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).


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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 $+6.$






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