Question

The correct order of the oxidation states of nitrogen in $NO$ , ${{N}_{2}}O,N{{O}_{2}}\text{ and }{{\text{N}}_{2}}{{O}_{3}}$ is:
[A] $N{{O}_{2}}$ < ${{N}_{2}}{{O}_{3}}$ < $NO$ < ${{N}_{2}}O$
[B] $N{{O}_{2}}$ < $NO$ < ${{N}_{2}}{{O}_{3}}$ < ${{N}_{2}}O$
[C] ${{N}_{2}}O$ < ${{N}_{2}}{{O}_{3}}$ < $NO$ < $N{{O}_{2}}$
[D] ${{N}_{2}}O$ < $NO$ < ${{N}_{2}}{{O}_{3}}$ < $N{{O}_{2}}$

Answer 1

To find the oxidation state of nitrogen in the given oxides, firstly notice that all the molecules are in a neutral state here. Each oxygen atom is in -2 oxidation state. Use this to find out the oxidation states of nitrogen and arrange them accordingly.

Complete step by step answer:
Before finding the oxidation number of nitrogen in the given oxides, we will discuss the few common rules that we should keep in mind while trying to find the oxidation numbers of different elements.
- We should know that the oxidation number of a free element is 0.
- If we have a monatomic ion then its oxidation number is equal to the charge on it. It can be positive, negative or neutral.
- Oxidation number of oxygen is generally -2 but in some cases it can be -1 or $\left( -\dfrac{1}{2} \right)$ too.
- The sum of oxidation numbers of all the atoms in a neutral compound is 0. If the compound has an overall charge, the sum of all the oxidation numbers will be equal to that.
- Oxidation number of hydrogen is +1 in most cases but with less electronegative elements it can be -1 too.
- There are some general oxidation numbers that are more or less accurate in all compounds like for group 1 it is +1, for group 2 it is +2 and for the halogens in a binary compound it is -1.
Now, let us try to find the oxidation number of nitrogen in the given oxides.

- Firstly, we have NO. From the above rule we know that the oxidation number of oxygen is -2 generally. Since this is a neutral compound, so the sum of oxidation states of all the atoms must be 0. For that oxidation number of nitrogen must be +2.
- Then we have ${{N}_{2}}O$. We have an oxidation number of oxygen as -2 and we have 2 nitrogen atoms so each nitrogen atom must have a charge of +1.
- Then we have $N{{O}_{2}}$. Here, we have 2 oxygen atoms with an oxidation number of -2 on each and -4 in total. So, for the oxidation number of the molecule to be neutral, the oxidation state of nitrogen is +4.
- And lastly we have ${{N}_{2}}{{O}_{3}}$. We have 3 oxygen atoms i.e. total oxidation number on the oxygen atoms is -6 and we have 2 nitrogen atoms. So each nitrogen atom has an oxidation number of +3.
We can see from the above discussion that nitrogen in $N{{O}_{2}}$ is in +4 oxidation state and that in ${{N}_{2}}{{O}_{3}}\text{,}{{\text{N}}_{2}}O$ and $NO$ is +3, +1 and +2.
So the order will be ${{N}_{2}}O$ < $NO$ < ${{N}_{2}}{{O}_{3}}$ < $N{{O}_{2}}$.
So, the correct answer is “Option D”.

Note: Oxidation number is the degree of oxidation i.e. number of electrons the atom of the considered element has lost in a chemical compound. We also refer to the oxidation number as oxidation state. It can be positive, negative or even zero.
According to IUPAC, oxidation state is defined as- “Oxidation state of an atom is the charge of this atom after ionic approximation of its hetero-nuclear bonds.”