Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

Question

Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem.

Answer 1

TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot.

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First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting:

The natural ionicity, $i_\ce{SF}$, of each $\ce{S-F}$ bond [in $\ce{SF6}$] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$, resulting in a sulfur center of charge $+2.69$. [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$. The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization.

If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$. So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram.

You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way:

The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation, although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak.

(An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910; see also refs. 12 and 13 in that article.)

Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$, then those "correct" resonance forms that we drew would involve $\ce{H-}$. I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable.

Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$, the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised.

In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$. Is there anything better? Well, there's the species $\ce{CH6^2+}$, which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$, and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258. While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$.

Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much. Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details.

Answer 2

The reason that they do not exist (or at least are not the most stable form) is because the decomposition reaction is exothermic.

\begin{aligned}\ce{ (1) && SF6 &-> SF4 + F2\\ (2) && SH6 &-> SH2 + 2 H2 }\end{aligned}

Reaction $(2)$ is much more exothermic than $(1)$, and it can be argued in two ways: either there is something very bad about the reactants ($\ce{SX6}$) or something very good about the products.

If you look at the bond strength of $\ce{F-F}$ vs. the bond strengths of $\ce{H-H}$ (well-known quantities) you will see that $\ce{H2}$ has a much stronger bond than $\ce{F-F}$.

There is every reason that $\ce{SH6}$ should exist if you are looking at how many "slots" sulfur has. If it has six slots for fluorine, then surely it could accommodate six hydrogens.

I don't really want to get into it, but I tend to think that "hypervalency" is borderline quackery, preoccupation with non-explanation. Either the atoms want to be there because their interaction is favourable electrostatically, or they would rather be in some other configuration.

Answer 3

This is caused by the molecule $\ce{SF6}$ being hypervalent, which means that the main element (in this case sulfur) has more then 8 valence electrons.

The reason why this can happen is extremely complex and, to be honest, I am not even sure whether it is a fully solved issue. I do know that the effect is related to the electronegativity of the ligands, which is very high for $\ce{O}$, $\ce{F}$ and $\ce{Cl}$ atoms and somewhat lower for $\ce{H}$-atoms. This might explain why it doesn't happen for hydrogen, but this is just speculation.

An incredibly extensive explanation of hypervalency (and related phenomena ) can be found in this post

Answer 4

As michielm said, it is because of electronegativity. The bond $\ce{S-F}$ is strongly polarized toward the fluorine (~more electrons are near fluorine), while the $\ce{S-H}$ bond is polarized toward the sulfur.

In $\ce{SH6}$ molecule there would be very high electron density around sulfur. This would increase both the electrostatic repulsion of electron and also the kinetic energy connected with occupation of higher atomic orbitals (this is hand-waving argument).

More rigorously it can be explained by a linear combination of atomic orbitals (LCAO), Hartree-Fock method (HF) or density functional theory (DFT). The increased electrostatic repulsion corresponds to Hartree term in HF or DFT. But even more important is the kinetic energy due to Pauli exclusion principle and requirement of orthogonalization of orbitals of each electron (resp. spin pairs). The orthogonalization requires a certain number of nodes in the wavefunction. Each such node increases the orbital energy (similar to states of particle in the box).

From the LCAO point of view this can be viewed as the contribution of higher atomic orbitals (d-orbitals, for example) to the bonding molecular orbitals.

In the case of $\ce{SF6}$ the contribution of sulfur atomic orbitals to the bonding states is lower (because most of the electrons are localized on fluorines). Because of that the energy of bonding molecular orbitals is not increased much by either Coulomb repulsion or by kinetic energy of higher atomic orbitals of sulfur.

A very rough argument can be given also using Thomas-Fermi model where the kinetic energy of electrons is proportional to $\rho^{5/3}$. However, Thomas-Fermi model is not appropriate for molecules.

Answer 5

Hypervalency is a bit of a touchy subject since there are strong opposing proponents for both the hybridization camp and the 3-center-4-electron bond camp.

While not directly related to your question's compounds, I shall point you to a fun book by Errol Lewars on the computational analysis of uncommon molecules: Modeling Marvels. While some of it isn't displayed on Google Books, a good amount of the relevant chapter is there. Indeed it even mentions the theoretical higher hydrides of the second-period elements.

Answer 6

While other answers describe the reasons hypercoordination is more difficult with hydrogen than with fluorine, the former has been achieved — and it may have an impact on superconducting materials applications.

Ordinary hydrogen sulfide ($\ce{H2S}$) decomposes under high pressure to produce $\ce{H3S}$, which at 155 GPa forms a body-centered cubic dpuble perovskite structure in which each sulfur atom is octahedrally coordinated with six hydrogen atoms, and the hydrogen atoms are linearly coordinated with two sulfur atoms1. Within a body-centered cubic cell the central sulfur atom is coordinated to hydrogen atoms at the face centers, and sulfur atoms at the corners are coordinated to hydrogen atoms along the face centers.

Figure from Ref. 1

$\ce{H3S}$ has an excess valence electron per formula unut, which goes into metallic bonding to make this material a metallic conductor. Mozzaffari et al.[2] explore the resistance characteristics of this cubic $\ce{H3S}$ phase, identifying a critical temperature of approximately 200 Kelvins: greater than that of any known cuprate superconductor and easily accessible with liquid nitrogen cooling.

References

1. Duan, Defang; Yu, Hongyu; Xie, Hui; Cui, Tian (2019). "Ab Initio Approach and Its Impact on Superconductivity". Journal of Superconductivity and Novel Magnetism. 32. https://doi.org/10.1007/s10948-018-4900-8.

2. Mozaffari, S., Sun, D., Minkov, V.S. et al. "Superconducting phase diagram of H3S under high magnetic fields". Nat Commun 10, 2522 (2019). https://doi.org/10.1038/s41467-019-10552-y

Answer 7

$\ce{F}$ is much more electronegative compared to $\ce{H}$. Thus it causes a contraction in the $d$ orbital of $\ce{S}$. The $d$ orbital of $\ce{S}$ becomes much more energetically stabilised. It can then participate in bonding with the filled $p$ orbital of $\ce{F}$. For $\ce{H}$ this is not possible. The cost of hybridisation is much higher in the latter case. Thus $\ce{SH4}$ does not exist but $\ce{SF4}$ does.

Answer 8

As sulfur belongs to group 16 in periodic table, its electronic configuration is ns²np⁴, it can show +2,+4 , +6 and -2 oxidation state. SF6 exists but SH6 doesn't because fluorine is the most electronegative element in the periodic table, its size is extremely small so it has greater polarising power, but if we see in SH6 then we get to know that the electronegativity of sulfur is much more than hydrogen, hydrogen doesn't have sufficient nuclear charge (effective nuclear charge) and its electronegativity is also not much, so SH6 cannot be formed.

Answer 9

Notice also how the Group 1 metals form peroxides much like hydrogen. Why does potassium form peroxides but sodium does not?

Alkalai Metals bond with Sulfur/Chalcogens in a 2:1 ratio and Hydrogen is behaving like an Alkalai Metal.

Lithium Sulflide = Li2S

Sodium Sulfide = Na2S

Potassium Sulfide = K2S

Rubidium Sulfide = Ru2S

Cesium Sulfide = Ce2S

Francium Sulfide = Fr2S

. . .

Hydrogen Oxide = H2O

Hydrogen Sulfide = H2S

Hydrogen Selenide = H2Se

Hydrogen Telluride = H2Te

If Hydrogen were like a Halogen, it would form H4S, and H6S. However Hydrogen doesn't obey the octet rule like halogens as it only needs to get rid of one electron, and yet it shares 2 electrons in a bond with Sulfur. That is exactly what the alkali metals do: they need to get rid of one electron and they share 2 with sulfur.

Na2S adopts the antifluorite structure, which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+. -https://en.wikipedia.org/wiki/Sodium_sulfide

Thus an atoms lacking or exceeding by 2 tends to form 2:1 ratios with atoms lacking or exceeding by 1.

Oxygen is really the problematic chalcogen: we are used to the Halogens behaving like hydrogen: See Dioxygen Diflouride and Hydrogen Peroxide. And then Hydrogen goes out and acts like an alkali metal.