1. Nov 29, 2003

### siriuswishbear

hey guys!
I was just wondering if anyone could help me out a bit on this.
I have the following radii (in angstroms...not that it really matters)
Ca: 1.74 Ca 2+: .99
Zn: 1.31 Zn 2+: .74
Does anyone have any idea why the difference between the atomic radii of the two elements is greater than the ionic radii of the two ions? (atomic difference is .43 and the ionic diff is .25)
I've asked a bunch of people (friends in my class, parents, parents of friends etc.) and no one has any ideas
Can you help?
thanks!
lar

2. Nov 29, 2003

### Monique

Staff Emeritus
That is not so difficult, since both ions have a positive charge of 2, they lost two electrons and thus are smaller.

I'd have to know the electron configuration to comment why the loss of two electrons for one results in a different size difference than the other.

3. Nov 29, 2003

### siriuswishbear

Ca= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2
Ca 2+= 1s^2 2s^2 2p^6 3s^2 3p^6 (aka [Ar])
Zn= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10
Zn 2+= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8
Does that help?

4. Nov 29, 2003

### Monique

Staff Emeritus
20Ca: [Ar] 4s2
30Zn: [Ar] 4s24d10

I guess the best explanation in the $$\inline\Delta$$ radius would be that Zn has a larger nuclear charge than Ca, it thus pulls the electrons in its shells stronger inward.

5. Nov 29, 2003

### siriuswishbear

Thanks Monique! I've been trying to find the answer for a couple of days, you're the first person that's come up w/ a logical reason (i got a lot of "just cuz..." and "i have no idea, go ask ___"
Thanks!

6. Nov 29, 2003

### Monique

Staff Emeritus
Thanks sirius :) I was thinking that the answer would lie in the suborbitals, since they have a different shape and all, but it also depends on the number of protons that pull the electrons towards the nucleus.