Question

Match List-I (Compounds) with List-II (Oxidation states of Nitrogen) and select answer using the codes given below the lists.

	List - I		List - II
$\left(a\right)$	$Na{N}_3$	$\left(1\right)$	$+5$
$\left(b\right)$	$N_2{H}_2$	$\left(2\right)$	$+2$
$\left(c\right)$	$NO$	$\left(3\right)$	$-\frac{1}{3}$
$\left(d\right)$	$N_2{O}_5$	$\left(4\right)$	$-1$

• A. $a-3b-4c-2d-1$
• B. $a-4b-3c-2d-1$
• C. $a-3b-4c-1d-2$
• D. $a-4b-3c-1d-2$

Answer 1

The correct option is A $a-3b-4c-2d-1$

$a-3$
The oxidation state of $N$ in $Na{N}_3$ is $-1/3$.
The oxidation state of $Na$ is $+1$. So total oxidation state of $3N$ atoms will be $-1$.
$b-4$
The oxidation state of $N$ in $N_2{H}_2$ is $-1$.
The oxidation state of $H$ is $+1$. So oxidation state of $N$ atom will be $-1$.
$c-2$
The oxidation state of $N$ in $NO$ is $+2$.
The oxidation state of $O$ is $-2$. So oxidation state of N atom will be $+2$.
$d-1$
The oxidation state of $N$ in $N_2{O}_5$ is $+5$.
The oxidation state of $O$ is $-2$.
Total oxidation state of $5O$ atoms will be $5\times (-2)=-10$.
Total oxidation state of $2N$ atoms will be $+10$.
So oxidation state of $N$ atom will be $+5$.