Why does potassium bifluoride exist whereas bichloride does not?

Why does the reaction \eqref{rxn:R1Q} take place, but the reaction \eqref{rxn:R2Q} does not?


$$
\begin{align}
\ce{HF(aq) + KF(aq) &-> KHF2(aq)}\label{rxn:R1Q}\tag{R1} \\
\ce{HCl(aq) + KCl(aq) &-> KHCl2}\label{rxn:R2Q}\tag{R2}
\end{align}
$$


I found that $\ce{[HF2]-}$ forms due to hydrogen bonding among $\ce{HF}$ molecules. How *exactly* is the hydrogen bonding responsible for the formation of bifluoride ions?




The bichloride ion $\ce{HCl2^-}$ does exist, but it requires other environment than usual solutions in water. E.g. solution of $\ce{[N(CH3)4]Cl}$ salt and *undissociated* $\ce{HCl(solv)}$ in nitrobenzene solvent, as mentioned in the ACS link of the Oscar's answer. It is the scenario that makes the partial hydrogen charge in $\ce{H^{(+)}-Cl{(-)}}$ quite favourable for chloride anions.




---


For water solutions scenario, the key is the presence of undissociated $\ce{HF}$ and the ability of $\ce{F-}$ to make hydrogen bond with $\ce{HF}$. Neither is available for $\ce{HCl}$ and $\ce{Cl-}$. There are just very minor traces of undissociated $\ce{HCl}$ in water and $\ce{Cl-}$ would not have bound to it even if it had been there.


There are no $\ce{HCl}$ or $\ce{KCl}$ in water as molecules. There are hydrated ions $\ce{H+(aq)},$ $\ce{K+(aq)}$ and $\ce{Cl-(aq)}.$ Hydrated molecules $\ce{HF(aq)}$ are there in hydrofluoric acid or in acidified fluoride solutions, reacting:


$$\ce{F-(aq) + HF(aq) <=> HF2^-(aq)}\tag{R1}$$


Hydrofluoric acid is weak mainly because of forming the stable ionic pair, so ion $\ce{H3O+(aq)}$ is mostly not free:


$$\ce{HF(aq) + H2O(l) <=> H3\overset{+}{O}\bond{...}F-(aq) <=> H3O+(aq) + F-(aq)}\tag{R2}$$


The structure of $\ce{HF2-}$ is $\ce{[F\bond{...}H-F]-}.$ There is also $\ce{[H-F\bond{...}H]+}$ in concentrated [hydrofluoric acid](https://en.wikipedia.org/wiki/Hydrofluoric_acid) or liquid  [hydrogen fluoride](https://en.wikipedia.org/wiki/Hydrogen_fluoride):


$$\ce{3 HF <=> H2F+ + HF2^-}\tag{R3}$$


Liquid hydrogen fluoride has such a high boiling point due $\ce{HF}$ chain linked by hydrogen bonds:


$$\ce{[\bond{...}H-F\bond{...}H-F\bond{...}H-F\bond{...}H-F\bond{...}]}$$


 13

The bichloride ion, including isolated salts, has in fact been known since 1954 [[1]](https://doi.org/10.1021/ja01644a066); but it appears to require relatively bulky counterions whereas bifluoride has a broader range of stable salts.


To form a complex ion of the type $\ce{HX2^-}$ the extra bond strength derived from a three-center four-electron bond (in the complex ion) must exceed the more favorable electrostatic attraction of the counterion to a more compact monatomic halide ion or solvent dipoke (which favors dissociation into $\ce{X^- + HX}$), a situation similar to that for [homoatomic trihalide ions $\ce{X3^-}$](https://chemistry.stackexchange.com/a/122036/17175). But unlike homoatomic trihalide ions, $\ce{HX2^-}$ has the central atom fixed at hydrogen which favors strong covalent bonding with lighter, more compact atoms. Thereby $\ce{HF\_2^-}$ is more favored than its heavier halide ion counterparts; whereas (at least in water solution and isolated compounds) triiodide ion is favored over lighter homoatomic halogen counterparts.


**Reference**


1.
Harry F. Herbrandson, Richard T. Dickerson Jr., and Julius Weinstein (11954). "The Bichloride Ion". *J. Am. Chem. Soc.* **76**(15), 4046.
<https://doi.org/10.1021/ja01644a066>


 8