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

Question

Professor PC Thomas Classes · Answer

(c) : $\mathrm{N}_{2} \mathrm{H}_{4} \Rightarrow 2 x+4(+1)=0$\ $\Rightarrow \quad 2 x+4=0 \Rightarrow x=-2$\ $\mathrm{NH}_{3} \Rightarrow x+3(+1)=0 \Rightarrow x=-3$\ $\mathrm{N}_{3} \mathrm{H} \Rightarrow 3 x+1(+1)=0$\ $\Rightarrow 3 x+1=0 \Rightarrow x=-1 / 3$\ $\mathrm{NH}_{2} \mathrm{OH} \Rightarrow x+2+1(-2)+1=0$\ $\Rightarrow x+1=0 \Rightarrow x=-1$\ Thus, highest oxidation state is $-1 / 3$.