Answer
$1.10\times{10}^{-8}N$
Work Step by Step
The equilibrium separation distance is the sum of the atomic radii, so we have:
$r_{0}=r_{Mg^{2+}}+r_{F^{-}}$
Using equation 2.14, and letting $Z_{1}=+2 $ for $Mg^{2+}$ and $Z_{1}=-1 $ for $F^{-}$,
$F_{A}=\frac{(2.31\times{10}^{-28}N.m^{2})(|+2|)(|-1|)}{(0.205\times{10}^{-9}m)^{2}}= 1.10\times{10}^{-8}N
$
