As we go down the group from 3d to 5

d, size increases.

Vanderwaal radius of Cu and Au is 140pm and 166pm respectively. Since Cu is smaller in size, ionization energy of Cu must be greater.

Then why is the ionization energy of Au greater than that of copper?

  • $\begingroup$ First off, the 4d element in the group is not gold but silver. $\endgroup$ Commented Apr 23, 2017 at 10:17
  • $\begingroup$ It doesn't​ matter although I made the edits. What I want to establish is that even though copper has less size relative to Au, its ionization energy is more. $\endgroup$
    – Arishta
    Commented Apr 23, 2017 at 10:26

1 Answer 1


Smaller size does not necessarily correlate with higher ionisation energy. The trends in 1st IE going across Period 2 offer a simple counterexample.

A better (but not 100% foolproof) guideline is to use the orbital energies (see: Koopmans' theorem). The higher the energy of the orbital from which the electron is removed, the lower the IE.

In the case of gold versus copper, the first ionisation corresponds to the removal of a 6s or 4s electron respectively. In gold, the 6s orbital is relativistically stabilised, leading to a larger than expected ionisation energy. See also Geoff Hutchison's answer here.


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