The oxidation state of nitrogen is highest in:

N_{3} H

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N H_{3}

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N H_{2} O H

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N_{2} H_{4}

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Solution

The correct option is A $N_{3} H$
(i) $N_{3} H$
$O . S . \Rightarrow 3 x + 1 = 0$
$x = - 1 / 3$

(ii) $N H_{3}$
$O . S . \Rightarrow x + 3 = 0$
$x = - 3$

(iii) $N H_{2} - O H$
$O . S . \Rightarrow x + 2 - 1 = 0$
$x = - 1$

(iv) $N_{2} H_{4}$
$O . S . \Rightarrow 2 x + 4 = 0$
$x = \frac{- 4}{2} = - 2$

$∴$ Highest oxidation state of nitrogen is $= - 1 / 3$

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