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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.

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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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    $\begingroup$ 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? $\endgroup$
    – bon
    Feb 1, 2016 at 10:17
  • $\begingroup$ 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. $\endgroup$ Feb 1, 2016 at 19:06
  • $\begingroup$ 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. $\endgroup$
    – bon
    Feb 1, 2016 at 19:29
  • $\begingroup$ 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. $\endgroup$ Feb 1, 2016 at 20:01
  • $\begingroup$ 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. $\endgroup$ Jan 15, 2017 at 17:24
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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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