# Why copper(I) is unstable in aqueous medium?

I am expecting that $$\ce{Cu+}$$ attains a $$\mathrm d^{10}$$ 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 ($$\mathrm d^9$$ configuration). Few examples from Doc Brown's Science Revision Website — The chemistry of copper:

\begin{align} \ce{\overset{+1}{Cu}_2O(s) + H2SO4(aq) &-> \overset{0}{Cu}(s) + \overset{+2}{Cu}SO4(aq) + H2O(l)} \tag{R1} \\ \ce{\overset{+1}{Cu}_2O(s) + 2 H+(aq) &-> \overset{0}{Cu}(s) + \overset{+2}{Cu}^2+(aq) + H2O(l)} \tag{R2} \end{align}

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.

• curiousbrain, I added a sentence to your answer -- please review and edit/rollback if you don't like it. Dec 28, 2015 at 16:17
• @Brian thanks for adding the part. It should be part of the answer according to the question. Thanks a lot! :) Dec 28, 2015 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.

Notes:

[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.

• Soft base ligands stabil7ze copper(I) over copper (II), and in many cases these produce more ligand field splitting than hard bases like water. So I am not sure this argument holds up. Apr 21 at 14:45
• @OscarLanzi This has been a while but off the top of my head I think the explanation would be that the bonds between thiourea and copper(I) or chloride and copper(I) would be stronger or 'more co-valent' leading to more energy gained by lowering the donating electrons' orbitals while the d electrons' slight destabilisation is less loss. For the harder ligands, the donating bond is centred more on the ligands meaning the energy gain is less and the d orbital shenanigans matter more, hence why the water complex is tipped in favour of copper(II).
– Jan
Apr 22 at 13:02
• As usual, there are a number of partial explanations, all of which are true in themselves but all need to be considered to get the true picture -- or so I interpret this right here, right now! ;)
– Jan
Apr 22 at 13:04

Quite the contrary, copper(I) is actually relatively stable compared with most fourth-period transition metals in the +1 oxidation state. As discussed here, copper in the +1 oxidation state can be stabilized, even in water solvent, by complexing with a soft base such as thiourea or chloride ion. That does not work with the +1 oxidation state in earlier transition metals in the series.

What happens with most fourth-period transition metals is an imbalance between ionization energy and solvation energy: the second ionization energy is small enough so that the additional solvation energy of a +2 ion versus a +1 ion by water outweighs it, therefore the +1 ion goes on to +2 (or more, especially early in the transition series).

With copper, however, the filled $$d$$ valence shell whose electrons only partially shield each other from the nuclear charge raises the second ionization energy of the copper. We still favor going to the +2 oxidation state when copper is complexed by water. But a ligand with more covalent bonding character and weaker electrostatic attraction, meaning a soft base, might now allow the +1 oxidation state to remain stabilized given copper's slightly higher second ionization energy. Then we get a stable copper(I) complex with the soft base, which in some cases is also soluble in water.

In the above, copper is contrasted with earlier transition elements in the fourth period. What about a comparison with zinc and later metals in the period? In these cases, the second ionization energy again drops off, until we reach the metalloidal elements arsenic and selenium, because the second ionization no longer comes from the $$3d$$ subshell (it comes from $$4s$$ or $$4p$$). Thus the post-transition metals in Period 4 also tend to go on to a +2 or higher oxidation state while copper, given the right environment, can be kept at +1.