N_(2)H_(4) loses 10 mol e^(-) and form new compound Y number of N does not change so what is the oxidation number of N in Y atom ?

N_(2)H_(4) loses 10 mol e^(-) and form new compound Y number of N does

1.	N_(2)H_(4) loses 10 mol e^(-) and form new compound Y number of N does not change so what is the oxidation number of N in Y atom ?
Answer» -1 -3 `+3` `+5` Solution :Oxidation NUMBER of N in `N_(2)H_(4)` is -4. TOTAL oxidation number is -4. `Y=-4+10=+6` `therefore` Oxidation number of N in Y is = `(+6)/2=+3`