In which of the following compounds, an element exhibits two different oxidation states ?

N H_{2} O H

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N H_{4} N O_{3}

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N_{2} H_{4}

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N_{3} H

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Solution

The correct option is C $N H_{4} N O_{3}$
In $N H_{2} O H$ oxidation of N is -1
$N H_{4} N O_{3}$ exists as $N H_{4}^{+} . N O_{3}^{-}$ ,
thus the oxidation state of N in $N H_{4}^{+}$ is -3 while in $N O_{3}^{-}$ is +5.
In $N_{2} H_{4}$ oxidation of N is -2
In $N_{3} H$ oxidation of N is $\frac{- 1}{3}$

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