I'm given a question:

Oxidation number of $\ce{O}$ in $\ce{BaO2}$ is $x$ and in $\ce{OF2}$ is $y$; then the value of $x+y$ is what?

Now my main question is that if $\ce{F}$ has $-1$ valency in $\ce{OF2}$ then $\ce{O}$ must have valency of $+2$. But is that possible since $\ce{O}$ mainly has oxidation states of $-1$ & $-2$ or am I making any mistake?

  • 1
    $\begingroup$ That is possible, and O can take any value from -2 to +2. $\endgroup$ – DHMO Oct 15 '16 at 6:33
  • $\begingroup$ one more thing - oxidation number includes the +- signs with it? $\endgroup$ – Gourabo Oct 15 '16 at 6:34
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – DHMO Oct 15 '16 at 6:38
  • $\begingroup$ Does valency include the signs too? $\endgroup$ – Gourabo Oct 15 '16 at 6:39
  • 1
    $\begingroup$ No.${}{}{}{}{}{}$ $\endgroup$ – DHMO Oct 15 '16 at 6:43

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegative atom acquire a positive charge. Depending on this hypothesis oxygen have 5 oxidation states.

  1. In all the oxides,oxygen has an oxidation state of $-2$. Eg. $\ce{CO2,CO}$

  2. In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of $-1$. For example, consider $\ce{H2O2}$, here $\ce{H}$ is less electronegative so it will acquire a charge of $+1$ and to balance the $2$ positive charge of 2 H-atoms,each oxygen atom will acquire a charge of $-1$.

  3. In all superoxides ($\ce{KO2,CsO2,RbO2}$), oxygen has an oxidation state of $-\frac{1}{2}$,this is because $\ce{K,Cs,Rb}$, being elements of the first group and less electronegative than oxygen acquire a charge of $+1$, to balance it, each oxygen atom acquires a charge of $-\frac{1}{2}$.

  4. In one of the exceptions $\ce{OF2}$, the fluorine being more electronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 fluorine atoms oxygen acquires a charge of $+2$.

  5. As last, there is $\ce{O2F2}$, similarly here to balance the $-2$ charge on 2 $\ce{F}$-atoms each oxygen atom acquire a charge of $+1$.

  • $\begingroup$ You’re forgetting $\pm 0$ which is oxygen’s oxidation state in $\ce{O2}$. $\endgroup$ – Jan Oct 15 '16 at 13:09
  • $\begingroup$ Thanks @Jan to mention it but i thought it is too simple that everyone who have studied redox must be aware of it. $\endgroup$ – Vidyanshu Mishra Oct 15 '16 at 13:13
  • $\begingroup$ If you’re going to list all, then list all imho ;) $\endgroup$ – Jan Oct 15 '16 at 13:13
  • $\begingroup$ i knew about 0 but can you please tell me about the sign + and - $\endgroup$ – Vidyanshu Mishra Oct 15 '16 at 13:14
  • $\begingroup$ $\pm$ gives $\pm$. $\endgroup$ – Jan Oct 15 '16 at 13:14

Oxygen can have 6 different oxidation states. Which oxidation state oxygen is in depends on which element oxygen is bonded to and what ratio the two elements are at:

  • -2: This occurs in oxides e.g. $\ce{OsO4}$ and $\ce{RuO4}$
  • -1: This occurs in peroxides e.g. $\ce{H2O2}$
  • -0.5: This occurs in superoxides e.g. compounds that contain the $\ce{O^-2}$ ion such as $\ce{KO2}$
  • 0: This occurs in $\ce{O2}$
  • +1: This occurs in $\ce{O2F2}$ as $2 \cdot -1+2 \cdot 1=0$ and fluorine has a greater electronegativity than oxygen.
  • +2: This occurs in $\ce{OF2}$ as $+2+2 \cdot -1=0$

There can also be a $0$ oxidation state in compounds, not just elemental oxygen. Hypofluorous acid is known. This molecule with the structural formula $\ce{H - O - F}$ has (in the oxidation state formalism) oxygen gaining an electron from hydrogen but losing one to fluorine.


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