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JEE Main
Exam
JEE Main
Subject
Chemistry
Chapter
Redox Reactions

JEE Main · Chemistry · Redox Reactions

One mole of N_2 H_4 loses 10 moles of electrons to form a new...

One mole of N2H4\mathrm{N}_2 \mathrm{H}_4 loses 10 moles of electrons to form a new compound ZZ. Assuming that all the nitrogens appear in the new compound, what is the oxidation state of nitrogen in ZZ ? (There is no change in the oxidation state of hydrogen.)

Options

  1. A−1-1
  2. B−3-3
  3. C+3+3
  4. D+5+5

Correct answer

C. +3+3

Step-by-step solution

N2H4−10e−⟶X10+\mathrm{N}_2 \mathrm{H}_4-10 e^{-} \longrightarrow X^{10+} (with two N\mathrm{N} atoms) Total oxidation number of two N\mathrm{N} atoms in N2H4\mathrm{N}_2 \mathrm{H}_4, ⇒2x+4=02x=−4\begin{aligned} \Rightarrow \quad 2 x+4 & =0 \\ 2 x & =-4\end{aligned}\mathrm{N}_2 \mathrm{H}_4loses10electrons,sototaloxidationnumberoftwoloses 10 electrons, so total oxidation number of two\mathrm{N}−atomsincreasesby10,i.e.,thetotaloxidationnumberoftwo-atoms increases by 10 , i.e., the total oxidation number of two\mathrm{N}−atomsin-atoms inY=-4+10=+6\thereforeOxidationnumberofeachOxidation number of each\mathrm{N}atominatom inX^{10+}=+3$
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