3 deleted 1 character in body edited Nov 29 '16 at 0:49 Mithoron 3,84788 gold badges2929 silver badges4747 bronze badges 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. 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 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 $$-1$$. 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 $$-\frac{1}{2}$$. 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 $$+2$$. 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. 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 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 $$-1$$. 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 $$-\frac{1}{2}$$. 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 $$+2$$. 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. 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 $$+1$$. 2 Added MathJax, removed clutter. edited Oct 15 '16 at 13:09 Jan 51.1k77 gold badges128128 silver badges273273 bronze badges 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. 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 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 $$-1$$. 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 $$-\frac{1}{2}$$. 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 $$+2$$. 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. 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 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 $$-1$$. 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 $$-\frac{1}{2}$$. 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 $$+2$$. 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 $$+1$$. 1 answered Oct 15 '16 at 7:26 Vidyanshu Mishra 89111 gold badge1010 silver badges2929 bronze badges 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.