$\ce{Cu(II)}$ and $\ce{Ag(II)}$ complexes are known to exist while $\ce{Au(II)}$ is unstable and disproportionates to $\ce{Au(I)}$ and $\ce{Au(III)}$.

My guess as to possible reasons to this effect is that $\ce{Au(II)}$ like $\ce{Cu(II)}$ and $\ce{Ag(II)}$ cannot undergo Jahn-Teller Distortion, the cause of their stability, because of the extremely diffuse nature of $\mathrm{d}$ orbitals of $\ce{Au(II)}$ owing to relativistic effects. Could this be the right reasoning?

Also, what is the cause of the stability of $\ce{Au(I)}$ and $\ce{Au(III)}$ over $\ce{Au(II)}$?

  • $\begingroup$ Somewhat related: chemistry.stackexchange.com/questions/87830/… $\endgroup$ Apr 16, 2019 at 11:26
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    $\begingroup$ My suspicion is that it has nothing to do with the +1 oxidation state, but rather the inaccessibility of the +3 oxidation state for Cu and Ag (whereas for Au, it is quite easy and common to reach +3). This can partly be attributed to the increase in principal quantum number (in general heavier congeners are easier to oxidise than lighter ones), but there is maybe also a relativistic consideration (Au 5d orbitals are destabilised, so it's unusually easy to remove electrons from them). $\endgroup$ May 17, 2019 at 15:35
  • 1
    $\begingroup$ Actually Ag(II) also disproportionates - AgO is actually $\ce{Ag^{I}Ag^{III}O2}$ $\endgroup$
    – Mithoron
    Oct 23, 2019 at 15:55
  • $\begingroup$ @orthocresol, I would be thankful if your comment could be made an answer. $\endgroup$
    – Maan
    Nov 14, 2019 at 3:29
  • $\begingroup$ @Mithoron true of the oxide. Silver(II) fluoride does seem to be actually silver(II). $\endgroup$ Nov 9, 2023 at 2:35

3 Answers 3


I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:

Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.

However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.

$$\ce{Cu2O + H2SO4 -> Cu + CuSO4 + H2O}$$

However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.


  1. Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
  2. Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).
  • $\begingroup$ I can’t find any Au(II) in your quoted abstract. If this is a counter-answer, can you highlight it? Or if you’re not countering the tendency of Au(II) to disproportionate, what are you countering? $\endgroup$
    – Jan
    Dec 20, 2019 at 8:18
  • $\begingroup$ I want to show that at least one of two metals ($\ce{Ag(II)}$) is behaving against what is claimed for. Also, I hope that for $\ce{Cu(I)}$ (instead of $\ce{Cu(II)}$) as well. I did not counter $\ce{Au(II)}$, because it is the comparison. That's why I included ref.2 to agree with OP. $\endgroup$ Dec 20, 2019 at 8:33
  • $\begingroup$ The claim in the question’s first line is that silver(II) complexes exist, which your abstract proves outright by stating that $\ce{Ag^IAg^{III}F4}$ transforms exothermically to $\ce{AgF2}$ (a silver(II) complex) below $\pu{0^\circ C}$. $\endgroup$
    – Jan
    Dec 20, 2019 at 8:36
  • $\begingroup$ The question ask for Au(II) but you described for Ag(II) with no mention of Au(II). Is there something I am missing? $\endgroup$ Nov 8, 2023 at 4:48
  • 1
    $\begingroup$ Ok got it. But maybe if you get some info. on Au(II) that would be good. I tried to close this question as dupe but OP also mentions that this post didn't explain the underlying mechanism/reason on why Au(II) is unstable, so you can take this opportunity. $\endgroup$ Nov 8, 2023 at 11:58

I don't think it makes sense to compare gold with silver and copper. In the 6th line of the periodic table, relativistic effects become important. And one of the relativistic effect is a contraction of the lengths. This effect is proportional to the sum of the two first quantum numbers. Gold outer orbitals are 4f, 5d and 6s. Its configuration is [Xe] 4f14 5d10 6s1. These 4f, 5d and 6s orbitals are defined by n=4, l=4, then n=5, l=3, then n=6, l=1. The total n+l is equal to 8, 8, and 7. It is lot. So the 4f, 5d and 6s are compressed and disappear among the other electrons. The Au atom looks as if its outer shell has 10+14+2 = 26 electrons "lost". As Au has a total of 79 electrons, it looks as if it has only 79 - 26 = 53 electrons. The atom with 53 electrons is iodine. So The chemical properties of gold looks like iodine. Iodine is known to have -1, +1, +3 as possible oxidation states. Like gold. In fact it is even known that gold may be reduced to -1, in the compound CsAu!

I may also add that the next atom after gold is mercury, which looks like the next atom after iodine: xenon. It looks like a "heavy xenon", since it boiling point is relatively low, much lower than the usual metals. It is nearly a noble gas. Same things for the next atoms, Tl and Pb. Thallium is similar to an alkaline metal, and lead to barium: PbSO4 is as insoluble as BaSO4.


It is seen that Au(III) and Au(I) complexes are more stable than Au(II).I have an explanation but that is my observation-

Let's consider their electronic configuration.

Au → 5d10 6s1

Au+ → 5d10

Au2+ → 5d9

Au3+ → 5d8

So, if we split d-orbital into t2g and eg, we get

  • For Au


  • For Au+

For Au

  • For Au2+


  • For Au3+


So, As you can see,

  • Au (I) - Fully Filled Configuration
  • Au (III)- Half-Filled Configuration

Therefore, both of them are more stable than Au(II) because of the fact that fully filled and half-filled orbitals render more stability than any other configuration.

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    $\begingroup$ This could happen to Ag and Cu too, but that is not generally the case in reality. I am more interested in the comparison between Cu, Ag and Au. Also yours could be only part of the complete answer. $\endgroup$
    – Maan
    Oct 27, 2019 at 8:08
  • $\begingroup$ I know in case of Cu Stability of Cu+2 is more than Cu+1 because of a higher amount of Hydration Energy of Cu+2 than Cu+1 which compensates the energy difference.(Even if Cu+1 > Cu+2 by stability if we consider electronic configuration ionly) Perhaps, you could think about your question in this angle. $\endgroup$
    – Sristy
    Oct 27, 2019 at 16:31

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