Animated Solution for Chemistry - s and p-Block Elements: Heating of an aqueous solution of aluminium chloride to dryness will give

When you dissolve a simple salt like sodium chloride ($\text{NaCl}$) in water and heat it to dryness, you get your sodium chloride back. It feels intuitive, almost like a universal rule of chemistry.

But chemistry is rarely that simple, and the universe loves exceptions.

When we deal with an aqueous solution of aluminium chloride (${\text{AlCl}}_3$), the rules of the game change entirely. If you think heating it will just leave behind solid ${\text{AlCl}}_3$, you are falling into a classic, beautifully designed trap. Let's break down exactly why this happens and uncover the hidden reactions taking place in that beaker.

To understand the fate of our solution, we must first look at the nature of the aluminium ion (${\text{Al}}^{3+}$).

Aluminium is a small ion, but it carries a hefty $+3$ charge. This creates an incredibly high charge density. When you drop ${\text{AlCl}}_3$ into water, the ${\text{Al}}^{3+}$ ions don't just peacefully coexist with the water molecules. They aggressively pull on the oxygen atoms of the water molecules, weakening the $\text{O-H}$ bonds.

This leads to a process called hydrolysis. The water molecules split, and the aluminium ions react with the hydroxide (${\text{OH}}^{-}$) ions, while the chloride ions pair up with the hydrogen (${\text{H}}^{+}$) ions.

The reaction looks like this:

Because hydrochloric acid ($\text{HCl}$) is a strong acid and aluminium hydroxide (${\text{Al(OH)}}_3$) is a weak base, the resulting solution is distinctly acidic.

Now, we introduce heat. We place our beaker on a burner and watch the temperature rise.

As the solution boils, water begins to evaporate. But water isn't the only thing leaving the beaker. The hydrochloric acid ($\text{HCl}$) formed during hydrolysis is highly volatile. As the temperature climbs, the $\text{HCl}$ escapes as a gas alongside the water vapor.

This is where Le Chatelier's Principle takes the wheel.

According to this principle, if you remove a product from a system at equilibrium, the system will shift forward to replace it. Because the $\text{HCl}$ gas is continuously escaping into the atmosphere, the hydrolysis reaction is relentlessly driven in the forward direction.

By the time all the liquid has evaporated, there is no ${\text{AlCl}}_3$ left to crystallize. It has all been converted into a solid residue of aluminium hydroxide (${\text{Al(OH)}}_3$).

Thermal Decomposition

If we stopped heating the moment the liquid disappeared, we might be left with aluminium hydroxide. But the question specifies heating to dryness, which implies strong, continuous heating.

Aluminium hydroxide is not a thermally stable compound. When subjected to the intense heat of a dry beaker, it undergoes thermal decomposition. The heat energy breaks the bonds within the hydroxide, forcing it to shed water molecules.

The decomposition reaction is:

The water formed in this step immediately vaporizes, leaving behind a highly stable, white, powdery solid.

This final survivor of our thermal gauntlet is aluminium oxide, commonly known as alumina (${\text{Al}}_2{\text{O}}_3$).

What started as a seemingly straightforward evaporation turned into a two-step chemical transformation.

First, the high charge density of aluminium forced the water to hydrolyze the salt, creating a volatile acid. Second, the heat drove off that acid, locking the aluminium into a hydroxide form, which then thermally decomposed into a tough, heat-resistant oxide.

Therefore, heating an aqueous solution of aluminium chloride to dryness will not give you ${\text{AlCl}}_3$, nor will it leave you with ${\text{Al(OH)}}_3$. The final, undeniable product is ${\text{Al}}_2{\text{O}}_3$.

This is why the correct option is (b).