Animated Solution for Chemistry - s and p-Block Elements: The reaction of $\text{NO}$ with ${\text{N}}_2{\text{O}}_4$ at $250\text{ K}$ gives

Imagine you are in a chemistry lab, and you have two distinct gases in front of you.

On one side, you have Nitrogen monoxide ($\text{NO}$), a colorless gas that plays a fascinating role in both atmospheric chemistry and human biology.

On the other side, you have Dinitrogen tetraoxide (${\text{N}}_2{\text{O}}_4$), another colorless gas, which is actually the dimer of the infamous brown gas, nitrogen dioxide (${\text{NO}}_2$).

Now, what happens when we bring these two gases together?

In chemistry, temperature is like the conductor of an orchestra. It dictates how the molecules will dance.

Our problem states that the reaction takes place at $250\text{ K}$. This is significantly below room temperature (which is around $298\text{ K}$).

At such low temperatures, the kinetic energy of the molecules is reduced. Instead of bouncing around chaotically, they prefer to come together and condense.

When $\text{NO}$ and ${\text{N}}_2{\text{O}}_4$ are mixed at $250\text{ K}$, they undergo a beautiful combination reaction.

The odd-electron molecule $\text{NO}$ finds a perfect partner in the fragments of ${\text{N}}_2{\text{O}}_4$.

They react to form a completely new compound: Dinitrogen trioxide (${\text{N}}_2{\text{O}}_3$).

The balanced chemical equation for this elegant transformation is:

But the magic doesn't stop at the chemical formula.

At this low temperature, ${\text{N}}_2{\text{O}}_3$ doesn't stay a gas. It condenses into a bright blue solid.

This striking blue color is a signature characteristic of ${\text{N}}_2{\text{O}}_3$ in its solid and liquid states, making it a favorite among chemists for visual demonstrations.

Therefore, the correct answer to our question is option (c), ${\text{N}}_2{\text{O}}_3$.

Let's take a deeper look at the hidden mechanics of this reaction.

Chemistry is often a game of electrons, and tracking oxidation states helps us understand who is giving and who is taking.

In Nitrogen monoxide ($\text{NO}$), oxygen has an oxidation state of $-2$. To balance this, nitrogen must be in a $+2$ oxidation state.

In Dinitrogen tetraoxide (${\text{N}}_2{\text{O}}_4$), we have four oxygen atoms contributing a total of $-8$. To balance this across two nitrogen atoms, each nitrogen must be in a $+4$ oxidation state.

Now, let's look at our product, Dinitrogen trioxide (${\text{N}}_2{\text{O}}_3$).

Three oxygen atoms contribute $-6$. Dividing this by two nitrogen atoms gives us an oxidation state of $+3$ for nitrogen.

Do you see what just happened?

We started with nitrogen in $+2$ and $+4$ states, and they met exactly in the middle to form a $+3$ state!

This specific type of redox reaction, where an element in two different oxidation states reacts to form a product with a single intermediate oxidation state, is called a comproportionation reaction.

It is the exact opposite of disproportionation.

Finally, let's talk about the chemical nature of our blue solid.

Is it acidic, basic, or neutral?

As a general rule of thumb in inorganic chemistry, the acidic character of an oxide increases with the oxidation state of the central atom.

A higher positive charge on the central atom means it will pull electron density away from the oxygen atoms.

When such an oxide dissolves in water, this strong pull weakens the $\text{O}-\text{H}$ bonds of the water molecules, making it easier to release ${\text{H}}^{+}$ ions into the solution.

Since nitrogen in ${\text{N}}_2{\text{O}}_3$ is in a relatively high $+3$ oxidation state, it acts as an acidic oxide.

In fact, when you dissolve ${\text{N}}_2{\text{O}}_3$ in water, it forms nitrous acid (${\text{HNO}}_2$).

Understanding these underlying principles—from temperature dependence to oxidation states—transforms a simple memory-based question into a profound exploration of chemical behavior.