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In which of the following compounds, nitrogen exhibits highest oxidation state?
A
$NH_2OH$
B
$N_2H_4$
C
$NH_3$
D
$N_3H$
Detailed Solution
$NH_2OH$: $x + 2(+1) + (-2) + 1 = 0 \Rightarrow x = -1$
$N_2H_4$: $2x + 4 = 0 \Rightarrow x = -2$
$NH_3$: $x + 3 = 0 \Rightarrow x = -3$
$N_3H$: $3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$
$-\frac{1}{3}$ is the highest of these, so nitrogen has its highest oxidation state in $N_3H$.
$N_2H_4$: $2x + 4 = 0 \Rightarrow x = -2$
$NH_3$: $x + 3 = 0 \Rightarrow x = -3$
$N_3H$: $3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$
$-\frac{1}{3}$ is the highest of these, so nitrogen has its highest oxidation state in $N_3H$.