Final answer:
The resulting gas phase ion after losing three electrons will have the electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵, which corresponds to option D among the given choices.
Step-by-step explanation:
When a gas phase atom with the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ loses three electrons to form a cation, the ion's electron configuration is determined by the removal of electrons starting from the outermost shell. In this case, the atom is a transition metal, so electrons are removed from the 4s orbital before the 3d orbitals. Hence, the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ after losing three electrons, which matches option D in the given choices. Therefore, the correct electron configuration of the resulting gas phase ion is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵.
The electron configuration of a gas phase atom with the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ losing three electrons would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d. In this case, the atom is losing electrons from the highest energy level, which is the 4s orbital. Therefore, the resulting ion will have the noble gas configuration of neon, 1s² 2s² 2p⁶.