JEE Main 2021 — Redox Reactions Question with Solution

To determine the oxidation states of nitrogen in each compound, we consider the overall charge of the compound and the known oxidation states of other elements involved, typically oxygen.

NO (Nitric Oxide):

Oxygen generally has an oxidation state of $-2$.

Let the oxidation state of nitrogen in NO be $x$.

The equation: $x+(-2)=0$.

Solving for $x$, we find $x=+2$.

NO${}_2$ (Nitrogen Dioxide):

Oxygen has an oxidation state of $-2$.

Let the oxidation state of nitrogen be $x$.

The equation: $x+2(-2)=0$.

Solving for $x$, we find $x=+4$.

N${}_2$O (Dinitrogen Monoxide or Nitrous Oxide):

Oxygen has an oxidation state of $-2$.

Let each nitrogen atom have an oxidation state of $x$.

The equation: $2x+(-2)=0$.

Solving for $x$, we find $x=+1$.

NO${}_3^{-}$ (Nitrate Ion):

Oxygen has an oxidation state of $-2$.

The overall charge of the ion is $-1$.

Let the oxidation state of nitrogen be $x$.

The equation: $x+3(-2)=-1$.

Solving for $x$, we find $x=+5$.

Therefore, the oxidation states of nitrogen are:

NO: $+2$

NO${}_2$: $+4$

N${}_2$O: $+1$

NO${}_3^{-}$: $+5$

The order of oxidation states from highest to lowest is:

NO${}_3^{-}$ ($+5$) > NO${}_2$ ($+4$) > NO ($+2$) > N${}_2$O ($+1$)

Thus, the correct option is:

Option D

NO$${}_3^{-}$$ > NO${}_2$ > NO > N${}_2$O