Of the compounds below, ________ has the smallest ionic separation.

Question

A) RbCl

B) KF

C) SrBr2

D) RbF

E) K2S

I know the answer, but I would like to know why it is.

Answer 1

SrB2 would be my guess.

Why? Sr has a 2+ charge whereas all the other cations have a 1+ charge. This will make the attraction of the anion greater and thus make the interionic distance shorter. In addition, Sr is smaller than the other cations also making the two ions closer.

Answer 2

Hi Sebastian P.,

Assuming the distance between adjacent, oppositely charged nuclei is what's meant by ionic separation, let's consider this problem.

1) Ions will get as close to each other as they can

2) Arrangement in space may matter -- if you are talking about anything other than the assumption above.

3) So what do you know about the sizes of ions? You should have a table of ion sizes, as well as a "rule of thumb" to remember in case you DON'T have a table someday! The table should show you that: A) ion sizes decrease greatly from left to right across a row on the periodic table (why? You should be able to explain this thoroughly in terms of increasing nuclear charge and incomplete shielding by valance electrons in the same shell!), and B) ionic sizes increase considerably moving down the periodic table; (the outer electron shells can't scrunch the inner shells out of the way, but must have a fixed spatial relationship with them, in order to preserve orthogonality (non-interference) of the electrons. -- although, the inner shells are somewhat contracted in response to increasing nuclear charge, this is in no way sufficient to counteract the effect of increasing quantum number n).

So, where does that leave you? You should be able to discard the non-optimal compounds, maybe have to look up radii for A vs. C. I'd bet on C, though -- you're trading the effect of the gain of a whole row "down" vs. the effect of the loss of a single position "across".

-- Cheers, Mr. d.