Final answer:
The electron configuration of V²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³, which is derived by removing two electrons from the 4s subshell of the neutral vanadium atom.
Step-by-step explanation:
The question asks for the electron configuration of the vanadium ion with a 2+ charge (V²⁺). The electron configuration is the arrangement of electrons in an atom's orbitals and involves the quantum mechanical model of the atom, which utilizes quantum numbers to describe the position and energy level of electrons. To determine the electron configuration of V²⁺, we first find the electron configuration of a neutral vanadium atom, which is atomic number 23, and then remove two electrons accounting for the 2+ charge. The electron configuration of neutral vanadium is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s². When two electrons are removed for V²⁺, they are taken from the highest energy level, which is the 4s subshell. Therefore, the electron configuration for V²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³.