Structure of the bifluoride anion [duplicate]








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[Is potassium bifluoride an ionic or a covalent compound?](/questions/40729/is-potassium-bifluoride-an-ionic-or-a-covalent-compound)

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I came across this structure of $\ce{HF2-}$ and realized that the bond shown between H and F is a hydrogen bond. 


$$\Large\ce{[F-H-F]-}$$


HF hydrogen bond are the strongest hydrogen bonds we know, but still they are not real bonds.
Writing the molecular formula as $\ce{HF2-}$ will give anyone an idea that the bonds between H and F are only covalent bonds.


Why do we write the molecular formula of $\ce{HF2-}$ as this and can you throw more light on its structure?
Have I correctly understood the structure?




Quoting from a comment of your’s:



> 
> Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present.
> 
> 
> 


This sentence is correct for simple two-electron-two-centre bonds; i.e, if each of the two bonds going to hydrogen were to represent a 2e2c bond, hydrogen would have a double negative charge.


However, the $\ce{HF2-}$ anion does not contain two 2e2c bonds. Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$. This can be understood by the following graphic:


[![Scheme of the HF2- 4-electron-3-centre bond](https://i.stack.imgur.com/PMRLn.gif)](https://i.stack.imgur.com/PMRLn.gif)  

**Figure 1:** Scheme of the 4-electron-3-centre bond in $\ce{HF2-}$. Image taken from Professor Klüfers’ [web scriptum of his general and inorganic chemistry course](http://www.cup.lmu.de/ac/kluefers/homepage/L_ac1.html).


The lowest molecular orbital, $\sigma\_1$, is a fully bonding $\ce{F-H-F}$ orbital with two electrons. The second, $\sigma\_2$ is nonbonding; no (accessable) hydrogen orbital exists that could complete it. The final one, $\sigma\_3$, is fully antibonding. This is similar to the orbital scheme of extended linear π systems, where the number of nodal planes increases by 1 with every higher orbital. And it should be understood as such: the nodal plane of $\sigma\_2$ traverses the hydrogen atom. Thus, while formally nonbonding, $\sigma\_2$ is actually weakly bonding for both $\ce{H-F}$ bonds. If we calculate the overall bond order, we still count the electrons in $\sigma\_2$ as nonbonding, giving a bond order of $0.5$ for each $\ce{H-F}$ bond.


The molecule can be understood in 2e2c Lewis terminology by using the following two resonance structures:


$$\ce{F-H\bond{...}F- <-> ^-F\bond{...}H-F}$$


‘Multiple bonds’ going away from a single hydrogen is not restricted to the 4-electron-3-centre bond known in $\ce{HF2-}$; they are also known and well-studied in boranes, although they are typically 2-electron-3-centre bonds in those electron-deficient compounds.


 5

There is no such distinction as *"real"* vs *"not-so-real"* bond. Hydrogen bonds are as real as any other bonds (although weaker than many), and this particular one, as you already know, is the strongest of all hydrogen bonds.


As to why we write the formula this way, the answer is simple: because **it is this way**. We fool around with some fluoride compounds, and then we find out this one; we analyze it, and it seems to contain K, H, and F, and the composition turns out to be $\ce{KHF2}$. Before we know anything about hydrogen bonding, or chemical bonding whatsoever, we already have the reason to write the formula the way it is. Then X-ray crystallography comes along, so we can actually *look* at it, and see it is **really** this way; it really does contain those $\ce{HF2-}$ anions.


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