The electronic configuration of Cu(II) is 3 d⁹ whereas that of Cu(I) is 3 d¹⁰. Which of the following is correct?

Solution

Chemistry ma’am · 4 steps

A full 3d¹⁰ looks stable, but in water hydration energy decides, and it favours Cu²⁺.

1. Why d10 misleads. A completely filled 3d¹⁰ subshell makes Cu⁺ look like the stable ion. But that only describes the isolated gaseous ion. In water or in a crystal, the energy released when the ion is surrounded matters just as much.
2. Hydration energy. Cu²⁺ is smaller and carries double the charge, so water molecules hold it far more tightly. Its hydration enthalpy is much more negative than that of Cu⁺. This extra energy more than pays for the second ionisation enthalpy of copper.
3. Disproportionation check. Because of this, Cu⁺ in water turns into Cu²⁺ and copper metal: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s). Using E°(Cu⁺/Cu) = +0.52 V and E°(Cu²⁺/Cu⁺) = +0.15 V, the cell potential is positive.

   E°_(cell) = 0.52 - 0.15 = +0.37 V ⇒ Δ G° < 0

4. So. The reaction is spontaneous, so Cu⁺ does not survive in water and Cu²⁺ does. Cu(II) is more stable, option (a).

The idea to keep. Do not judge ion stability from configuration alone. In water, hydration enthalpy can outweigh ionisation enthalpy, and a small, highly charged ion like Cu²⁺ gains the most from it.

Answer from the NTA final answer key. Solution by PrepWiser.