Question

The correct order of N-compounds in its decreasing order of oxidation states is

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• A. $HN{O}_3,NO,N_2,N{H}_{4}Cl$
• B. $HN{O}_3,NO,N{H}_{4}Cl,N_2$
• C. $N{H}_{4}Cl,N_2,NO,HN{O}_3$
• D. $HN{O}_3,N{H}_{4}Cl,NO,N_2$

Answer 1

The correct option is A $HN{O}_3,NO,N_2,N{H}_{4}Cl$

For $HN{O}_3$:
Let oxidation state of $N=x$
$1+x+3(-2)=0$
$\Rightarrow x=+5$

For $NO$:
Let oxidation state of $N$ = $x$
$x+(-2)=0$
$\Rightarrow x=+2$

For $N_2$:
Let oxidation state of $N$ = $x$
$x=0$

For $N{H}_{4}Cl$:
Let oxidation state of $N$ = $x$
$4\left(1\right)+x+(-1)=0$
$\Rightarrow x=-3$

$H\stackrel{+5}{N}{O}_3,\stackrel{+2}{N}O,{\stackrel{0}{N}}_2,\stackrel{-3}{N}{H}_{4}Cl$

Hence, option (b) is correct.