In NH_4NO_2,the oxidation number of nitrogen | vedclass

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(C) The compound $NH_4NO_2$ dissociates into ions as $NH_4NO_2 \rightleftharpoons NH_4^+ + NO_2^-$. For the ammonium ion $(NH_4^+)$,let the oxidation

Vedclass · Accepted Answer

$-3$ and $+3$

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(C) The compound $NH_4NO_2$ dissociates into ions as $NH_4NO_2 \rightleftharpoons NH_4^+ + NO_2^-$. For the ammonium ion $(NH_4^+)$,let the oxidation number of $N$ be $x$. Thus,$x + 4(+1) = +1$,which gives $x = -3$. For the nitrite ion $(NO_2^-)$,let the oxidation number of $N$ be $x$. Thus,$x + 2(-2) = -1$,which gives $x = +3$. Therefore,the oxidation numbers of nitrogen in $NH_4NO_2$ are $-3$ and $+3$.

In $NH_4NO_2$,the oxidation number of nitrogen will be

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