Question

Find the oxidation number of nitrogen in each case :

$N$ in $N{H}_3,N{H}_{2}OH,Na{N}_3$ and $N{H}_{4}N{O}_2$.

• A. $-3,-1,-\frac{1}{3},+3\&-3$
• B. $-3,-2,-\frac{1}{3},+1$
• C. $-3,-1,1,+2$
• D. None of the above

Answer 1

The correct option is A $-3,-1,-\frac{1}{3},+3\&-3$

The oxidation number of $N$ in $N{H}_3,N{H}_{2}OH,Na{N}_3$ 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 $N{H}_{2}OH$.
$X+3-2=0$
$X=-1$

Let X be the oxidation number of $N$ in $Na{N}_3$.
$1+3X=0$
$X=\frac{-1}{3}$

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 $\left({N{O}_2}^{-}\right)$. 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 ${N{O}_2}^{-}$:

$X-4=-1$

$x=+3$