Oxidation states of oxygen

Question

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?

one more thing - oxidation number includes the +- signs with it? — crimsonKn1ght, Oct 15 '16 at 06:34
it is one of the exceptional oxidation state of Oxygen.another exception is O2F2 where oxidation state of o is +1 — Vidyanshu Mishra, Oct 15 '16 at 07:06

score 10 · Accepted Answer · edited Nov 29 '16 at 00:49

10

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.

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

edited Nov 29 '16 at 00:49

Mithoron

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answered Oct 15 '16 at 07:26

Vidyanshu Mishra

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You’re forgetting $\pm 0$ which is oxygen’s oxidation state in $\ce{O2}$. – Jan Oct 15 '16 at 13:09
Thanks @Jan to mention it but i thought it is too simple that everyone who have studied redox must be aware of it. – Vidyanshu Mishra Oct 15 '16 at 13:13
If you’re going to list all, then list all imho ;) – Jan Oct 15 '16 at 13:13
i knew about 0 but can you please tell me about the sign + and - – Vidyanshu Mishra Oct 15 '16 at 13:14
$\pm$ gives $\pm$. – Jan Oct 15 '16 at 13:14
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Add $+(1/2)$ for the compound often (a little oversimplified) rendered as $\ce{O_2^+PtF_6^-}$, formed by combining oxygen with platinum hexafluoride. This reaction heralded the similar oxidation of xenon which busted the notion of "inert gases". – Oscar Lanzi Aug 08 '18 at 09:38

score 3 · Answer 2 · answered Jan 28 '19 at 08:36

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.

score 2 · Answer 3 · edited Jan 28 '19 at 08:28

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 $+2+2 \cdot -1=0$

score 1 · Answer 4 · answered Jan 08 '23 at 04:11

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Oxygen can have a range of oxidation states including -2,-1,0,+1,+2

So, yes oxygen can have positive oxidation states when present with more electronegative elements than it in a compound.

answered Jan 08 '23 at 04:11

Based GamerX

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Oxidation states of oxygen

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