One mole of N₂H₄ looses 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)

The oxidation state of N in hydrazine is -2.

1 mole of hydrazine contains 2 moles of N and loses 10 moles of electrons. Hence, 1 N atom will lose 5 electrons. Hence, its oxidation number will increase by 5.

Hence, the oxidation number of N in compound A will be −2+5=+3.