Balance the following equation by oxidation number method : `I_(2)+HNO_(3)toHIO_(3)+NO_(2)+H_(2)O`
Calculate the oxidation number of CI in `HCIO_(4)`

Explanation

Second part : Let the oxidation number of P in `Na_(2)PO_(4)`be x
`thereforeunderset(("for Na"))(3xx(+1)+x)underset(("for O"))(+4xx(-2))=0thereforex=+5`
Therefore , the oxidation number of P in `PO_(4)`is +5
In this reaction `I_(2)` undergoes oxidation while `HNO_(3)`undergoes reduction .the oxidation number of 1 increases by 5 units (0 to 5) and that of N decreases by 3 units (+5 to +2) .For 2 atoms of I, total increases in oxidation number =2xx5 10 units.
Therefor balancing the increase and decrease and decrease in oxidation states requires 10 molecule of `HNO_(3)` for 3 molecule of `I_(2)` .Hence the balanced equation for the given reaction will be-
`3I_(2)+10HNO_(3)to6HIO_(3)+10NO+2H_(2)O`
Second part : Let the oxidation state of CI in `HCIO_(4)` be x t
`therefore underset(("for Na"))(1xx(+1)+x)underset(("for O"))(+4xx(-2))=0 therefore x=+7`
So, oxidation number of CI in `HCIO_(4)` is +7.