Electron Configuration of Ions

How does the energy released in lattice formation compare to the energy required to strip away a second electron from the Na⁺ ion? Since the Na⁺ ion has a noble gas electron configuration, stripping away the next electron from this stable arrangement would require more energy than what is released during lattice formation (Sodium I₂ = 4,560 kJ/mol). Thus, sodium is present in ionic compounds as Na⁺ and not Na²⁺. Likewise, adding an electron to fill a valence shell (and achieve noble gas electron configuration) is exothermic or only slightly endothermic. To add an additional electron into a new subshell requires tremendous energy - more than the lattice energy. Thus, we find Cl^- in ionic compounds, but not Cl^2-

This amount of energy can compensate for values as large as I₃ for valence electrons (i.e. can strip away up to 3 valence electrons). Because most transition metals would require the removal of more than 3 electrons to attain a noble gas core, they are not found in ionic compounds with a noble gas core. A transition metal always loses electrons first from the higher 's' subshell, before losing from the underlying 'd' subshell. (The remaining electrons in the unfilled d subshell are the reason for the bright colors observed in many transition metal compounds!) For example, iron ions will not form a noble gas core:

• Fe: [Ar]4s²3d⁶
• Fe²⁺: [Ar] 3d⁶
• Fe³⁺: [Ar] 3d⁵

Some metal ions can form a pseudo noble gas core (and be colorless), for example:

• Ag: [Kr]5s¹4d¹⁰ Ag⁺ [Kr]4d¹⁰ Compound: AgCl
• Cd: [Kr]5s²4d¹⁰ Cd²⁺ [Kr]4d¹⁰ Compound: CdS

The valence electrons do not adhere to the "octet rule" in this case (a limitation of the usefulness of this rule). Note: The silver and cadmium atoms lost the 5s electrons in achieving the ionic state.

When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest principle quantum number