# What are the oxidation states exhibited in the SF_4 molecule?

Do you want the $\text{excited state or oxidation states..........??}$
If we gots $S {F}_{4}$, the fluorine is definitely more electronegative than sulfur, and we thus assign oxidation numbers of $\stackrel{+ I V}{S}$, and $\stackrel{- I}{F}$. As always the weighted sum of the oxidation numbers is equal to the charge on the ion or molecule.
For $S {F}_{4}$, we gots a neutral molecule, and $4 \times - I + I V = 0$ AS REQUIRED. For $S {F}_{6}$ we gots $\stackrel{+ V I}{S}$, and $\stackrel{- I}{F}$. And for the interhalogen we gots $\stackrel{+ V I I}{I}$, and $\stackrel{- I}{F}$.