Question

Which of the following shows nitrogen in its increasing order of oxidation number?

• A. $N_{2}O<NO<N{O}_2<N{O}_3^{-}<N{H}_4^{+}$
• B. $N{H}_4^{+}$<$N_{2}O$
• C. $N{H}_4^{+}<N_{2}O<N{O}_2<N{O}_3^{-}<NO$
• D. None of these

Answer 1

The correct option is D None of these

Oxidation state of Nitrogen in $N_{2}O$

$\Rightarrow 2x-2=0$

$\Rightarrow x=-1$

$NO$

$\Rightarrow x-2=0$

$\Rightarrow x=+2$

$N{O}_2$

$\Rightarrow x-4=0$

$\Rightarrow x=+4$

$N{O}_3^{-}$

$\Rightarrow x-6=-1$

$\Rightarrow x=+5$

$N{H}_4^{+}$

$\Rightarrow x+4=+1$

$\Rightarrow x=-3$

So, increasing order of Nitrogen is:$N{H}_4^{+}<N_{2}O<NO<N{O}_2<N{O}_3^{-}$