Question #ea078

1 Answer
Aug 7, 2015

Copper does not have one valence electron.


As I'm sure you know, copper is one of those elements that have "anomalous" electron configurations, in the sense that it does not match the configuration predicted by the Aufbau principle.

More specifically, copper's electron configuration is - I'll use the noble gas shorthand version - looks like this

#"Cu": ["Ar"] 3d^(10) 4s^(1)#

Now, when the copper (I) ion is formed, the copper atom loses the electron found in the 4s-orbital. The electron configuration of the copper (I) ion is

#"Cu"^(+): ["Ar"] 3d^(10)#

Even though this is a particularly stable energetic configuration, these 10 electrons located in the 3d-subshell are not actual core electrons, which is simply another way of saying that they can act as valence electrons if the proper cicumstances arise.

Remember, for transition metals, the electrons that are located outside the noble gas core are potential valence electrons.

When copper forms the copper (II) ion, it loses two electrons, one from the 4s-orbital as shown above, and one from the 3d-subshell. The electron configuration of the copper (II) ion is

#"Cu"^(2+): ["Ar"] 3d^9#

And you could go on, since copper can also exist in a +3 and +4 oxidation states, albeit these are rare and require the presence of very strong oxidizing agents.

The bottom line is that the electron configuration of the copper (I) ion allows for further oxidation to take place. Those electrons, i.e. the ones located in the 3d-subshell, are not core electrons.

This means that they can act as valence electrons when copper is in the presence of oxidizing agents. Hydration energy plays a role here, too.

Anyway, the idea is that copper has more than one valence electron, which automatically implies that it can form multiple cations, i.e. exist in multiple oxidation states.