Question #972a1

1 Answer
anor277
Jan 23, 2017

Consider the charge density of fluoride ion with respect to iodide ion......

Explanation:

This site reports the ionic radius of #F^-# is #1.19xx10^-10*m#, versus #2.06xx10^-10*m# for #I^-#. Since they both have the unit electronic charge, the fluoride ion should be a much stronger base. This is also reflected in the acidities of #HF# and #HI#, i.e. #pK_a(HF)=3.14# versus #pK_a(HI)<0#.

As far as I know this is an entropy effect rather than an enthalpy effect, however, unless you are an undergrad you don't have to consider this.

Related questions
See all questions in Nucleophile vs. Base Strength
Impact of this question
1628 views around the world
You can reuse this answer
Creative Commons License