Q. In which of the following compounds, nitrogen exhibits highest oxidation state?

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A

N₂H₄

B

NH₃

C

N₃H

D

NH₂OH

Solution:

$\left(\text{N}\right)_{2} \left(\text{H}\right)_{4} ⇒ 2 \text{x} + 4 \left(+ 1\right) = 0 ⇒ 2 \text{x} + 4 = 0$
$⇒ \text{x} = - 2$
$\left(\text{NH}\right)_{3} ⇒ \text{x} + 3 \left(+ 1\right) = 0 ⇒ \text{x} = - 3$
$\left(\text{N}\right)_{3} \text{H} ⇒ 3 \text{x} + 1 \left(+ 1\right) = 0 ⇒ 3 \text{x} + 1 = 0$
$⇒ \text{x} = - 1 / 3$
$\left(\text{NH}\right)_{2} \text{OH} ⇒ \text{x} + 2 + 1 \left(- 2\right) + 1 = 0 ⇒ \text{x} + 1 = 0$
$⇒ \text{x} = - 1$
Thus, highest oxidation state is -1/3.

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