ronthepon Posted October 8, 2006 Report Posted October 8, 2006 I'm not sure if all the theory behind this is outdated yet or not, but here goes. Copper has an outer electronic configuration of '3d10 4s1' . So, I'd expect it to lose one electron, and get to the seemingly stable 3d10 state. Why on earth does it lose another electron so easily? Why is the 3d9 configuration so much more stable and prevalent than 3d10? Why is the existence Cu+2 ion favoured more than the Cu+1? Quote
mir Posted October 9, 2006 Report Posted October 9, 2006 Copper(I) exists often as the oxide Cu2O. The oxide is the reason why copper is red, not yellow (which is the true color of the metal). But the oxidation number is correctly quite rare for copper. Most of the reason is because the aquesous chemistry of copper is dominated by the +2 state because it is more stable (copper(II) complexes easily with water, chloride, hydroxide and ammonia). All three metals in the coinage metalgroup (gold, silver and copper), has the oxidation number +1. Only for silver +1 is common. I guess your keyword for your answer is the Jahn-Teller effect. An energy benefit is won by the splitting of the d(x^2-y^2) and d(x^2) orbitals into two new orbitals, one electron is high in energy, two electrons low in energy. Quote
ronthepon Posted October 9, 2006 Author Report Posted October 9, 2006 I'll look into the direction you've given. BTW, gold shows +3 state a lot... Quote
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