Animated Solution for Chemistry - s and p-Block Elements: The correct order of the oxidation states of nitrogen in $\text{NO}$, ${\text{NO}}_2$, ${\text{N}}_2\text{O}$ and ${\text{N}}_2{\text{O}}_3$ is

Oxidation states are a fundamental concept in chemistry, acting as a bookkeeping tool to keep track of electrons during chemical reactions. In this problem, we are tasked with finding the correct increasing order of the oxidation states of nitrogen in four different oxides: $\text{NO}$, ${\text{NO}}_2$, ${\text{N}}_2\text{O}$, and ${\text{N}}_2{\text{O}}_3$.

Before we dive into the calculations, let's establish our ground rules. For neutral molecules, the sum of the oxidation states of all the atoms must equal zero.

Furthermore, oxygen is a highly electronegative element (second only to fluorine). In almost all of its compounds (with a few exceptions like peroxides or when bonded to fluorine), oxygen aggressively pulls two electrons towards itself to complete its octet. Therefore, we can confidently assign an oxidation state of $-2$ to each oxygen atom in these nitrogen oxides.

Let's calculate the oxidation state of nitrogen (let's call it $x$) in each molecule step-by-step.

1. Nitrous Oxide (${\text{N}}_2\text{O}$) Here, we have two nitrogen atoms and one oxygen atom. Setting up our equation: So, the oxidation state of nitrogen in ${\text{N}}_2\text{O}$ is $+1$.

2. Nitric Oxide ($\text{NO}$) This molecule has one nitrogen and one oxygen atom. The equation is straightforward: The oxidation state of nitrogen in $\text{NO}$ is $+2$.

3. Dinitrogen Trioxide (${\text{N}}_2{\text{O}}_3$) With two nitrogen atoms and three oxygen atoms, our equation becomes: The oxidation state of nitrogen in ${\text{N}}_2{\text{O}}_3$ is $+3$.

4. Nitrogen Dioxide (${\text{NO}}_2$) Finally, for one nitrogen and two oxygen atoms: The oxidation state of nitrogen in ${\text{NO}}_2$ is $+4$.

Now that we have all the values, let's arrange them in increasing order:

Substituting the corresponding molecules back into this inequality, we get:

This perfectly matches option (b).

A Pro-Tip for the Exam: Notice the ratio of oxygen atoms to nitrogen atoms in each formula. As the $\text{O}:\text{N}$ ratio increases, the highly electronegative oxygen atoms pull more electron density away from nitrogen, resulting in a higher positive oxidation state. You can often use this intuition to quickly verify your calculated order!