Write the Valence-Shell Ground-State electron configuration for each species: Cl-, P3-, Mg2+, Mo3+, V3+
The Correct Answer and Explanation is:
- Cl⁻ (Chloride ion): [Ne] 3s² 3p⁶
- P³⁻ (Phosphide ion): [Ne] 3s² 3p⁶
- Mg²⁺ (Magnesium ion): [Ne]
- Mo³⁺ (Molybdenum(III) ion): [Kr] 4d³
- V³⁺ (Vanadium(III) ion): [Ar] 3d²
Explanation
Electron configurations describe the distribution of electrons in atomic orbitals and are crucial for understanding the chemical behavior of elements and ions. The valence-shell configuration refers specifically to the outermost shell of electrons involved in bonding and reactivity.
For Cl⁻, the neutral chlorine atom has 17 electrons. Adding one electron to form the anion yields 18 electrons, the same as argon. Thus, the valence-shell configuration mirrors a filled third shell: 3s² 3p⁶.
P³⁻ has gained three electrons compared to its neutral phosphorus state (15 electrons), reaching 18 electrons as well. Its configuration also becomes identical to argon, featuring a filled third shell.
For Mg²⁺, magnesium starts with 12 electrons. Losing two yields the electron configuration of neon (10 electrons). Since the valence electrons (3s²) are removed, no electrons remain in the third shell, and the configuration simplifies to [Ne].
Transition metals are more complex. Mo³⁺ (42 protons) loses three electrons. Normally, molybdenum’s neutral configuration is [Kr] 5s¹ 4d⁵. When it loses three electrons, it preferentially sheds the 5s and two 4d electrons, resulting in [Kr] 4d³. The 5s orbital is depopulated first due to its slightly higher energy.
For V³⁺, neutral vanadium is [Ar] 4s² 3d³. It loses the two 4s electrons and one 3d electron when forming V³⁺, leading to the 3d² configuration. The 4s electrons are lost prior to 3d in ion formation because they are higher in energy once the atom is ionized.
These configurations help explain properties like oxidation states, ionic radii, and coordination behavior in compounds.
