Question

Which of the following has been arranged in order of increasing oxidation number of nitrogen?

• A. $N{H}_3<N_2{O}_5<NO<N_2$
• B. $N{H}_4^{+}<N_2{H}_4<N{H}_2{O}_H<N_{2}O$
• C. $N{O}_2^{+}<N{O}_3^{-}<N{O}_2^{-}<N_3^{-}$
• D. $N{O}_3<Na{N}_3<N{H}_4^{+}<N_{2}O$

Answer 1

The correct option is B $N{H}_4^{+}<N_2{H}_4<N{H}_2{O}_H<N_{2}O$

Analysing the options

Option (A):
(i)$N{H}_3$
$x+1\times 3=0$
$x=-3$

(ii) $N_2{O}_5$
$2x+5\times (-2)=0$
$2x=10$
$x=+5$

(iii)$NO$
$x+(-2)=0$
$x=+2$

$N_2$
Oxidation state of $N_2$ is 0 because it is present in its most stable elemental form.
Thus, the increasing order order of oxidation number of 𝑁 is
$N{H}_3<N_2<NO<N_2{O}_5$

Option (B)

(i)$N{O}_2^{+}$
$x+2\times (-2)=+1$
$x=+5$

(ii)$N{O}_3^{-}$
$x+3\times (-2)=-1$
$x=+5$

(iii) $N{O}_2^{-}$
$x+2\times (-2)=-1$
$x=+3$

(iv)$N_3^{-}$
$3x=-1\Rightarrow x=-\frac{1}{3}$
Thus, increasing order of oxidation number of N :
$N_3^{-}<N{O}_2^{-}<N{O}_2^{+}=N{O}_3^{-}$
$-\frac{1}{3}+3+5+5$

Option (C) :
(i) $N{H}_4^{+}$
$x+1\times 4=+1$
$x=-3$

(ii)$N_2{H}_4$
$2x+4\times \left(1\right)=0$
$2x+4=0$
$x=-2$

(iii)$N{H}_{2}OH$
$x+1\times 2+(-2)+1=0$
$x+2-2+1=0$
$x=-1$

(iv) $N_{2}O$
$2\times x+(-2)=0$
$x=+1$

Thus , increasing order of oxidation number of N
$N{H}_4^{+}<N_2{H}_4<N{H}_{2}OH<N_{2}O$
$-3-2-1+1$
Option (D):
(i) $N{O}_3^{-}$
$x+3\times (-2)=-1$
$x=+5$

(ii) $Na{N}_3$
$1+3x=0$
$x=-\frac{1}{3}$

(iii) $N{H}_4^{+}$
$x+4\times 1=+1$
$x=-3$

(iii) $N_{2}O$
$2\times x-2=0$
$x=+1$
Thus, increasing order of oxidation number of N :
$N{H}_4^{+}<Na{N}_3<N_{2}O<N{O}_3^{-}$
$-3-\frac{1}{3}+1+5$
So,the correct answer is option (c)