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).