Question #972a1

1 Answer
Jan 23, 2017

Answer:

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.