What are the oxidation states exhibited in the SF_4 molecule?

1 Answer
anor277
Sep 13, 2017

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.

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