This was one of the question that was given in the school exam.

But what I am expecting is that $\ce{Cu+}$ attains a $d^{10}$ configuration by losing one electron from s-subshell. Since it has fully filled d-orbital, it should be stable. But it is found that it is unstable and tends to disproportionate in aqueous medium or with moist air to form $\ce{Cu^2+}$ where $\ce{Cu^2+}$ has incompletely filled d-orbital ($\ce{d^9}$ configuration) . For example:

$$\ce{Cu2O(s) + H2SO4(aq) -> Cu(s) + CuSO4(aq) + H2O(l)}$$

$$\ce{Cu2O(s) + 2H+(aq) -> Cu(s) + Cu^2+(aq) + H2O(l)}$$

Oxidation number changes: $\ce{2Cu(I) -> Cu(0) + Cu(II)}$


Why is this so?


You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t.

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^2+}$ ion has a greater charge density than the $\ce{Cu+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu+}$ ion loses an electron to become $\ce{Cu^2+}$ which can then release this hydration energy. A nearby $\ce{Cu+}$ ion is the most facile reduction target for the removed electron, which is why $\ce{Cu(s)}$ is also formed.

Hence, $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in aqueous medium.

  • $\begingroup$ curiousbrain, I added a sentence to your answer -- please review and edit/rollback if you don't like it. $\endgroup$ – hBy2Py Dec 28 '15 at 16:17
  • $\begingroup$ @Brian thanks for adding the part. It should be part of the answer according to the question. Thanks a lot! :) $\endgroup$ – curiousbrain Dec 28 '15 at 16:43

While I agree generally with curiousbrain’s answer, I don’t think that the charge density alone is the culprit.

Rather, $\ce{Cu+}$ is a $\mathrm{d^{10}}$ ion which therefore has no real preference for any ligand shell — much like zinc(II). All 10 d-electrons will always populate antibonding orbitals with respect to the $\ce{M-OH2}$ coordinate bond, weakening them and creating a highly labile ligand sphere.

On the contrary, $\ce{Cu^2+}$ has a rather well defined strongly Jahn-Teller distorted octahedral ligand sphere. While the lower $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{z^2}$ orbitals are antibonding[1] and fully populated, the very strongly antibonding $\mathrm{d}_{x^2 - y^2}$ orbital is only populated by a single electron, strengthening the copper-water interactions and making the ligand sphere less labile. This may well be a reason why the displacemen of an electron is favourable in aquaeous media.


[1]: If one looks at the orbital scheme of a typical octahedral complex, additionally includes π interactions between ligands and metal, and finally also considers the Jahn-Teller distortion, it becomes evident that all metal-centred orbitals are antibonding to a certain extent. However, $\mathrm{d}_{x^2-y^2}$ is much more strongly antibonding than all other d-orbitals.


Not the answer you're looking for? Browse other questions tagged or ask your own question.