NH_(2)NH_(2) compound loses 10 mole e^(-) and form new compound x then calculate oxidation number of N_(2) in x compound. (Here oxidation number of H does not change.)

NH_(2)NH_(2) compound loses 10 mole e^(-) and form new compound x then

1.	NH_(2)NH_(2) compound loses 10 mole e^(-) and form new compound x then calculate oxidation number of N_(2) in x compound. (Here oxidation number of H does not change.)
Answer» `-3` `+3` `-1` `+5` Solution :`N_(2)H_(4)tox+10bare` `thereforeN_(2)H_(4)-10baretox` `therefore2x+4=0` `therefore2x=-4` Here, 10 electrons release during reaction. So, such electrons ADDED in this reaction. `2x=-4+10` `thereforex=+3`