The given configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹. For a ground state atom, the 4s orbital is filled before the 3d. The expected configuration would be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^9$. However, a half-filled or fully-filled d subshell is more stable. A 3d¹⁰ configuration is exceptionally stable, so an electron from the 4s orbital is promoted to achieve it. This is the stable ground state configuration for copper (Cu).

Final Answer: Ground state of atom