Cu(II) and Ag(II) complexes are known to exist while Au(II) is unstable and disproportionates to Au(I) and Au(III).

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

Also, what is the cause of the stability of Au(I) and Au(III) over Au(II)?

  • $\begingroup$ Somewhat related: chemistry.stackexchange.com/questions/87830/… $\endgroup$ – Nilay Ghosh Apr 16 at 11:26
  • 3
    $\begingroup$ 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). $\endgroup$ – orthocresol May 17 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.