Chapter 2 - Atomic Structure and Interatomic Bonding - Questions and Problems - Page 48: 2.15

$5.91\times10^{-10}N$

Using equation 2.14, we obtain: $Z_{1}=+2$ for $Ca^{2+}$ and $Z_{2}=-2$ for $Zn^{2-}$ and $r=1.25 nm = 1.25 \times 10^{-9} m $ So the force of attraction between the two ions is: $F_{A}=\frac{(2.31\times10^{-28})(|+2|)(|-2|)}{{(1.25 \times 10^{-9} m)}^{2}}= 5.91\times10^{-10}N $