# Structure of the bifluoride anion [duplicate]

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?

## marked as duplicate by Todd Minehardt, ringo, Klaus-Dieter Warzecha, bon, ronDec 10 '16 at 17:40

• How did you get the idea that those are hydrogen bonds? – Karl Dec 10 '16 at 7:38
• Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present. Looking at the structure, it was easy to say that the other bond must have been hydrogen bond. Hydrogen bonds don't require orbital overlap between the two concerned atoms, they are more of electrostatic interactions caused by the difference in electronegativity. – Arishta Dec 10 '16 at 9:09
• chemistry.stackexchange.com/questions/40729/… – Nilay Ghosh Dec 10 '16 at 13:34
• chemistry.stackexchange.com/questions/8549/… – Mithoron Dec 10 '16 at 16:10
• Hydrogen bonds have a significant covalent component. That's why they're so strong. – bon Dec 10 '16 at 17:34

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.

• I thought we say hydrogen bonds are not real bonds because they are not formed by orbital overlap like other covalent bonds, they are a result of electrostatic interaction between some highly electronegative atoms and hydrogen...so i was not very convinced with the idea that these electrostatic interactions could result in real bond formation... – Arishta Dec 10 '16 at 8:58
• We also observe hydrogen bonding in HF molecule, but we dont consider that hydrogen bonding in the actual composition of HF. This does not seem intuitive to me, maybe because its the first time i am actually seeing hydrogen bond being a part of molecular formula of a compound. – Arishta Dec 10 '16 at 9:01
• Hydrogen bond, as well as covalent bond or any other kind of bond, is never a part of molecular formula. Again, we write the formula this way because the compound is composed this way, even before we know anything about bonding. – Ivan Neretin Dec 10 '16 at 9:17

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:

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.

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.

• Re: "Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$" But, sir, in $\ce{ClF3}$, we can "excite" the p-electrons of chlorine atom into its vacant d-orbital. (I am using VBT) This makes three unpaired electrons in chlorine's orbitals, which can then form covalent bonds with the three fluorine atoms. I was taught that such an excitation in not possible in the fluorine atom. Then, how can you say that $\ce{HF2-}$ has a 4c-3e bond similar to $\ce{ClF3}$, when such an excitation is not possible in the former case? Thanks for your answer! – Gaurang Tandon Feb 4 '18 at 6:36
• @GaurangTandon Surely you can excite an electron into chlorine’s 3d orbital—if you blast a significant amount of energy (probably visual light range) onto it. However, the d orbitals have been shown to not take part in bonding at all because of how much higher their energy is (remember that even 4s comes before them according to the aufbau principle). There are enough answers on this site that debunk this myth. – Jan Feb 4 '18 at 16:35
• Understood. You meant the hypercoordinated valency, while I was confusing myself with the traditional VSEPR $sp^3d$ thing. Thanks! – Gaurang Tandon Feb 5 '18 at 2:05
• I have drawn a (not so neat) diagram in ChemSketch - is this a valid Lewis diagram for $\ce{HF2-}$? I have based it on $\ce{ClF3}$'s structure. Three lone pairs are in the $sp^2$ hybrid orbitals of $F$ atom, and the two $H$ atoms in its pure $p$-orbital. Is it correct? – Gaurang Tandon Feb 5 '18 at 2:18
• @GaurangTandon No, fluorine should be unhybridised. One s-type lone pair, two p-type lone pairs. ther than that, it’s okay. – Jan Feb 7 '18 at 1:17