How does one find a most likely charge for ions for a given element?

[Nathan Nakaguchi 1G]

Currently working on Homework 2A #15, essentially it asks to find a most likely charge for a given element if it were to become an ion (like for S or Ga). I'm not sure how to approach this, do I focus on ionization energy and electron affinity? How does one determine how many electrons gained or lost as well? Or is this just a memorization thing?

[Michelle Chan 1J]

The atom would lose its unpaired electrons to become an ion or gain e- to fill its shell. For example, H would lose 1 e- to become H+1 and F would gain an e- to fill its shell to come F-1.

[Kaitlyn Ang 1J]

In addition, if the element is in the d blocks, they are more likely to lose their "s" electrons first in order to fill a full or half shell of "d" electrons.

[Haley Dveirin 1E]

Count the valence electrons and if the number is closer to 8, it gains however many it needs to reach 8. If it is closer to 0, it loses however many necessary to get to 0. For d block elements, valence electrons in the s orbit move down to the d orbit and then it will gain however many electrons needed to finish filling the d block.

[Junxi Feng 3B]

It depends on the valence electrons of the atom. For instance, if the atom has 6 valence electrons in its outer shell, it tends to gain 2 e- to form its most stable form, which means its ion will -2. With the same idea, if the atom has only 2 valence electrons, it tends to lose 2 e- and form ion with +2.

[Julia Mazzucato 4D]

There are so exceptions with the transition metals (they may have different ionic charges depending on what they react with), but in general the elements in groups 1, 2, and 3 will form cations with charges 1+, 2+, and 3+ respectively, while the elements in groups 4,5,6,7 will form anions with charges 4-,3-,2-,1-. Noble gases will not form ions.

[Alan Wu]

Another way to determine the most likely charge for an ion of a given element is to look at that element's successive ionization energies. For example, Mg's 1st IE = 737.7 KJ/mol, 2nd IE = 1450.6 KJ/mol, and 3rd IE = 7732.6 KJ/mol. The huge increase from the second to third IE for Mg means that after removing 2 e- from Mg, it becomes extremely difficult to remove the third electron. This is because Mg has lost all of its valence electrons (Mg has 2 of them), and the 3rd electron to remove would be from an inner shell. Thus, Mg commonly forms 2+ ions. For Na, the 2nd IE would increase dramatically. For Al, the 4th IE would increase dramatically.