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One mole of $\mathrm{N}_2 \mathrm{H}_4$ loses 10 moles of electrons to form a new compound $Z$. Assuming that all the nitrogens appear in the new compound, what is the oxidation state of nitrogen in $Z$ ? (There is no change in the oxidation state of hydrogen.)
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Verified Answer
The correct answer is:
$+3$
$\mathrm{N}_2 \mathrm{H}_4-10 e^{-} \longrightarrow X^{10+}$ (with two $\mathrm{N}$ atoms)
Total oxidation number of two $\mathrm{N}$ atoms in $\mathrm{N}_2 \mathrm{H}_4$,
$\begin{aligned} \Rightarrow \quad 2 x+4 & =0 \\ 2 x & =-4\end{aligned}$
$\mathrm{N}_2 \mathrm{H}_4$ loses 10 electrons, so total oxidation number of two $\mathrm{N}$-atoms increases by 10 , i.e., the total oxidation number of two $\mathrm{N}$-atoms in
$Y=-4+10=+6$
$\therefore$ Oxidation number of each $\mathrm{N}$ atom in
$X^{10+}=+3$
Total oxidation number of two $\mathrm{N}$ atoms in $\mathrm{N}_2 \mathrm{H}_4$,
$\begin{aligned} \Rightarrow \quad 2 x+4 & =0 \\ 2 x & =-4\end{aligned}$
$\mathrm{N}_2 \mathrm{H}_4$ loses 10 electrons, so total oxidation number of two $\mathrm{N}$-atoms increases by 10 , i.e., the total oxidation number of two $\mathrm{N}$-atoms in
$Y=-4+10=+6$
$\therefore$ Oxidation number of each $\mathrm{N}$ atom in
$X^{10+}=+3$
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