In which of the following compounds is the ox | vedclass

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(A) To find the oxidation state of $N$ in each compound:$1$. In $N_3H$: Let the oxidation state of $N$ be $x$. $3x + 1 = 0 \implies 3x = -1 \implies x

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$N_3H$

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

In which of the following compounds is the oxidation state of $N$ the highest?

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