Question

When ammonium nitrate is gently heated, an oxide of nitrogen is formed. What is the oxidation state of nitrogen in this oxide?
A.$ + 4$
B.$ + 2$
C.$ + 3$
D.$ + 1$

Answer 1

Hint: Oxidation number of an element: It is defined as the total number of electrons that an atom accepts or loses in order to make a chemical bond which results in the formation of a chemical compound.

Complete step by step answer:
When ammonium nitrate is gently heated, an oxide of nitrogen is formed. The reaction is as follows:
$N{H_4}N{O_3}\xrightarrow{\Delta }{N_2}O + 2{H_2}O$. From the reaction it is clear that the oxide formed is ${N_2}O$.
Let us first talk about the oxidation number.
Oxidation number of an element: It is defined as the total number of electrons that an atom accepts or loses in order to make a chemical bond which results in the formation of a chemical compound.
Here we are given with the compound ${N_2}O$. And we know that generally oxygen shows the oxidation number as $ - 2$ because it belongs to the group $16$ in periodic table. The atomic number of oxygen is $8$ so it has electronic configuration as: $1{s^2},2{s^2},2{p^4}$. From here it is clear that if oxygen has two more electrons in its $p - $ orbital then it will be stable (because the elements which have eight electrons in their outermost shell are stable). So oxygen can gain two electrons to attain stability. Hence the general oxidation number of oxygen in the compounds is $ - 2$.
Now the compound ${N_2}O$ is a neutral compound. So the sum of the oxidation numbers of all the elements should be zero. Here in this compound the elements are nitrogen and oxygen. And we already know the oxidation number of oxygen i.e. $ - 2$. Now let us assume that the oxidation number of nitrogen is $x$. So the equation of sum of the oxidation numbers will be as:
$
  2x + ( - 2) = 0 \\
  x = + 1 \\
 $.
So the oxidation number of nitrogen in this compound i.e.${N_2}O$, will be $ + 1$.

Hence option D is correct.

Note:
If a compound has net charge either positive or negative then the sum of all charges in the compound will be equal to that net charge on the compound. By using this concept we can calculate the oxidation number of elements in charged species.