# Determine the oxidation number of sulfur in SF6

I've just started learning about redox chemistry and oxidation numbers. I can work out most of the oxidation number questions, but I don't know how to determine the oxidation number of sulfur in $\ce{SF6}$. This is not an ionic compound, and doesn't contain $\ce{H}$ or $\ce{O}$, so it seems as if I have no starting point.

If you could, please answer the question and provide a simple explanation.

In the case of $\ce{SF6}$, sulfur would have the oxidation number of +6 because the charge being applied to the fluorine is +6. Similarly fluorine would consequently have an oxidation number of -1 since $$6x + 6 = 0 \Rightarrow x = -1$$

(the right-hand side is equal to zero since that happens to be the net charge on the overall chemical formula).

• Although this is the correct answer, it doesn't really explain why fluorine is negative and sulphur positive. Why couldn't sulphur be -6 and fluorine +1?
– bon
Commented Feb 1, 2016 at 10:17
• This has to do with electronegativity, as the more electronegitive an atom is the more likely it is to gain electrons and in our case Flourine is a lot more electronegitive then Sulfur. Thus, Fluorine is more likely to gain electrons from Sulfur and retain a -1 charge. Commented Feb 1, 2016 at 19:06
• Oxidation number doesn't really have much to do with gaining or losing electrons. The sulphur doesn't actually have a +6 charge. Oxidation numbers are just a formalism to help with balancing reactions.
– bon
Commented Feb 1, 2016 at 19:29
• That maybe true, however electronegativity helps predict weather the oxidization number will be positive or negative irrespective of weather that ion's charge actually retains that magnitude. Commented Feb 1, 2016 at 20:01
• In case of covalent compound oxidation state is determined by cleavage of bonds. During cleavage of a bond between two different assaign +1 charge to less electronegative atom and -1 to more electronegative atom for single bond. Similarly +2 and -2 for double bond and so on. F is most electronegative atom hence it always show -1 oxidation state because it form only single bond. Commented Jan 15, 2017 at 17:24

Being the most electronegative element in periodic table (placed in right-upper corner) fluorine always has negative oxidation state in compounds besides $\ce{F2}$ and due to the fact the fluorine is a halogen (group $17$) the only negative oxidation state is $-1$.

So fluorine in $\ce{SF6}$ has an oxidation state $-1$. Then we can calculate o. s. for sulfur:
$x + 6×(-1) = 0$
$x = +6$

So sulfur in this compound has an o. s. of $+6.$