For the reaction,NH_4NO_2 rightarrow N_2 + 2H | vedclass

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(C) In the reactant $NH_4NO_2$,the nitrogen atoms are present in two different oxidation states: $1$. In the ammonium ion $(NH_4^+)$,the oxidation sta

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Oxidised as well as reduced

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(C) In the reactant $NH_4NO_2$,the nitrogen atoms are present in two different oxidation states: $1$. In the ammonium ion $(NH_4^+)$,the oxidation state of $N$ is $x + 4(+1) = +1$,so $x = -3$. $2$. In the nitrite ion $(NO_2^-)$,the oxidation state of $N$ is $x + 2(-2) = -1$,so $x = +3$. In the product $N_2$,the oxidation state of $N$ is $0$. Since the nitrogen atom in $NH_4^+$ $(-3)$ increases its oxidation state to $0$ in $N_2$,it undergoes oxidation. Since the nitrogen atom in $NO_2^-$ $(+3)$ decreases its oxidation state to $0$ in $N_2$,it undergoes reduction. Therefore,nitrogen is both oxidised and reduced in this reaction.

For the reaction,$NH_4NO_2 \rightarrow N_2 + 2H_2O$,which of the following phenomena is true regarding nitrogen?

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