Q.Why is $Cu^{2+}$ ion coloured while $Zn^{2+}$ ion is colourless in aqueous solution?

Why magnetism and electron configuration matter

Magnetic behaviour in transition metal compounds comes down to one thing: unpaired electrons. A substance with any unpaired electrons is paramagnetic (attracted to a magnetic field); one with all electrons paired is diamagnetic (weakly repelled). So to judge whether cuprous salts are diamagnetic, we need to know the electron configuration of the cuprous ion, Cu⁺.

Copper’s atomic number is 29. The neutral atom has the configuration $[\text{Ar}]\,3d^{10}4s^1$ — that one 4s electron is what gives copper its typical +1 and +2 oxidation states. When copper loses one electron to form Cu⁺, it loses that 4s electron first (not a 3d electron, because the 4s orbital is higher in energy once occupied). So Cu⁺ becomes $[\text{Ar}]\,3d^{10}$.

That’s a completely filled d-subshell. All ten 3d electrons are paired up in five orbitals. No unpaired electrons means diamagnetic.

Now let’s check the Assertion and Reason carefully.

1. Assertion (A): “Cuprous salts are diamagnetic.”

   True. As we just saw, Cu⁺ has no unpaired electrons. Every cuprous salt — CuCl, Cu₂O, Cu₂S, etc. — contains Cu⁺, so the salt as a whole is diamagnetic. (The anion doesn’t contribute unpaired electrons either, so the whole compound is diamagnetic.)

2. Reason (R): “Cuprous ion has completely filled 3d-orbitals.”

   True. The configuration is $3d^{10}$, which is indeed completely filled.

3. Is (R) the correct explanation of (A)? …

The Concept: Colour and Electronic Structure

Colour in transition metal compounds is a direct consequence of their electronic configuration. When white light falls on a substance, certain wavelengths are absorbed, and the complementary colour is transmitted or reflected. For a transition metal ion in solution, the absorption typically occurs in the visible region due to d-d transitions — electrons jumping from a lower-energy d-orbital to a higher-energy d-orbital within the same subshell.

This is only possible when the d-subshell is partially filled. If the d-orbitals are completely empty ($d^0$) or completely filled ($d^{10}$), no d-d transition can occur because there is either no electron to excite or no vacant orbital to receive it. Such ions appear colourless.

Step-by-Step Reasoning

1. Electronic configuration of the atoms

   Copper (Cu, atomic number 29): $[Ar]\,3d^{10}4s^1$

   Zinc (Zn, atomic number 30): $[Ar]\,3d^{10}4s^2$

2. Formation of the +2 ions

   When forming $Cu^{2+}$, the atom loses the 4s electron and one 3d electron:

   $Cu^{2+}: [Ar]\,3d^9$

   When forming $Zn^{2+}$, the atom loses both 4s electrons:

   $Zn^{2+}: [Ar]\,3d^{10}$

3. d-orbital occupancy in aqueous solution

   In aqueous solution, water molecules act as ligands and create a crystal field around the metal ion. For octahedral complexes (common for both ions in water), the five d-orbitals split into two sets: the lower-energy $t_{2g}$ set ($d_{xy}, d_{xz}, d_{yz}$) and the higher-energy $e_g$ set ($d_{z^2}, d_{x^2-y^2}$).

   • $Cu^{2+}$ ($3d^9$): Nine electrons occupy the d-orbitals. The configuration is $(t_{2g})^6(e_g)^3$, with one unpaired electron in the $e_g$ level. There is a vacant orbital in the $e_g$ set, and an electron from the filled $t_{2g}$ set can be excited into it by absorbing visible light. This d-d transition gives $Cu^{2+}$ its characteristic blue colour in aqueous solution.

   • $Zn^{2+}$ ($3d^{10}$): All five d-orbitals are completely filled. There is no vacant d-orbital to accept an excited electron. The only possible electronic transitions would involve much higher energy levels (like 4s or 4p), which require ultraviolet light, not visible. Hence, no visible light is absorbed, and the solution appears colourless. …