Question

$1$ mol of $N_2{H}_4$ loses $10$ mol of electrons to form a new compound $Y$. Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in $Y$ (no change in the oxidation state of $H$)?

Answer 1

$\underset{\uparrow }{N_2{H}_4}⟶\underset{\uparrow }{N_2}\left(X\right)+10e^{-}$
O.N of two $N$ atoms of $N_2{H}_4=-4$
O.N of two $N$ atoms in oxidised species $=-4+10=+6$ as 10 electrons are lost.
Thus, oxidation state of $N$ in new compound $=+3$.