Ionization and its effect on other orbitals.

[That Neuron]

Just hit a mental wall here, I was thinking of simple formation of an ionic compound, where a metal with a small number of outer electrons loses the outer electrons to a electronegative atom. Simple right? I know!

But it got me thinking, even though the outer electrons are easily ripped off due to the electropositivity raising their potential, and then when these "loose" electrons are gone and the underlying low potential nonreactive shell is exposed the increased positivity probably also lowers the potential of the atom. making it even more nonreactive.

But what about the anion that is formed... It has to gain an electron, now I understand the electronegativity is very high, and it is logical that an electron would be attracted and bond to the anion and the free electron's potential would be lowered by bonding (and perhaps by exchange interaction as well), but would the other orbitals in the outer shell have higher potentials due to the increased distance from the nucleus and smaller distribution of electric charge?

Im just soooooo confused, could anyone clarify this for me?

 

[jbsteele]

Thanks in advance.The answer is yes, the other orbitals in the anion's outer shell do have higher potentials due to the increased distance from the nucleus and smaller distribution of electric charge. This is because when the electron is added to the anion, it increases the overall negative charge of the atom, which causes the orbitals to become more diffuse. This means that the electrons are farther away from the nucleus, which increases their potential energy. Additionally, the smaller distribution of electric charge also contributes to the higher potentials of the other orbitals in the anion's outer shell.