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)

Solution:

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

Oxidation state nitrogen in product 'X' = +6/2
                                                       = +3