Oxidation number is the charge left on the atom of interest, when all the bonding pairs of electrons are broken with the charge devolving to the most electronegative atom.
We have #ClO_2^-#; the individual oxidation numbers of the elements #(2xxO + Cl)# must sum (by definition) to the charge on the chlorite ion, which is #-1#. Oxygen gets the charge when we break the #Cl-O# bonds, so it is #-II# (oxygen is more electronegative than #Cl#). So #-2-2+Cl_(ON) = -1#. This gives #Cl# in chlorite a formal oxidation number of #+III#. What is the oxidation number of #Cl# in perchlorate anion, #ClO_4^-#? Please post your answer here.
Note that the oxidation number of #O# in is compounds is normally #-II#, as here. Elemental oxygen is #0#, BUT peroxide, #O_2^(2-)#, is #-I#. See here.
