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The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetiveelectronegative 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 electronegetiveelectronegative 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 electronegetiveelectronegative 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 flourinefluorine being more electronegetiveelectronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 flourinefluorine atoms oxygen acquires a charge of $+2$.

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

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetive 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 electronegetive 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 electronegetive 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 flourine being more electronegetive acquires a charge of $-1$ and to balance the $-2$ charge of 2 flourine atoms oxygen acquires a charge of $+2$.

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

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 $+1$.

2 Added MathJax, removed clutter.
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The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetive 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.CO2,CO

2.In all peroxides(oxygen-oxygen linkage),oxygen has an oxidation state of -1.forr example, consider H2O2,here h is less electronegetive 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 (KO2,CsO2,RbO2),oxygen have an oxidation state of -1/2,this is because K,Cs,Rb,being element of first group and less electronegetive than oxygen acquire a charge of +1,to balance it,each oxygen atom acquire a charge of -1/2.

4.In one of the exceptions OF2,the flourine being more electronegetive acquire a charge of -1 and to balance the -2 charge of 2 flourine atom oxygen acquire a charge of +2.

5.As last,there is O2F2,similarily here to balance the -2 charge on 2 F-atoms each oxygen atom acquire a charge of +1. That was i knew about,forgive me if i forgot to add something here.

  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 electronegetive 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 electronegetive 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 flourine being more electronegetive acquires a charge of $-1$ and to balance the $-2$ charge of 2 flourine atoms oxygen acquires a charge of $+2$.

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

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetive 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.CO2,CO

2.In all peroxides(oxygen-oxygen linkage),oxygen has an oxidation state of -1.forr example, consider H2O2,here h is less electronegetive 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 (KO2,CsO2,RbO2),oxygen have an oxidation state of -1/2,this is because K,Cs,Rb,being element of first group and less electronegetive than oxygen acquire a charge of +1,to balance it,each oxygen atom acquire a charge of -1/2.

4.In one of the exceptions OF2,the flourine being more electronegetive acquire a charge of -1 and to balance the -2 charge of 2 flourine atom oxygen acquire a charge of +2.

5.As last,there is O2F2,similarily here to balance the -2 charge on 2 F-atoms each oxygen atom acquire a charge of +1. That was i knew about,forgive me if i forgot to add something here.

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetive 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 electronegetive 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 electronegetive 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 flourine being more electronegetive acquires a charge of $-1$ and to balance the $-2$ charge of 2 flourine atoms oxygen acquires a charge of $+2$.

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

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The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegetive 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.CO2,CO

2.In all peroxides(oxygen-oxygen linkage),oxygen has an oxidation state of -1.forr example, consider H2O2,here h is less electronegetive 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 (KO2,CsO2,RbO2),oxygen have an oxidation state of -1/2,this is because K,Cs,Rb,being element of first group and less electronegetive than oxygen acquire a charge of +1,to balance it,each oxygen atom acquire a charge of -1/2.

4.In one of the exceptions OF2,the flourine being more electronegetive acquire a charge of -1 and to balance the -2 charge of 2 flourine atom oxygen acquire a charge of +2.

5.As last,there is O2F2,similarily here to balance the -2 charge on 2 F-atoms each oxygen atom acquire a charge of +1. That was i knew about,forgive me if i forgot to add something here.