΔfH°(NF4+,g). Inorg. Chem., 42(9), 2886–2893. 10.1021/ic0206544.
  3. Gillespie, R. J., Ouchi, K., and Pez, G. P. (1969). Fluorosulfuric acid solvent system. VI. Solutions of phosphorus, arsenic, bismuth, and niobium pentafluorides and titanium tetrafluoride. Inorg. Chem., 8(1), 63–65. 10.1021/ic50071a015.
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  • $\begingroup$ @OscarLanzi the table is accurate. Even I had the same doubt. What happens, based on theoretical studies, is that the (Mulliken) charges on P is actually close to 2, increasing the radius ratio to accommodate the octahedral geometry, and as we go down the group the (Mulliken) charges and atomic radii increase, reducing the inter-fluoride repulsions and stabilizing the anions on the whole. The radius ratio is to show the most 'natural' geometries for the elements in (formal) V oxidation states. $\endgroup$
    – ananta
    Commented Jun 13, 2023 at 13:20
  • $\begingroup$ In a way, the highly electronegative fluorine has to provide lesser electron density to the central atom as we move down the group to achieve an octahedral geometry. $\endgroup$
    – ananta
    Commented Jun 13, 2023 at 13:23

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