A 2.5 mol sample of hydrazine, N2H4 loses 25 mole of electrons in being converted to a new compound X . Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound X ?

Question

A 2.5 mol sample of hydrazine, N2H4 loses 25 mole of electrons in being converted to a new compound X . Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound X ? (A)  -1  (B)  -2  (C)  +3  (D)  +4

Tardigrade Admin · Accepted Answer

Oxidation number of N in N2H4,-2 Number of electrons lost by each N atom =(25/2× 2.5)=5 Then, x-(-2)=5 or x=+3

Q. A 2.5 mol sample of hydrazine, $N_{2} H_{4}$ loses 25 mole of electrons in being converted to a new compound $X$ . Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound $X$ ?

$- 1$

$- 2$

$+ 3$

$+ 4$

Oxidation number of N in $N_{2} H_{4}, - 2$ Number of electrons lost by each N atom $= \frac{25}{2 \times 2.5} = 5$ Then, $x - (- 2) = 5$ or $x = + 3$

Q. A 2.5 mol sample of hydrazine, N2​H4​ loses 25 mole of electrons in being converted to a new compound X . Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound X ?

−1

−2

+3

+4