## ion charge

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
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### ion charge

Write the most likely charge for the ions formed by each of the following elements:
a. S
b. Te
c. Rb
d. Ga
I don't know where to start.

Kavee Dodampahala 4E
Posts: 51
Joined: Sat Jul 20, 2019 12:17 am

### Re: ion charge

You should look at the number of valence electrons the element has based on its position in the periodic table. For example, sulfur has 6 valence electrons and so would gain 2 more electrons to have a full octet. Gaining 2 electrons would then make a -2 ion.

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

### Re: ion charge

For d. Ga, I don't understand why the answer is Ga^3+?
the ground state configuration I got was [Ar]3d^10 4s^2 4p^1, so I assumed that it would be best to just get rid of the 4p^1, so that it would be [Ar]3d^10 4s^2 and the charge would be Ga^+.

Abhi Vempati 2H
Posts: 104
Joined: Fri Aug 09, 2019 12:17 am

### Re: ion charge

@Katherine Wu: Ga is a weird one because it's in Column/Family 3 of the periodic table, meaning it tends to form a 3+ charge. It confused me at first, but then I realized there's two ways to look at this:

1. Look at the family Ga is in. Ga is in Family 13 (or 3A). Then, I looked at the elements above it (B and Al) and used my prior knowledge to figure out what charges they form. Al usually forms a 3+ cation, and boron is a bit weird because it tends to form covalent bonds, so I ignored it. Then, I used logic and assumed that Ga forms a 3+ (based on aluminum).

2. Look at the principal quantum number. Ga has an n = 4 shell, meaning that it probably needs to get rid of all those electrons to form a more stable cation with up to only an n = 3 shell. In order to do this, you need to get rid of the 4s and 4p electrons. You can't just get rid of 4p because the entire n = 4 shell will still be present, resulting in less stability.

Hope this helps!