Question

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

Answer 1

Do you want the `"excited state or oxidation states..........??"`

Explanation:

Now we cannot really assess the excited state; it depends on the form of spectroscopy or form of excitation you use. Certainly we can assess the oxidation state of each species, which is the conceptual charge left on an atom of interest in a molecule when all the bonding pairs of electrons are BROKEN with the charge assigned to the MOST electronegative atom.....

If we gots `SF_4`, the fluorine is definitely more electronegative than sulfur, and we thus assign oxidation numbers of `stackrel(+IV)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 `SF_4`, we gots a neutral molecule, and `4xx-I+IV=0` AS REQUIRED. For `SF_6` we gots `stackrel(+VI)S`, and `stackrel(-I)F`. And for the interhalogen we gots `stackrel(+VII)I`, and `stackrel(-I)F`.