stability of Cu(II), Cu(I) and Zn(II) ion

Question

I would like to ask about the stability of $\ce{Cu^{2+}}$ and $\ce{Cu^{+}}$ and also $\ce{Cu^{2+}}$.

$\ce{Cu^{2+}}$ has a configuration of $(3\mathrm{d})^9$ but $\ce{Cu^{+}}$ has a configuration of $(3\mathrm{d})^{10}$. Why does $\ce{Cu^{2+}}$ are more stable and more common than $\ce{Cu^{+}}$. I am thinking of LFSE but I am not sure is it the correct way of thinking.

If $\ce{Cu^{2+}}$ $(3\mathrm{d})^9$ is more stable than $\ce{Cu^{+}}$ $(3\mathrm{d})^{10}$, then why we cannot see a $\ce{Zn^{3+}}$ where it is also $(3\mathrm{d})^9$?

Notice the difference between meaning of M(II) and M(2+). The former is the metal atom with oxidation number +II, the latter the true ion with the charge +2e. So better to say Cu(2+) than Cu(II) ion. M(II) may part of covalent molecules, can have charge ligands etc. — Poutnik, Feb 11 '22 at 07:32
Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic

For more, see Math SE MathJax tutorial. — Poutnik, Feb 11 '22 at 07:33
Does this answer your question? Why Cu+ is unstable in aqueous medium? — Nilay Ghosh, Feb 11 '22 at 07:43
@Poutnik u r right. I am a new user and I dont know how to type M(2+) neatly sorry for that I will change it after reading that passage — some one, Feb 11 '22 at 08:00
@someone Try to type $\ce{M^2+}$ to get $\ce{M^2+}$, or $\ce{C2O4^2-}$ to get $\ce{C2O4^2-}$. Or $\ce{M^{II}}$ to get $\ce{M^{II}}$ — Poutnik, Feb 11 '22 at 09:03
Filling subshells is not everything with transition metals. You also have differences in electrostatic interactions, especially with hard-base anions or dipoles. With softer bases there is a different outcome. — Oscar Lanzi, Feb 11 '22 at 10:05
The link given by Nilay Ghosh 5 hours ago does not yield an answer. It says simply that hydrated $\ce{Cu^{2+}}$ is more stable, because it is more charged than $\ce{Cu+}$, so that it attracts more strongly the water molecules. If this was the reason, it could be applied to the case of the hydrated zinc atom, which should be more stable at the oxidation number +III. And of course,$\ce{Zn^{+3}}$ does not exist — Maurice, Feb 11 '22 at 13:33
@someone It would be faster if the question was clear from the beginning, not giving space for misinterpretation. — Poutnik, Feb 11 '22 at 14:15
@Poutnik I am not a native speaker of english. Sorry for the misunderstanding. — some one, Feb 11 '22 at 14:18
No problem, neither am I. I have meant the note rather about the content than the language itself. If the question, its context and/or background are unclear, not receiving what has been expected is what was paid for by the question. // You e.g. do not specify that(or if) you mean hydrated ions. — Poutnik, Feb 11 '22 at 14:26

Oscar Lanzi · Answer 1 · 2022-07-11T16:10:06.243

It boils down to the balance between ionization energy and increased electrostatic attraction of a more highly charged ion to solvent molecules.

If we look at ionization energies here and compare these with the maximum charge seen for forming a water-solvated metal ion, we see a pattern: the maximum charge corresponds to the point where the ionization energy surpasses 3000 kJ/mol, which is about 31 electron volts, at least for metals through the fourth period. In effect, the additional attraction to water molecules imparted by adding one more electronic positive charge to a metal ion is enough to balance 3000 kJ/mol of ionization energy. For instance, the second ionization energy of magnesium is 1451 kJ/mol whereas the third ionization energy is about 7733 kJ/mol, so magnesium will form solvated ions up to +2 charge.

Applying this to copper, we find that thecsecond ionization energy is about 1958 kJ/mol, so copper can form water-solvated copper(II) in aqueous solution. For zinc this second ionization energy is about 1734 kJ/mol, decreased from the copper value by using an electron from the 4p subshell versus 3d. But then the third ionization energy of zinc is about 3833 kJ/mol, so zinc is limited to forming aqueous +2 ions like copper. The solvating power of water molecules is enough to break into the 3d subshell of copper, but not the more tightly bound 3d shell of zinc.

Computations suggest that zinc actually can reach the +3 oxidation state in a very specific, nonaqueous environment using highly stabilized counterions with three negative charges. See the entry for zinc in Wikipedia's list of oxidation states.

But why the second ionization of $Cu$ is $1958 kJ/mol$ while 3rd ionization of $Zn$ is $3833 kJ/mol$? It seems that the value is unusually large. Both of the ionization involves removing an electron in $3d$ subshell turning it from $3d^{10}$ to $3d^9$. Since Zn has 1 more nuclear charge, it is expected to have a slightly larger value. But in real case, it is almost a double. — some one, Feb 11 '22 at 15:30
But the net charges attracting the electron are different. More charge on the zinc ion = bigger ionization energy. — Oscar Lanzi, Feb 11 '22 at 17:03

stability of Cu(II), Cu(I) and Zn(II) ion

1 Answers1

Linked