Question

A mole of $N_2H_4$ loses $10$ mol of electrons to form a new compound $Y$. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $Y$? (There is no change in the oxidation number of hydrogen.)

• A. $-1$
• B. $-3$
• C. $+3$
• D. $+5$

Answer 1

Answer: (C)

$+3$

Answer 2

From the given information, we have $N_{2}H_{4} \to Y+10e^{-}$ The oxidation state of $N$ in $N_{2}H_{4} = -2$ {$\because 2x + 4 = 0$ or $x = -2$} The two nitrogen atoms in $Y$ (product) will balance the charge of $10\, e^{-}$. Hence the oxidation state of $N$ will increase by $+\, 5$ i.e. from $-2$ to $+\,3$.