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JEE Advanced levelSingle correct optionDifficulty 3 of 3pn012

Arrange the following in increasing order of the oxidation state of nitrogen: $NH X_{3}$ , $N X_{2} H X_{4}$ , $NH X_{2} OH$ , $HN X_{3}$ and $NO X_{2} X^{-}$ .

A $NH X_{3} < N X_{2} H X_{4} < NH X_{2} OH < HN X_{3} < NO X_{2} X^{-}$
B $NH X_{3} < HN X_{3} < N X_{2} H X_{4} < NH X_{2} OH < NO X_{2} X^{-}$
C $NH X_{3} < N X_{2} H X_{4} < HN X_{3} < NH X_{2} OH < NO X_{2} X^{-}$
D $N X_{2} H X_{4} < NH X_{3} < NH X_{2} OH < HN X_{3} < NO X_{2} X^{-}$

Answer: A

Taking H as +1 and O as -2:

NH X_{3}

gives -3;

N X_{2} H X_{4}

gives

2 x + 4 = 0

, so -2;

NH X_{2} OH

gives

x + 3 - 2 = 0

, so -1;

HN X_{3}

gives

3 x + 1 = 0

, so average -1/3;

NO X_{2} X^{-}

gives

x - 4 = - 1

, so +3. The order is -3 < -2 < -1 < -1/3 < +3.

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