[caeks]

Hello.
I would like to ask a quick question...

How can copper have oxidation states other than +I? For example CuO, where copper's oxidation state is +II; or KCuO2, where copper's oxidation state is +III.

The electrons are positioned: Cu (+29) | 2) 8} 18) 1)
Having a positive oxidation state of more than +I wouldn't be in accordance with the octet rule/18-rule (eight[een] electrons in the last shell).

I've been taught that an element's highest oxidation state is determined by the group number, doesn't matter if it's positive or negative (e.g. II A 2 for magnesium, IV A 4 for carbon etc). Maybe this isn't (entirely) true?

I even drew some subshell [spdf] graphs here and it seems random for it to happen.


Thanks in advance!
(Excuse any linguistic mistakes, I'm not a native English speaker.)

[caeks]

It's been a week.
What's up?

[nicknow123]

Forgive me if I am not entirely accurate; electron configuration is not my strong suit. First off, the octet rule has quite a few exceptions, especially higher up on the periodic table, so it cannot be trusted for every question of oxidation. +2 and +3 simply may not be entirely up to the copper. If an overpowering elemental force comes along that requires two electrons, (like O in CuO), the copper may simply end up in a less than ideal bond. Also, even if a six or seven electron valence shell isn't quite eight, it may be more stable than copper's previous position. That's really all I can think of, and I cannot confirm my correctness. Best of luck in your search.