Q66 / 200 · NEET 2012 (Screening) · Chemistry

In which of the following compounds, nitrogen exhibits highest oxidation state?

  • A
    $\mathrm{N}_2 \mathrm{H}_4$
  • B
    $\mathrm{NH}_3$
  • Correct answer
    C
    $\mathrm{N}_3 \mathrm{H}$
  • D
    $\mathrm{NH}_2 \mathrm{OH}$

Solution

Let the oxidation state of nitrogen in the given compounds be \(x\). (a) \(\mathrm{N}_2 \mathrm{H}_4\) \(\begin{aligned} & 2(x)+(+1) 4=0 \\ & 2 x=-4 \end{aligned}\) So, \(x=-2\). (b) \(\mathrm{NH}_3\) \(x+(+1)^3=0\) So, \(x=-3\) (c) \(\mathrm{N}_3 \mathrm{H}\) \(\begin{aligned} & (x) 3+(+1)=0 \\ & 3 x=-1 \end{aligned}\) So, \(x=-1 / 3\) (d) \(\mathrm{NH}_2 \mathrm{OH}\) \(\begin{aligned} & x+(+1) 2+(-2)+(+1)=0 \\ & x+2-2+1=0 \\ & x+1=0 \end{aligned}\) So, \(x=-1\) Thus, oxidation state of nitrogen is highest in \(\mathrm{N}_3 \mathrm{H}\).

Hence, the correct answer is option: 'C'.

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