The oxidation state of nitrogen is highest in

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(A) To find the oxidation state of nitrogen $(x)$ in each compound:$1$. In $N_3H$: $3x + 1 = 0$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3} \a

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$N_3H$

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(A) To find the oxidation state of nitrogen $(x)$ in each compound:$1$. In $N_3H$: $3x + 1 = 0$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3} \approx -0.33$$2$. In $NH_2OH$: $x + 2(1) + (-2) + 1 = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = -1$$3$. In $N_2H_4$: $2x + 4(1) = 0$ $\Rightarrow 2x = -4$ $\Rightarrow x = -2$$4$. In $NH_3$: $x + 3(1) = 0 \Rightarrow x = -3$Comparing the values: $-0.33 > -1 > -2 > -3$. The highest oxidation state is in $N_3H$.

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