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

In which of the following compounds, nitrogen exhibits highest oxidation state?
  1. $\mathrm{N}_2 \mathrm{H}_4$
  2. $\mathrm{NH}_3$
  3. $\mathrm{N}_3 \mathrm{H}$
  4. $\mathrm{NH}_2 \mathrm{OH}$

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 \(\mathrm{N}_3 \mathrm{H}\).

Asked in: NEET 2012 (Screening)

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