20
$\begingroup$

I saw somewhere (can't recall where) that KF is the most ionic compound. I expected CsF. Does the greater polarizability of Cs allow it to more easily form covalent bonds compared to K? Does this overcome the fact that K being in n = 4 should bond better with F in n = 2?

$\endgroup$
1
  • 1
    $\begingroup$ use hannay- smith equation and fajan rule togther,you will get your answer. $\endgroup$ Commented Oct 9, 2016 at 15:46

1 Answer 1

22
$\begingroup$

Yes, this is a subtle thing.

Using the Pauling electronegativities, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be "more ionic."

Unfortunately, an ionic bond requires separating charge, so $\ce{Cs+F-}$. The problem is that $\ce{Cs+}$ is much larger than $\ce{K+}$ and so the dipole moment for a fully ionic $\ce{Cs+F-}$ would be much larger.

We find that the charge isn't fully separated, likely because of the need to stabilize the large dipole moment.

So some people plot a curve of "% ionic character" as determined by the actual dipole moment vs. the expected dipole moment (i.e, full charge separation). We find that this curve reaches a limit ~75-85% ionic character.

IIRC, I think on that basis LiF is more Ionic than KF.

enter image description here

For more, I really like Bruce Robinson's lecture notes: http://courses.washington.edu/bhrchem/c152/Lec23.pdf via the Internet Archive

$\endgroup$
4
  • 1
    $\begingroup$ The bit about having to separate the dipole moment was very helpful. $\endgroup$ Commented Oct 4, 2014 at 20:59
  • 1
    $\begingroup$ I could go to hard-soft acid base theory since $\ce{Cs}$ is a softer cation than $\ce{K+}$ or $\ce{Li+}$ but I think the dipole moment is sufficient to explain the trend. $\endgroup$ Commented Oct 5, 2014 at 0:52
  • 1
    $\begingroup$ I always used HSAB theory to rationalize it but wanted more so I came here. $\endgroup$ Commented Oct 5, 2014 at 1:25
  • 2
    $\begingroup$ @GeoffHutchison This answer is quite good. May I add something that the OP might find useful: en.wikipedia.org/wiki/Fajans%27_rules Cheers! $\endgroup$
    – getafix
    Commented Feb 12, 2016 at 0:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.