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The electronic configuration for $\ce{Cu}$, adjusted for Hund's rule, is:

$[\ce{Ar}] 3d^{10} 4s^1$

So, shouldn't $\ce{Cu}$ have an oxidation number of $+1$? whereby it gives off its outermost 4s electron? Why can it also have a $+2$ charge?

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Copper can form compounds with an oxidation number of +1 but you mean why does it more commonly form compounds with an oxidation number of +2. – user2617804 May 25 '14 at 22:42
up vote 18 down vote accepted

Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$. By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced. While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the energy gained through bond formation and lattice energy (or hydration energy) then $\ce{Cu^{+2}}$ compounds will form. In fact $\ce{Cu^{+2}}$ is the most common oxidation state of copper so the energetics must generally work out that the energy gained by forming more than one bond to copper and gaining additional lattice (or hydration) stabilization, more than offsets the energy cost of removing that second electron. Removal of additional electrons can also occur with copper to form $\ce{Cu^{+3}}$ and $\ce{Cu^{+4}}$ compounds, for example, $\ce{K3CuF6}$ and $\ce{Cs2CuF6}$.

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