Electron Configuration of Ions
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7-7-2020
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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 Cl2-
Table 1: Lattice energies range from around 700 kJ/mol to 4000 kJ/mol:
| Compound |
Lattice Energy (kJ/mol) |
| LiF |
1024 |
| LiI |
744 |
| NaF |
911 |
| NaCl |
788 |
| NaI |
693 |
| KF |
815 |
| KBr |
682 |
| KI |
641 |
| MgF2 |
2910 |
| SrCl2 |
2130 |
| MgO |
3938 |
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
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