Question

One mole of ${\mathrm{N}}_2{\mathrm{H}}_4$ loses 10 moles of electrons to form a new compound A. Assuming that all the Nitrogen appears in the new compound. What is the oxidation state of Nitrogen in A?

Answer 1

Hydrazine:

• Hydrazine is a strong base and it is also called Nitrogen hydride.
• Hydrazine is dangerously unstable.
• The oxidation state of Hydrogen is $+1$.
• The oxidation state of $\mathrm{N}$ in Hydrazine $\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$ is $\mathrm{x}$

$$\begin{gathered} \mathrm{x}+\mathrm{x}+1+1+1+1=0 \\ 2\mathrm{x}+4=0 \\ 2\mathrm{x}=-4 \\ \mathrm{x}=\frac{-4}{2} \\ \mathrm{x}=-2 \end{gathered}$$

• The oxidation state of $\mathrm{N}$ in Hydrazine is $-2$.
• Here one mole of Hydrazine contains two moles of $\mathrm{N}$ and loses ten electrons.
• If one $\mathrm{N}$ atom loses five electrons, then its oxidation number will increase by 5.
• Therefore, the oxidation number of $\mathrm{N}$ in compound A will be $-2+5=3$.