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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.

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    $\begingroup$ related: chemistry.stackexchange.com/questions/444/… $\endgroup$ – ManishEarth Apr 20 '13 at 19:49
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    $\begingroup$ The simple answer? Fluorine wants its last valence electron so badly that it's willing to break many of the rules normally observed by other elements. Hydrogen is much more easygoing. $\endgroup$ – KeithS Apr 23 '13 at 14:32
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    $\begingroup$ You don't want $\ce S$ in $-6 \text{ or } -4$ oxidation state, do you? $\endgroup$ – evil999man May 14 '14 at 3:43
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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

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    $\begingroup$ Hypervalency isn't an answer. The comparison here is H vs. F not S vs. O so hypervalency has nothing to do with it. Ideas related to hypervalency may or may not be relevant. $\endgroup$ – Jeff Oct 20 '13 at 21:30
  • $\begingroup$ @Jeff - where do you read that I compare O to S? I compare highly electronegative ligands (O, F, Cl) to H which is less electronegative. $\endgroup$ – Michiel Oct 21 '13 at 5:13
  • $\begingroup$ Anyway, since I wrote the answer I have come to believe that hypervalency is just a phenomenological model that can explain some of the symptoms' of breaking of the octet rule' but doesn't explain the reasons at all $\endgroup$ – Michiel Oct 21 '13 at 5:24
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    $\begingroup$ This answer is very vague and there are a lot better ones given already. If you yourself came to the conclusion, that there are better theories, you should expand this post. Some people just look for the accepted answer and then leave, they might not read the better ones. $\endgroup$ – Martin - マーチン Mar 26 '14 at 10:25
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    $\begingroup$ Sulphur does not have more than eight valence electrons; all sulphur compounds abide by the octet rule. $\endgroup$ – Jan Oct 2 '15 at 18:05
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I'll present a LCAO-MO argument. But first let's debunk a myth.

$\ce{SF6}$ has "hypervalent" sulfur and the 3d orbitals on sulfur participate in bonding

No. This is not true. I would close one eye if it is taught in high school but really, this cannot be further from the truth. Here's one reference: J. Am. Chem. Soc. 1986, 108, 3586; there are many more. For those who want to find out more, there is a wealth of information on the d-orbital and octet rule controversy; a simple search on the ACS website for "hypervalency" will yield a number of results. Unfortunately I could not find a review on the topic.

Anyway, so how does one describe the structure of $\ce{SF6}$? Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually pretty similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

SF6 MO

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 are localised on 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:

Resonance forms

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 $e_\mathrm{g}$ 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 already much better 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. "Aha! You've been lying to us all this while! The d orbitals are important!" No. They do help out, but how much do they actually help out? Not very much at all, considering the large energy gap between the S3d and F2p orbitals - again, see Martin's comments for more details.

As I've shown, you don't need d orbitals to rationalise 6-coordinate sulfur, and you don't need d orbitals to explain 5-coordinate phosphorus or the strength of the $\ce{P=O}$ bond either. The only reason why one should ever draw Lewis structures as such is for simplicity.

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    $\begingroup$ (Un?-)Surprisingly this sent me on a rampage to take my newly obtained natural resonance theory algorithm for a spin. It took way more attempts than I would like to admit. The algorithm determines 16 (!) resonance structures. That includes the hypervalent one (for some reason it's totally in love with it). Not all of the remaining ones are symmetry equivalent though. There are three structures where the ionic bonds are on the same axis, $\ce{F- \bond{->}[SF4]^2+ \bond{<-} F- }$, which have lesser contribution than the structures where the ionic bonds are on different axis. $\endgroup$ – Martin - マーチン Apr 22 '16 at 6:34
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    $\begingroup$ @Martin-マーチン Ah, I see... Thanks for the comment! The hypervalent one definitely contributes a bit. From the analysis above I would assume that the weightage of that resonance structure is correlated with the amount of d orbital involvement, which has to be nonzero (exactly how much, I guess that is for another calculation!) Yes, the "cis" and "trans" resonance forms are different and I didn't spot that while writing it, I'll edit it later today when I'm free. $\endgroup$ – orthocresol Apr 22 '16 at 8:03
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    $\begingroup$ Ah yes, cis and trans, that's the words I was looking for. Exactly how much the hypervalent structure contributes is up to the method used. On BP86+D3/def2-SVP it is about 5%, on MP2/def2-SVP it is about 8%, and on HF/def2-SVP it is about 16% (not very surprisingly). There is probably also a basis set dependency. The bonds are 22% S (37%s, 60%p, 3%d) and 78% F (19%s, 81%p, >0.1%d) on MP2. So you remain correct with stating "Not very much at all [...]" contribution from the d-orbitals. (An answer well written btw.) $\endgroup$ – Martin - マーチン Apr 22 '16 at 8:33
  • $\begingroup$ @Martin-マーチン Interesting calculations! I don't know how easy or difficult it would be, but it would be interesting to see what $^1J(\ce{S,F})$ coupling constants the various calculations predict; particularly how the computed coupling constants approach the experimental value as a function of hypercoordinated structure "content". $\endgroup$ – ron Apr 27 '16 at 16:04
  • $\begingroup$ @ron unfortunately I'm not the one to talk to when it comes to calculations about NMR spectra. I'm not even sure how accurate those coupling constants would be. Certainly a good exercise though. $\endgroup$ – Martin - マーチン Apr 27 '16 at 16:45
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(I strongly disagree with the comprehensiveness of the accepted answer, but here I go...)

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.

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    $\begingroup$ I completely agree with you about hypervalency being near-quackery. All we have are approximate theories about what goes on in these molecules. I doubt anyone has an actual physical idea what's happeing in there. $\endgroup$ – sayantankhan May 4 '13 at 13:41
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    $\begingroup$ @Bolt64 Since I wrote my answer I have read up on QTAIM and related theory and I now indeed believe that hypervalency is more a phenomenological model than anything else. $\endgroup$ – Michiel May 11 '13 at 20:44
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    $\begingroup$ BTW, what do you mean by hypervalency being borderline quackery? I don't really understand what you mean there? $\endgroup$ – Tomcat Aug 29 '13 at 8:06
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    $\begingroup$ This is the accurate answer. The accepted one is not even trying to provide an answer to the question and would probably be wrong if it were. $\endgroup$ – Jeff Oct 20 '13 at 21:32
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    $\begingroup$ I do very much agree with your point about hypervalency, I also strongly disagree with some of your other statements. A reaction that is exothermic does not mean that the reactant is unstable. Even if that same reaction would be exergonic usually an activation barrier has to be overcome, and that would make the molecule meta-stable and still existing. $\endgroup$ – Martin - マーチン Mar 26 '14 at 9:35
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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.

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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.

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$\ce{F}$ 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.

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    $\begingroup$ Hybridisation is a mere mathematical concept, hence it will not change any energy levels at all. $\endgroup$ – Martin - マーチン Mar 26 '14 at 9:42
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    $\begingroup$ There is no (or negligible) d-orbital contribution to the bonding in SF6. $\endgroup$ – orthocresol Apr 20 '16 at 15:12
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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.

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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.

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