Verified answer

The e⁻ configuration of Cu(0) is in fact [Ar] 3d^10 4s^1 (Most people will put 4s first but that is incorrect.

http://en.wikipedia.org/wiki/Copper

Whatever, when an e⁻ is removed to give Cu(I) it is the 4s e⁻ that is lost so Cu(I) is [Ar] 3d^10.

the atomic number ofCopper (Cu) = 29. T 1s2 2s2 2p6 3s2 3p6 4s1 3d10. (OR) [Ar] 4s1 3d10 is electronic configuration observe the last shell 4s1 and 3d10 . here in 4s orbital there is one unpaired electron. So, it has definitely some magnetic moment. "The formula to find the magnetic moment is [ √(n)(n+2)) ]. {Here n= no. of unpaired electrons.}" So for Cu the magnetic moment is [√(1)(1+2) ] = √3 coming to Cu+ , here +1 or + indicates one electron is removed from 4s orbital of Cu. So, now the electronic configuration becomes [Ar] 4s0 3d10. So, now it has no unpaired electrons , then the magnetic moment of Cu+ becomes [√(0)(0+2)] = 0. This is the way 2 find the magnetic moment. OK. I have explained it clearly, still if u have any doubt ask ur Chemistry lecturer.