Question

# A mole of ${\mathrm{N}}_{2}{\mathrm{H}}_{4}$ loses 10 mol 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?

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Solution

## Step 1: Oxidation state:"The oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element."Step 2: Calculation of oxidation state:The oxidation state of $\mathrm{N}$ in Hydrazine $\left({\mathrm{N}}_{2}{\mathrm{H}}_{4}\right)$is shown below: $2\mathrm{x}+4=0\phantom{\rule{0ex}{0ex}}2\mathrm{x}=-4\phantom{\rule{0ex}{0ex}}\mathrm{x}=-2$.If one mole of hydrazine contains two moles of nitrogen and loses 10 moles of electrons, one nitrogen atom loses five electrons. Hence, its oxidation number will increase by $5$. Therefore, the oxidation state of $\mathrm{N}$ in compound A will be: $\mathrm{Oxidation}\mathrm{state}\mathrm{of}\mathrm{N}=\left(-2\right)+\left(5\right)=+3$Thus, the oxidation state of Nitrogen in compound A $+3$.

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