Chemistry - Redox Reactions Question with Solution | TestHub

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