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.

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