# Question #972a1

This site reports the ionic radius of ${F}^{-}$ is $1.19 \times {10}^{-} 10 \cdot m$, versus $2.06 \times {10}^{-} 10 \cdot m$ 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 $H F$ and $H I$, i.e. $p {K}_{a} \left(H F\right) = 3.14$ versus $p {K}_{a} \left(H I\right) < 0$.