In which of the following compounds does nitr | vedclass

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(A) To find the oxidation state of nitrogen $(x)$ in each compound:$1$. In $N_3H$: $3x + 1 = 0 \implies x = -1/3 \approx -0.33$$2$. In $NH_2OH$: $x +

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$N_3H$

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(A) To find the oxidation state of nitrogen $(x)$ in each compound:$1$. In $N_3H$: $3x + 1 = 0 \implies x = -1/3 \approx -0.33$$2$. In $NH_2OH$: $x + 2(+1) + (-2) + 1 = 0 \implies x + 2 - 2 + 1 = 0 \implies x = -1$$3$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$$4$. In $NH_3$: $x + 3(+1) = 0 \implies x = -3$Comparing the values: $-0.33 > -1 > -2 > -3$. Therefore,nitrogen exhibits the highest oxidation state in $N_3H$.

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

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