In which of the following compounds,does nitr | vedclass

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(C) To find the oxidation state of nitrogen $(N)$ in each compound,we assign hydrogen $(H)$ a value of $+1$ and oxygen $(O)$ a value of $-2$:$1$. In $

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$N_3H$

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(C) To find the oxidation state of nitrogen $(N)$ in each compound,we assign hydrogen $(H)$ a value of $+1$ and oxygen $(O)$ a value of $-2$:$1$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.$2$. In $NH_3$: $x + 3(+1) = 0 \implies x = -3$.$3$. In $N_3H$: $3x + 1(+1) = 0 \implies 3x = -1 \implies x = -1/3$.$4$. In $NH_2OH$: $x + 2(+1) + (-2) + 1(+1) = 0 \implies x + 1 = 0 \implies x = -1$.Comparing the values $-2, -3, -1/3, -1$,the highest value is $-1/3$.

In which of the following compounds,does nitrogen exhibit the highest oxidation state?

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