Question

Among $N{H}_3,HN{O}_3,Na{N}_3$ and $M{g}_3{N}_2$ the number of molecules having nitrogen in negative oxidation state is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

The oxidation state of hydrogen is $+1$ in all of its compounds (except in metallic hydride)
The oxidation state of $Na$ and $Mg$ is $+1$ and $+2$ respectively while the oxidation state of oxygen is $-2$ (except peroxides and $O{F}_2$ ).
The oxidation state of $N$ in $N{H}_3$ is
$x+3(+1)=0$
$x=-3$
The oxidation state of $N$ in $HN{O}_3$ is
$1+x+3(-2)=0$
$x=5$
The oxidation state of $N$ in $Na{N}_3$ is
$+1+3x=0$
$x=-1/3$
The oxidation state of $N$ in $M{g}_3{N}_2$ is
$3(2)+2x=0$
$6+2x=0$
$2x=-6$
$x=-3$
Hence, three molecules (ie, $N{H}_3,Na{N}_3,M{g}_3{N}_2)$
have negative oxidation state