Answer:
in both these compounds, sulphur is
expected to show oxidation state of +6. This can happen is case two electrons
from filled (3s and 3p) orbitals are promoted to vacant 3d orbitals. Only atoms
of high electro negativity such as fluorine (4.0) can cause the promotion of
electrons while the same is not possible with hydrogen with comparatively less
electro negativity (2.1). Thus, \[S{{F}_{6}}\] can exist while \[S{{H}_{6}}\]
cannot.
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