Q.Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

The key idea is that colour in transition metal compounds arises from d-d transitions of electrons between split d-orbitals, which requires at least one unpaired electron in the d-subshell.

Step 1 — Determine the d-electron count for each metal ion:

• $Ag_2SO_4$: Ag is in +1 state. Ag⁺ has $[Kr]\,4d^{10}$ — all d-orbitals fully filled, no unpaired electrons.
• $CuF_2$: Cu is in +2 state. Cu²⁺ has $[Ar]\,3d^{9}$ — one unpaired electron.
• $ZnF_2$: Zn is in +2 state. Zn²⁺ has $[Ar]\,3d^{10}$ — fully filled, no unpaired electrons. …

The key idea here is that colour in transition metal compounds is not simply about having a transition metal — it depends on the presence of unpaired electrons in the d-orbitals, and on the possibility of d–d transitions when light is absorbed. In the solid state, the crystal field splitting allows electrons to jump between d-orbitals, absorbing specific wavelengths and giving the compound its characteristic colour. If all d-electrons are paired (as in $d^{10}$ or $d^0$), no such transition is possible, and the compound is white or colourless.

Let’s examine each compound one by one.

1. Ag₂SO₄ — Silver is in the +1 oxidation state here. The electronic configuration of Ag is $[Kr]\,4d^{10}5s^1$. In Ag⁺, it loses the 5s electron, leaving $4d^{10}$. That’s a completely filled d-subshell. No unpaired electrons, no d–d transitions possible. So this salt is colourless in the solid state.

2. CuF₂ — Copper is in the +2 state. Cu has $[Ar]\,3d^{10}4s^1$; Cu²⁺ loses the 4s electron and one 3d electron, giving $3d^9$. That’s one unpaired electron. In a crystal field (here, fluoride ions create an octahedral environment), the d-orbitals split, and the single electron can absorb visible light to jump from a lower to a higher d-orbital. This gives CuF₂ a distinct colour (it is actually blue or blue-green). So this compound is coloured.

3. ZnF₂ — Zinc is in the +2 state. Zn has $[Ar]\,3d^{10}4s^2$; Zn²⁺ loses both 4s electrons, leaving $3d^{10}$. Again, a completely filled d-subshell. No unpaired electrons, no d–d transitions. ZnF₂ is white. …

Method: Unpaired Electron Check (d-orbital configuration)

This method uses the electronic configuration of the central metal ion to determine if unpaired electrons exist. Only ions with unpaired d-electrons can produce colour in solid state (via d-d transitions).

Steps:

Step 1: Identify the metal ion and its oxidation state

• (i) $Ag_2SO_4$ → Ag⁺
• (ii) $CuF_2$ → Cu²⁺
• (iii) $ZnF_2$ → Zn²⁺
• (iv) $Cu_2Cl_2$ → Cu⁺

Step 2: Write the d-orbital configuration for each ion

Common Mistakes & How to Avoid Them

Mistake 1: Assuming all copper compounds are coloured

Many students think all Cu compounds are coloured because Cu is a transition metal.

Reality: Colour depends on the oxidation state and electronic configuration.

• CuF₂ → Cu is in +2 state → $3d^9$ → unpaired electron → coloured ✓
• Cu₂Cl₂ → Cu is in +1 state → $3d^{10}$ → no unpaired electron → white/colourless ✗

How to avoid: Always check the d-electron count — not just the element name.

Mistake 2: Forgetting that Zn²⁺ and Ag⁺ are d¹⁰ systems

Students often mark ZnF₂ or Ag₂SO₄ as coloured because they are “transition metals.”

• Zn²⁺ → $3d^{10}$ → fully filled → no d–d transitions → colourless
• Ag⁺ → $4d^{10}$ → fully filled → colourless

How to avoid: Memorise that d¹⁰ ions (Zn²⁺, Cd²⁺, Ag⁺, Cu⁺) are always colourless in solid state (unless charge transfer occurs, which is rare in simple salts).

Mistake 3: Confusing Cu⁺ (d¹⁰) with Cu²⁺ (d⁹)

Students see “copper” and assume it’s coloured — but Cu₂Cl₂ has Cu⁺, not Cu²⁺.

How to avoid: Write the electronic configuration of the metal ion before deciding.

Mistake 4: Ignoring that colour arises from unpaired electrons in d-orbitals

Some students think any transition metal compound is coloured. …