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In the 7th edition question 2A.13 part D asks to predict the type of orbital from which an electron will need to be removed to form the +1 ion of Cu. The answer is supposed to be the 4s orbital. Can someone explain why? I would've thought it would lose it from the 3d orbital?
Copper has a full 3d orbital. Cu: [Ar]3d^104s^1 The way Gail explained it today is that, although 4s is at a higher energy level, there is an additional energy cost to disrupting that full orbital, and since the difference between energy levels get smaller and smaller, it ends up requiring less energy to remove the electron from the 4s orbital. The general rule of thumb is to maximize the number of half full and full orbitals at higher energy levels (i.e. d^5 or d^10 rather than d^4 or d^9)
The ground state of Cu is 3d^10 4s^1 because copper atoms have enough electrons to completely fill the 3d sub-shell, but only if one electron from the 4s sub-shell is used and this arrangement makes the atom more stable so it takes on this arrangement, so in order to get a +1 charge it would first lose the electron from the 4s orbital
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