One mole of N2H4 loses 10 moles of electrons to form a new compound 'X'. Assuming that all the nitrogen appears in the new compound, the oxidation state of nitrogen in X is (there is no change in the oxidation number of hydrogen)

1 mole N2H4 loses 10mole of electrons .(assume no change in Ox.state of hydrogen)In N2H42x + 4(+1) = 0x = -2Ox.state nitrogen = -2Total Ox.state number of 'Nitrogen/mole of N2H4 = -410 mole of electrons are lost it means N2H4 undergoes oxidation net increase in ox.number of nitrogen = 10Total oxidation number of two nitrogen atoms before reaction = -4Net increase in Ox.number of nitrogen = 10Total oxidation number of two nitrogen atom after reaction = -4 + 10                                                                                   = +6Oxidation state nitrogen in product 'X' = +6/2                                                       = +3