Animated Solution for Chemistry - d and f-Block Elements: The addition of dilute $\text{NaOH}$ to ${\text{Cr}}^{3+}$ salt solution will give

Imagine you are standing in a chemistry laboratory, holding a test tube filled with a vibrant green solution of Chromium(III) ions, ${\text{Cr}}^{3+}$. This is a classic setup for exploring the fascinating world of transition metal chemistry. When you introduce a few drops of dilute sodium hydroxide ($\text{NaOH}$) into this solution, a rapid chemical transformation occurs right before your eyes.

As the hydroxide ions (${\text{OH}}^{-}$) from the dilute $\text{NaOH}$ mix with the ${\text{Cr}}^{3+}$ ions, they react to form a gelatinous, green precipitate. Intuitively, you might write the formula of this precipitate as simple chromium(III) hydroxide, ${\text{Cr(OH)}}_3$. However, the reality of transition metal chemistry is often more nuanced.

In aqueous solutions, this precipitate is more accurately described as a hydrated oxide. The water molecules are intimately trapped within the crystal lattice of the chromium oxide. Therefore, the most chemically precise representation of this green solid is ${\text{Cr}}_2{\text{O}}_3({\text{H}}_2\text{O}{)}_n$.

The question specifically emphasizes the use of dilute $\text{NaOH}$. This is a critical detail! Chromium(III) oxide is an amphoteric substance, meaning it can react with both acids and bases.

If we were to add an excess of strong $\text{NaOH}$, the initial green precipitate of ${\text{Cr}}_2{\text{O}}_3({\text{H}}_2\text{O}{)}_n$ would dissolve. It would react further with the abundant hydroxide ions to form a soluble, complex ion, typically represented as the tetrahydroxochromate(III) ion, $[{\text{Cr(OH)}}_4{]}^{-}$, which yields a clear green solution.

Because we are strictly using dilute $\text{NaOH}$, the reaction stops at the precipitation stage. Thus, the correct product is the solid precipitate of ${\text{Cr}}_2{\text{O}}_3({\text{H}}_2\text{O}{)}_n$.