Why does SF6 exist but SH6 does not? Chemical Bonding NCERT class 11| Neet and Jee Important Questions

Why does SF6 exist whereas SH6 does not?

First of all, the order of increasing in electronegativites of Hydrogen, Sulfur and Fluorine is:

H (2.2) < S (2.6) < F (4.0)

Therefore, in SF6 you can imagine, F is “oxidizing” S so S atom exhibits a positive oxidation state, which is +6 in this case. However, when H and S are present things reverse: S is “oxidizing” H so H atom exhibits a positive oxidation state of +1 (the only way), and S atom is in its lowest, -2, oxidation state. This corresponds to a widely known compound, H2S, not SH6.

Viewing SH6 as S (+6) and H (-1) makes things even worse. Each of the atoms is at its “critical” oxidation state for S (+6) is powerfully oxidative and H (-1) is powerfully reductive (clarified thanks to Lukasz Golon), so one could imagine a drastic intramolecular oxidation-reduction reaction to take place, with H (-1) reducing S (+6). This is another reason why SH6 does not exist.

Moreover, the existence of SF6 also depends on the sizes of the atoms. Sulfur atom in its highest oxidation state possesses a smallest radius. Since Fluorine atoms are also small, they can have up to 6 of them surrounding S. You will see this trend clearer: as we go down the VIIA group, we will not find any SX6 compounds (for X = Cl, Br, I) due to their relatively larger sizes compare to F. This is called the “steric effect”.