Question

In a reaction one mole of ${{N}_{2}}{{H}_{4}}$ loses ten moles of electrons to form a new compound Y. Assuming that all nitrogen appears in the new compound. What is the oxidation number of N in Y?

Answer 1

We need to know the number of electrons which is needed to make the bond, in order to find the oxidation state of the nitrogen in ${{N}_{2}}{{H}_{4}}$ or hydrazine. Then by determining whether the nitrogen gains or loses electrons we can determine the oxidation state of nitrogen in Y.

Complete step by step answer:
-Let’s begin with the concept of oxidation number. It can be defined in various terms such as addition of oxygen, in terms of hydrogen removal, addition of electronegative elements or in term of change of electrons in an oxidation process.
-In the context of the given question we need to consider oxidation number in terms of the electron transfer. As we know in chemical reactions it’s important to know whether an atom loses or gain an electron. Oxidation number helps us to keep a track of these electron transfers.
- The given question can be represented as follows
${{N}_{2}}{{H}_{4}}\to Y+10{{e}^{-}}$

- As we can see one mole of hydrazine loses ten moles of electron to form the new compound Y. It’s given that all nitrogen appears in the new compound, hence the oxidation state of hydrogen remains unchanged and the electrons are lost from nitrogen.
- One mole of ${{N}_{2}}{{H}_{4}}$ contains two moles of nitrogen. That is two moles of nitrogen loses ten moles of electron. Therefore, one mole of nitrogen will lose five moles of electron.
-Thus, the initial oxidation state of nitrogen in hydrazine is −2. Then it loses five moles of electron and the final oxidation state of nitrogen can be calculated as follows

The final oxidation state of N =−2 + 5
=+3
Since electrons are getting removed, the charge should be positive.
Therefore the oxidation state of N in Y is +3.

Note: Keep in mind that for an uncombined element the oxidation number will be zero. Generally, oxygen has an oxidation number of $+2$ and Hydrogen has an oxidation number of $+1$. For the neutral compounds, the sum of oxidation numbers of its constituent atoms will be zero and in the case of polyatomic ions, the sum of oxidation numbers of atoms will be equal to the net charge of the polyatomic ion.