Can some1 explain to me why transition metals have more than one oxidation state?

Given their complex electronic structure, they can be stable in more than one configuration, however in general there is a single most stable state, for other you need to spend energy. Since as shell size increases, energy differences go down, the energy difference between two ionic states may not be much and stability persists. Plus, having empty d orbitals always helps.

I read somewhere that for some elements (like copper, silver and gold), there is a predicted and an accepted electron configuration. For example, for copper, the predicted electron configuration is 3d9 4s2 but the accepted structure is 3d10 4s1. Is this related to why copper has 2 oxidation states and can lose 1 electron or 2 electrons? If not, then how exactly does copper become stable by losing 2 electrons?

I read somewhere that for some elements (like copper, silver and gold), there is a predicted and an accepted electron configuration. For example, for copper, the predicted electron configuration is 3d9 4s2 but the accepted structure is 3d10 4s1. Is this related to why copper has 2 oxidation states and can lose 1 electron or 2 electrons? If not, then how exactly does copper become stable by losing 2 electrons?

This happens with those transition metals right next to the metaloids. The orbital energy's are so close together that the atom is more stable with the D orbital filled.

I don't think you'll predict 3d9 4s2 for copper, since fully filled sub-shells are very nice things. They have less energy amongst other things due to symmetry. Due to symmetry they have a high exchange energy which is always nice. Half and fully filled are always preferred.

The thing I don't get is that if copper has 3d10 4s1, then when it loses 2 electrons, doesn't it become 3d9 and isn't that unstable then?