Like $\ce{Fe}$, it can be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$, same with $\ce{Ni}$ and $\ce{Co}$, what determines whether the ionic compound will be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$?


1 Answer 1


An empty, half-filled or fully-filled electron shell is especially stable. $\ce{Fe}$ metal is [Ar] 3d6 4s2. $\ce{Fe^{2+}}$ emptiess the 4s-shell. $\ce{Fe^{3+}}$ give that plus a half-filled 3d-shell. Vanadium II, III, IV, V. Now try chloride, hypochlorite, chlorite, chlorate, and perchlorate.

The most stable oxidation state depends on the medium and added ligands, if any. Cu(I) is a Lewis soft acid. It is stabilized by soft bases like pyridine, $\ce{MeCN}$, olefins, $\ce{I^{-}}$. $\ce{Cu^{2+}}$ is a hard Lewis acid. It is stabilized by hard bases like ammonia, water, $\ce{Cl^{-}}$. Dissolve $\ce{CuCl2}$ in pyridine or $\ce{MeCN}$ and you have a powerful oxidizing agent. Dissolve a $\ce{Cu^{1+}}$ salt in water and it disproportionates into $\ce{Cu}$ metal and $\ce{Cu^{2+}}$. Put $\ce{Cu}$ metal and $\ce{Cu^{2+}}$ into $\ce{MeCN}$ and you get $\ce{Cu^{1+}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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