Skip to main content
Chemistry

Return to Question

Prettified Q, fixed formatting title
Source Link
Safdar Faisal
Safdar Faisal
  • 7.1k
  • 4
  • 28
  • 54

How can the Is there a difference between oxidation number be different fromstate and valency?

I just looked up Peroxideperoxide on Google, and I saw that the oxidation state of Ooxygen in R-O-O-R$\ce{R-O-O-R}$ is -1$–1$, and not -2$–2$ (like H2O$\ce{H2O}$), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong? 

I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each Ooxygen forms 2$2$ bonds so, if I remove the other atoms, won't the imaginary charge left still be -2?$-2$? Then how, does -1$-1$ come into play?

How can the oxidation number be different from valency?

I just looked up Peroxide on Google, and I saw that the oxidation state of O in R-O-O-R is -1, and not -2(like H2O), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong? I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each O forms 2 bonds so, if I remove the other atoms, won't the imaginary charge left still be -2?? Then how, does -1 come into play?

Is there a difference between oxidation state and valency?

I just looked up peroxide on Google, and I saw that the oxidation state of oxygen in $\ce{R-O-O-R}$ is $–1$, and not $–2$ (like $\ce{H2O}$), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong? 

I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each oxygen forms $2$ bonds so, if I remove the other atoms, won't the imaginary charge left still be $-2$? Then how, does $-1$ come into play?

Source Link
isaac_samuel
isaac_samuel
  • 31
  • 4

How can the oxidation number be different from valency?

I just looked up Peroxide on Google, and I saw that the oxidation state of O in R-O-O-R is -1, and not -2(like H2O), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong? I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each O forms 2 bonds so, if I remove the other atoms, won't the imaginary charge left still be -2?? Then how, does -1 come into play?