Question

Arrange the following in order of increasing oxidation number of nitrogen.
$N{H_3},\,{N_3}H,\,{N_2}O,\,NO,\,{N_2}{O_5}$

Answer 1

Hint: To solve this question firstly we should know about the concept of oxidation number. Oxidation number describes the degree of an atom in a chemical compound. Oxidation number also known as oxidation states the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. With this information, and applying our knowledge to it, we will be approaching our answer.

Complete step by step solution: Let x be the oxidation number of $N$ in $N{H_3}$
$x\, + \,3\, = \,0$
$x\, = \, - 3$
So, the oxidation number of Nitrogen in $N{H_3}$ is $ - 3$

In ${N_3}H$
Let x be the oxidation number of $N$ in ${N_3}H$
$
   \Rightarrow \,\,x\, + \,1\, = \,0 \\
   \Rightarrow \,\,{N_3}H\, = \,0 \\
   \Rightarrow \,\,3x\, + \,1\, = \,0 \\
   \Rightarrow \,\,\frac{1}{3} \\
$
So, the oxidation number of nitrogen in ${N_3}H$ is $\frac{1}{3}$

In ${N_2}O$
Let x be the oxidation number of nitrogen in ${N_2}O$
$
   \Rightarrow \,\,2x\, + \,\left( { - 2} \right)\, = \,0 \\
   \Rightarrow \,\,2x\, = \,2 \\
$
x= 1
So, the oxidation number of nitrogen in ${N_2}O$ is 1

In $NO$
Let x be the oxidation number in $NO$
$
   \Rightarrow \,\,x\, + \,\left( { - 2} \right)\, = \,0 \\
   \Rightarrow \,\,x\, - \,2\, = \,0 \\
   \Rightarrow \,x\, = \,2 \\
$
So, the oxidation number is 2

In ${N_2}{O_5}$
Let x be the oxidation number of $N$ in ${N_2}{O_5}$
$
   \Rightarrow \,\,2x\, + \,\left( { - 2 \times 5} \right)\, = \,0 \\
   \Rightarrow \,\,x\, = \,5 \\
$

Hence order of increasing oxidation number of nitrogen is $N{H_3} < {N_3}H < {N_2}O < NO < {N_2}{O_5}$

Note: With an increase in the number of electronegativity of O atoms, the oxidation number of nitrogen is also increased. Oxidation numbers could be positive, negative or zero.