My teacher told me that $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in solution, but why? I think $\ce{Cu+}$ is $\ce{[Ar] 3d^10 }$ ,$\ce{Cu^2+}$ is $\ce{[Ar] 3d^9}$, why is $\ce{Cu^2+}$ more stable then?
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1$\begingroup$ possible duplicate - chemistry.stackexchange.com/questions/42855/… $\endgroup$– Nilay GhoshMar 8, 2016 at 13:00
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Hund's rules, which are where we get the idea that half filled and completely filled orbitals are more stable than partially filled orbitals, are specifically about the ground states of atoms. Hund's rules do not apply to ions in solution which are complexed with ligands (like water).
What you fail to consider is that $\ce{Cu+}$ is larger than $\ce{Cu^{2+}}$ but with only half the charge, so its charge density is much lower. As a result, so its enthalpy of hydration for $\ce{Cu+}$ is much less, meaning it is much less stabilized by the formation of the metal-ligand bonds. As a result, in aqueous conditions $\ce{Cu+}$ tends to disproportionate to $\ce{Cu}$ and $\ce{Cu^{2+}}$.