Oxidation state of $ I $ in $ HI{{O}_{4}} $ is +7 as: $ 1+x+4(-2)=0 $ $ x=+7 $ Oxidation state of $ I $ in $ {{H}_{3}}I{{O}_{5}} $ is +7 as. $ 3+x+5(-2)=0 $ $ x=+7 $ Oxidation state of $ I $ in $ {{H}_{2}}I{{O}_{6}} $ is +7 as $ 5+x+6\,(-2)=0 $ $ x=+7 $