Find the oxidation number of nitrogen in each case :

$N$ in $N H_{3}, N H_{2} O H, N a N_{3}$ and $N H_{4} N O_{2}$ .

- 3, - 1, - \frac{1}{3}, + 3 & - 3

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- 3, - 2, - \frac{1}{3}, + 1

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- 3, - 1, 1, + 2

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None of the above

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Solution

The correct option is A $- 3, - 1, - \frac{1}{3}, + 3 & - 3$
The oxidation number of $N$ in $N H_{3}, N H_{2} O H, N a N_{3}$ and $N H_{4} N O_{2}$ are $- 3, - 1, - \frac{1}{3}, + 3 / - 3$ respectively.
Let X be the oxidation number of $N$ in $N H_{3}$ .
$X + 3 = 0$
$X = - 3$

Let X be the oxidation number of $N$ in $N H_{2} O H$ .
$X + 3 - 2 = 0$
$X = - 1$

Let X be the oxidation number of $N$ in $N a N_{3}$ .
$1 + 3 X = 0$
$X = \frac{- 1}{3}$

Oxidation state of $N$ in $N H_{4} N O_{2}$ ,
The given molecule is made up of the combination of two ions which are ammonium ion ${(N H_{4}}^{+})$ and nitrate ion $({N O_{2}}^{-})$ . In this, nitrate ion is the counter ion.

Now we have to determine the oxidation number of N in both the ions.

The oxidation number of N in ${N H_{4}}^{+}$ :
$X + 4 = 1$
$X = - 3$

The oxidation number of N in ${N O_{2}}^{-}$ :
$X - 4 = - 1$
$x = + 3$

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