What is the distinction between #"oxides"#, #"peroxides"#, and #"superoxides"#?

2 Answers
Mar 23, 2017

Answer:

Oxygen generally assumes #-II# #"oxidation state"#.

Explanation:

While oxidation state is a formalism, oxygen generally is conceived to accept 2 electrons in its compounds to give a #-II# oxidation state. This is certainly true of water, #OH_2#, i.e. #O^(2-) + 2xxH^+#.

And in metal oxides: #Na_2O, MgO, Fe_2O_3#, the metal oxidation states are #Na(+I), "i.e. " Na^+#, #Mg(+II), "i.e. " Mg^(2+)#, and #Fe(+III), "i.e. " Fe^(3+)# respectively.

Mar 26, 2017

Answer:

I have been asked to expand this answer, and explain the difference between #"oxides"#, #"peroxides"#, and #"superoxides"#, and my attempt follows.

Explanation:

Now hydrogen peroxide is #HO-OH#, and it clearly contains an #O-O# bond.  Because our definition of oxidation number is #"the charge left on the central atom when all the bonding pairs"# #"of electrons are BROKEN, with the charge assigned"# #"to the most electronegative atom,"#

this exercise results in the sharing of the electrons (because the oxygen atoms have equal electronegativity):

i.e. #HO-OHrarr2xxdotOH#

(#dotOH# is the so-called hydroperoxyl radical, a NEUTRAL radical species.) The oxygens in hydrogen peroxide thus have a formal oxidation state of #-I#. Why? Because we consider the #OH# species, where clearly hydrogen is LESS electronegative than oxygen.

And when we write peroxide salts, i.e. sodium peroxide, we use a formula of #Na_2O_2#; i.e. a salt of #""^(-)O-O^-#.

And for #"superoxides"#, to continue the formalism, we write #O_2^-#, that is a mixed oxidation state dioxide of #O^(0)# and #O^(-)#, to give an average oxidation number of #O^(-1/2)#.  In very old literature, this goes by the label #"hyperoxide"#