Coulomb's law for Be2+ with S2- and O2- ions

[Westlund2]

I am doing a challenge problem. Can you tell me if I am correct in my reasoning, and help me out with what I am asking?
Thanks!

I used Coulomb's law to find the force of attraction between Be2+ and S2- in BeS and between Be2+ and O2- in BeO.

Radius(Be) = 59 pm
Radius(S) = 170 pm
Radius(O) = 140 pm

F(BeS) = -6.856x10^29 J/m
F(BeO) = -9.078x10^29 J/m

1. I was asked to find the radius of the ion, S. I looked it up to be 170, and it works in Coulomb's equation to get the F, but how do I determine the radius?

2. Is it correct that by this info, that there is a stronger force of attraction between BeO than between BeS?

3. If BeO does have the stronger force of attraction, it should line up with the other ways to determine attraction. The bond length between BeO is smaller than between BeS, which means it has a stronger attraction. Also, O is more electronegative than S, so it will have a stronger ability to attract Be, right? Which is the more important atomic property in determining Coulombic force?

 

[Negatratoron]

You're in chem 109 right? i think I'm in the same exact class :]

1: Anyway, so you just need to solve the coulomb's law equation for d and find the total distance between the nuclei. You know the atomic radius of the Be ion, so if you subtract that from d you should get the atomic radius of the S ion.

2: Definitely correct.

3: I'm not sure on this one, but I think that the atomic radius is far more important. I might be completely wrong on this. We'll see once my quiz is graded. Anyway, how I imagine this is that the S ion takes two electrons from the Be, so q1 and q2 are approximately 2 and -2. However Be, a positive ion, attracts those electrons back toward itself a little bit, so q1 is just a little tiny bit less than 2 and q2 is a little bit more than -2. Oxygen is more electronegative that sulfur, so q1 and q2 are closer to 2 and -2 in BeO than they are in BeS, for oxygen does a better job keeping its electrons from trying to scurry over toward the positively-charged Be atom.