Question

One mole of ${N_2}{H_4}$ loses $10moles$ 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 compound $Y$ ?
[There is no change in the oxidation state of hydrogen.]
A. $- 1$
B. $- 3$
C. $+ 3$
D. $+ 5$

Answer 1

In order to solve this question, first of all we must have to determine the balanced chemical equation for the reaction that takes place. The oxidation number of the compound on the reactant side must be equal to the oxidation number of the compounds on the product side.

Complete step by step answer:
As per the question, the balanced chemical equation for the above reaction can be written as:
${N_2}{H_4} \to Y + 10{e^ - }$
As it has been already mentioned that there is no change in the oxidation number of hydrogen, we can deduce that the oxidation number of hydrogen will remain $+ 1$ throughout the reaction.
Let the oxidation number of nitrogen atoms in the nitrogen containing compound on the product side be$x$ . Thus, based on the reaction, we can write the mathematical equation as:
$2x + 4 \times ( + 1) + 10 \times ( - 1) = 0$
The above equation is determined by rewriting the chemical equation as:
${N_2}{H_4} - 10{e^ - } \to Y$
Now, solving the above mathematical equation, we have:
$2x - 6 = 0$
Thus, $x = + 3$
Hence, the oxidation number of the nitrogen atom in the compound $Y$ will be equal to $+ 3$ .

The correct option is C. $+ 3$ .

Note:
The oxidation number of an atom or an element is defined as the number of electrons lost or gained by the atom/ element during the course of the complete reaction. The oxidation number of a freely occurring element is considered zero such as the noble gases. The oxidation number of a compound is equal to the net charge present on it.