Cu exist as Cu^+ and Cu^2+ ion. Which of the following statements is…

The correct answer is Option (2) → \( \text{Cu}^{+}\) is \(3d^{10}\) and \( \text{Cu}^{2+}\) is \(3d^9\) but \( \text{Cu}^{2+}\) is more stable.

Copper (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \). Losing the 4s electron gives \( \ce{Cu+} \): \( \mathrm{3d^{10}} \); losing one 3d electron as well gives \( \ce{Cu^2+} \): \( \mathrm{3d^{9}} \). From the configurations alone, the completely filled 3d¹⁰ of \( \ce{Cu+} \) would be expected to be more stable.

In aqueous solution, however, the stability of an ion also depends on its hydration enthalpy, and the small, doubly charged \( \ce{Cu^2+} \) is hydrated far more strongly. So \( \ce{Cu^2+(aq)} \) is the more stable ion, and \( \ce{Cu+} \) in water disproportionates into \( \ce{Cu^2+} \) and copper metal:

    \( \displaystyle \ce{2Cu+(aq) -> Cu^2+(aq) + Cu(s)} \)

So the correct statement is that \( \ce{Cu+} \) is 3d¹⁰ and \( \ce{Cu^2+} \) is 3d⁹, but \( \ce{Cu^2+} \) is more stable.

Why the other options are incorrect

• Option (1) \( \text{Cu}^{2+}\) in aqueous solution undergoes disproportionation reaction — Incorrect. It is \( \ce{Cu+} \), not \( \ce{Cu^2+} \), that disproportionates in water; \( \ce{Cu^2+(aq)} \) is the stable product of that change.

• Option (3) stability of \( \text{Cu}^{+}\) and \( \text{Cu}^{2+}\) depends on nature of copper salt — Incorrect. The stability of the two ions in water is decided by the balance of hydration enthalpy and ionisation enthalpy, which is the same whichever copper salt is dissolved. A few copper(I) compounds such as CuI and \( \ce{Cu2O} \) are stable solids only because they are insoluble and keep \( \ce{Cu+} \) out of solution.

• Option (4) \( \text{Cu}^{+}\) and \( \text{Cu}^{2+}\) are equally stable — Incorrect. If they were, \( \ce{Cu+} \) would not disproportionate in water.

NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – “… The stability of Cu²⁺ (aq) rather than Cu⁺(aq) is due to the much more negative Δ_hydH^o of Cu²⁺ (aq) than Cu⁺, which more than compensates for the second ionisation enthalpy of Cu.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: