Assertion (A): The copper(I) ion, Cu+, is not stable in aqueous solution and undergoes disproportionation.
Reason (R): The hydration enthalpy of Cu+(aq) is much more negative than that of Cu2+(aq).
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) is true but (R) is false.
In disproportionation, one oxidation state changes into a higher and a lower state at the same time. Whether Cu+ or Cu2+ is the stable ion in water is decided by the energy released on hydration against the energy needed for ionisation.
Assertion (A): The copper(I) ion, Cu+, is not stable in aqueous solution and undergoes disproportionation. — True. In water, Cu+ changes into Cu2+ (+1 to +2, oxidation) and copper metal (+1 to 0, reduction):
\( \displaystyle \ce{2Cu+(aq) -> Cu^2+(aq) + Cu(s)} \)
Reason (R): The hydration enthalpy of Cu+(aq) is much more negative than that of Cu2+(aq). — False. It is the other way round: the hydration enthalpy of Cu2+(aq) is much more negative than that of Cu+. This large release of energy more than compensates for the second ionisation enthalpy of copper, which is why Cu2+(aq), not Cu+(aq), is the stable ion.
NCERT (2026-27 edition):
4.3.7 Trends in Stability of Higher Oxidation States – “…However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.”
“2Cu+ → Cu2+ + Cu”
“The stability of Cu2+ (aq) rather than Cu+(aq) is due to the much more negative ΔhydHo of Cu2+ (aq) than Cu+, which more than compensates for the second ionisation enthalpy of Cu.”