Animated Solution for Chemistry - d and f-Block Elements: Which one of the following when dissolved in water gives coloured solution in nitrogen atmosphere?

Imagine you are in a chemistry lab, holding four different white or pale-coloured salts. Your task is to dissolve them in water and find out which one magically turns into a beautifully coloured solution. But there is a twist—you must do this inside a chamber filled with nitrogen gas! Let's break down the chemistry behind this fascinating phenomenon.

For any transition metal compound to exhibit colour when dissolved in water, it must strictly obey two golden rules.

First, the salt must actually be soluble in water. If it just sits at the bottom of the beaker as a precipitate, it won't form a coloured solution.

Second, the metal ion formed upon dissolution must possess unpaired d-electrons. These unpaired electrons can absorb specific wavelengths of visible light and jump to higher energy d-orbitals—a process known as the d-d transition. The colour we see is the complementary colour of the light absorbed.

Let's examine our first candidate, copper(II) chloride, or ${\text{CuCl}}_2$. When we drop it into water, it readily dissolves to release ${\text{Cu}}^{2+}$ ions.

If we look at the electronic configuration of copper (atomic number 29), the ${\text{Cu}}^{2+}$ ion has a configuration of $[\text{Ar}]3d^9$. This means it has exactly one unpaired electron in its d-subshell. Because of this unpaired electron, it undergoes d-d transitions and imparts a brilliant blue colour to the solution.

Now, what about zinc chloride, ${\text{ZnCl}}_2$? It also dissolves easily in water, giving ${\text{Zn}}^{2+}$ ions.

However, zinc has an atomic number of 30, making the configuration of ${\text{Zn}}^{2+}$ exactly $[\text{Ar}]3d^{10}$. All of its d-orbitals are completely filled, and all electrons are perfectly paired. Without any unpaired electrons, d-d transitions are impossible. Thus, the solution remains completely colourless.

We are left with silver chloride ($\text{AgCl}$) and cuprous chloride (${\text{Cu}}_2{\text{Cl}}_2$). Here is where many students make a silly mistake!

Both of these salts are notoriously insoluble in water. They will simply form a white precipitate at the bottom of the beaker.

Furthermore, even if we hypothetically considered their ions, ${\text{Ag}}^{+}$ has a $4d^{10}$ configuration and ${\text{Cu}}^{+}$ has a $3d^{10}$ configuration. Both lack unpaired electrons, meaning they would be colourless regardless of their solubility.

So, our clear winner is ${\text{CuCl}}_2$. But why did the examiner specifically mention a "nitrogen atmosphere"? This is a classic JEE trap!

If we performed this experiment in normal air, the oxygen present could react with the colourless ${\text{Cu}}^{+}$ ions from the insoluble ${\text{Cu}}_2{\text{Cl}}_2$. Oxygen is a strong oxidizing agent and would oxidize the ${\text{Cu}}^{+}$ ions into ${\text{Cu}}^{2+}$ ions.

Suddenly, your colourless mixture would start turning blue, leading to immense confusion! By conducting the experiment in an inert nitrogen atmosphere, we completely prevent this oxidation, ensuring that we only observe the true, intrinsic properties of the original salts.