What is the change in the oxidation number of | vedclass

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(A) In $NO_3^{-}$,let the oxidation number of $N$ be $x$. Since the oxidation number of $O$ is $-2$,we have: $x + 3(-2) = -1 \implies x - 6 = -1 \impl

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$+5$ to $-3$

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(A) In $NO_3^{-}$,let the oxidation number of $N$ be $x$. Since the oxidation number of $O$ is $-2$,we have: $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.In $NH_4^{+}$,let the oxidation number of $N$ be $y$. Since the oxidation number of $H$ is $+1$,we have: $y + 4(+1) = +1 \implies y + 4 = +1 \implies y = -3$.Therefore,the oxidation number of nitrogen changes from $+5$ to $-3$.

What is the change in the oxidation number of nitrogen when $NO_3^{-}$ is converted to $NH_4^{+}$ ion?

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